Slow hierarchical queries

Hello
I wrote an application that allows the user to filter the data displayed based on several parameters, for example one of the parameters (b) is hierarchical, each element (with the exception of the root element) has a father and several sons, the hierarchy is managed by a specific table (LINKS), I need when the user selects b the result will include also all his descendants.

This is the query:

SELECT *.
R
JOIN IN-HOUSE RT ON R.A = RT. A
WHERE (R.B IN (select sub_id
the beginning of LINKS with father_id =: id
Connect prior sub_id = father_id)
or R.B =: id or: id = 0)
AND (R.A =: GOLD: a = 0)
ORDER OF R.B, R.A

The problem is that the query is so slow, it allows the application of stuck, when I've omitted the line: ' or R.B =: id or: id = 0 ' he not stuck but this line is necessary because the user may also fill the filter field or try to filter b himself and not his descendants.

Is it possible to improve the performance of this query or write it better?

You can move the logic on r.b in the subquery.
There is a small chance that it could accelerate the entire query.
not tested

SELECT *
FROM R
JOIN RT ON R.A = RT.A
WHERE R.B IN (select sub_id
                   from LINKS
                   start with father_id = :id
                   connect by prior sub_id= father_id
                   UNION ALL
                   select :id sub_id from dual
                   UNION ALL
                   select sub_id
                   from LINKS
                   where :id = 0
                   )
AND (R.A=:a or :a=0)
ORDER BY R.B, R.A

Also see the ecexution use it and explain how to call the select statement.
A goal of optimization often used is wo writing several queries and call the right according to the parameters.
Different queries may be optimized independently of each other and hence faster query where everything is condensed into one big.

pls/sql example

if :a=0 and :b=0 then
   SELECT *
   into...
   FROM R
   JOIN RT ON R.A = RT.A
   ORDER BY R.B, R.A;
elsif :b=0 then
   SELECT *
   into...
   FROM R
   JOIN RT ON R.A = RT.A
   where R.A=:a
   ORDER BY R.B, R.A;
elsif :a=0 then
   SELECT *
   into...
   FROM R
   JOIN RT ON R.A = RT.A
   WHERE R.B IN (select sub_id
                   from LINKS
                   start with father_id = :id
                   connect by prior sub_id= father_id
                   UNION ALL
                   select :id sub_id from dual)
   ORDER BY R.B, R.A;
else /* both parameters are selected */
   SELECT *
   into...
   FROM R
   JOIN RT ON R.A = RT.A
   WHERE R.B IN (select sub_id
                   from LINKS
                   start with father_id = :id
                   connect by prior sub_id= father_id
                   UNION ALL
                   select :id sub_id from dual)
   and R.A=:a
   ORDER BY R.B, R.A;
end;

Tags: Database

Similar Questions

  • Hierarchical queries - problem with condition "begins by".

    Hi people

    I play with the connection by and start with clause and am faced with a particular problem which I can not solve...

    My data set is:

    Create table dates_q
    (start_date date,
    end_date date)
    /
    

    REM INSERTING into dates_q
    Insert into dates_q ("START_DATE","END_DATE") values (to_date('01-JAN-14','DD-MON-RR'),to_date('10-JAN-14','DD-MON-RR'));
    Insert into dates_q ("START_DATE","END_DATE") values (to_date('11-JAN-14','DD-MON-RR'),to_date('20-JAN-14','DD-MON-RR'));
    Insert into dates_q ("START_DATE","END_DATE") values (to_date('10-MAR-14','DD-MON-RR'),to_date('20-MAR-14','DD-MON-RR'));
    Insert into dates_q ("START_DATE","END_DATE") values (to_date('21-MAR-14','DD-MON-RR'),to_date('31-MAR-14','DD-MON-RR'));
    Insert into dates_q ("START_DATE","END_DATE") values (to_date('01-APR-14','DD-MON-RR'),to_date('10-APR-14','DD-MON-RR'));
    

    Now I basically just want to get your hands on hierarchical queries and working with the syntax of various...

    What I now want is, start with the date of April 1 as my start date and work backward to build my 'tree '. The condition of my tree is between two rows; my start and end dates differ from 1 day. If they do not; I don't want these records in my tree.

    And using sys_connect_by_path, I want to get all the way from the root.

    Thus, for example,.

    SELECT a.*,
           sys_connect_by_path(start_date, '|'),
           LEVEL lvl
      FROM dates_q a
     CONNECT BY PRIOR end_date = (start_date - 1)
    

    I get the following output

    START_DATEEND_DATESYS_CONNECT_BY_PATH(START_DATE,'|')LVL
    01.01.201410.01.2014| 1 JANUARY 141
    11.01.201420.01.2014| 1 JANUARY 14 | JANUARY 11, 142
    11.01.201420.01.2014| JANUARY 11, 141
    10.03.201420.03.2014| MARCH 10, 141
    21.03.201431.03.2014| MARCH 10, 14. MARCH 21, 142
    01.04.201410.04.2014|10-MAR-14|21-MAR-14|01-APR-143
    21.03.201431.03.2014| MARCH 21, 141
    01.04.201410.04.2014| MARCH 21, 14. 1 APRIL 142
    01.04.201410.04.2014| 1 APRIL 141

    But for the moment I did not have any starting point... Now comes the FUN part...

    When I give the State of departure; I get a single row :-(

    Example of

    SELECT a.*,
           sys_connect_by_path(start_date, '|'),
           LEVEL lvl
      FROM dates_q a
     CONNECT BY PRIOR end_date = (start_date - 1)
     START WITH start_date = To_Date('01-apr-2014','dd-mon-yyyy');
    

    The result is

    START_DATEEND_DATESYS_CONNECT_BY_PATH(START_DATE,'|')LVL
    01.04.201410.04.2014| 1 APRIL 141

    Just a line...!

    I'm unable to understand this and work more and need help.

    The formation of the tree works only in a 'sense' and I'm going the other way around? Don't know what it means but just something that comes to mind. :/

    Thank you

    K

    P.S. - database is 10g R2.

    Hello

    Thanks for the display of the data of the sample; It is very useful.

    What do you expect the result will be and why?

    LEVEL = 1 contains all rows that meet the condition to START WITH.  The rows that meet the condition

    start_date = To_Date('01-apr-2014','dd-mon-yyyy')

    in this case?  Only the line you actually obtained.

    LEVEL = N (where N > 1) contains all rows that meet the conditions regarding some FRONT CONNECT BY rank level = N - 1.  Since the only line level = 1 to end_date = To_Date('10-apr-2014','dd-mon-yyyy'), lines satisfy the condition

    End_date PRIOR = (start_date - 1).

    ? None.  End_date PREREQUISITE is April 10, while rows with start_date April 11 would fulfill this condition, there is no line on LEVEL = 2 and the query stops there.

    You would have expected this from your previous results.  The line with the start_date April 1 had no children in the previous application, so there no children in any application that has the same State of CONNECT BY.

    Maybe you meant the CONNECT BY condtion to be

    End_date = BEFORE (start_date - 1).

  • Reg: Prior in hierarchical queries:

    Hi Experts,

    A little doubt - I've been working on the requests, but it still haunts me... Hierarchical queries
    One thing I am more confused about is the prior clause.

    I'll try to explain my concern below.
    Here, the tree yesterday gets reversed by changing the clause PRIOR to 'Manager Id' to 'Employee Id' and vice versa.

    Exactly how to understand where to put FORWARD... which column?
    I can get my results just double check... but I just want to know the concept. So stupid right... but do not want to take this doubt more far. ;)
    ranit@XE11GR2>> select * from
      2  emp;
    
         EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO
    ---------- ---------- --------- ---------- --------- ---------- ---------- ----------
          7369 SMITH      CLERK           7902 17-DEC-08        800                    20
          7566 NORGAARD   MANAGER         7839 01-APR-07       2975                    20
          7654 MARTIN     SALESMAN        7698 28-SEP-07       1250       1400         30
          7698 BLAKE      MANAGER         7839 01-MAY-09       2850                    30
          7782 FOOTE      MANAGER         7839 09-JUN-08       2450                    10
          7788 SCOTT      ANALYST         7566 19-APR-87       3000                    20
          7839 ELLISON    PRESIDENT            17-NOV-06       5000                    10
          7844 TURNER     SALESMAN        7698 08-SEP-08       1500          0         30
          7876 ADAMS      CLERK           7788 18-JUN-07       1100                    20
          7900 JAMES      CLERK           7698 03-DEC-07        950                    30
          7902 LOFSTROM   ANALYST         7566 04-DEC-08       3000                    20
          7934 MILLER     CLERK           7782 23-JAN-08       1300                    10
    
    12 rows selected.
    
    Elapsed: 00:00:00.04
    ranit@XE11GR2>> select * from
      2  emp
      3  connect by empno  = prior mgr
      4  start with empno = 7698;
    
         EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO
    ---------- ---------- --------- ---------- --------- ---------- ---------- ----------
          7698 BLAKE      MANAGER         7839 01-MAY-09       2850                    30
          7839 ELLISON    PRESIDENT            17-NOV-06       5000                    10
    
    Elapsed: 00:00:00.01
    ranit@XE11GR2>> select * from
      2  emp
      3  connect by prior empno = mgr
      4  start with empno = 7698;
    
         EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO
    ---------- ---------- --------- ---------- --------- ---------- ---------- ----------
          7698 BLAKE      MANAGER         7839 01-MAY-09       2850                    30
          7654 MARTIN     SALESMAN        7698 28-SEP-07       1250       1400         30
          7844 TURNER     SALESMAN        7698 08-SEP-08       1500          0         30
          7900 JAMES      CLERK           7698 03-DEC-07        950                    30
    
    Elapsed: 00:00:00.07
    Please help guys.
    Oracle Database 11g Express Edition Release 11.2.0.2.0 - Production
    PL/SQL Release 11.2.0.2.0 - Production
    "CORE     11.2.0.2.0     Production"
    TNS for 32-bit Windows: Version 11.2.0.2.0 - Production
    NLSRTL Version 11.2.0.2.0 - Production
    Thank you
    Vanessa B.

    I know the feeling!

    START WITH gives you the first line of the result set.

    CONNECT BY gets you the next row. At this point, you already have is the "before" line and the line that you get is the current line.

    If the current line is supposed to be the Manager of the previous line, this means that Bishop PREREQUISITE = empno.

    If the previous line is supposed to be the Manager of the current line, which means PRIOR empno = mgr.

    Think AHEAD as referring to the line that you just received and "without any preconditions" as the line you want.

    Another way of thinking that is to ask "what new information I receive?

    If you say "Bishop PREREQUISITE = empno", you already know the Director of the previous row. New information will be the Director of the current line. As you continue, you will learn step by step who are managers.

    If you say "PRIOR empno = mgr", you already know that the employee of the previous row. New information will be the new empno. So step by step, you will find the empnos.

    You always ' move to before current. "Of mgr to empno" descends from the string, "empno to Bishop" goes to the top of the chain.

    Published by: stew Ashton on February 16, 2013 11:57

  • Gites join vs hierarchical queries

    Hello

    Please tel me who must use a 1?
    I have to get job list of the simple Manger, should I use self-join or hierarchical queries (CONNECT BY and earlier versions)?

    Yours sincerely

    Hello

    944768 wrote:
    Hello

    Please tel me who must use a 1?
    I have to get job list of the simple Manger, should I use self-join or hierarchical queries (CONNECT BY and earlier versions)?

    It depends on your data and your needs.

    Whenever you have a question, please post a small example of data (CREATE TABLE and INSERT statements) for all the tables involved, so people who want to help you can recreate the problem and test their ideas. Also post the results you want from this data, as well as an explanation of how you get these results from these data.
    Explain, using specific examples, how you get these results from these data.
    If you show what the problem using commonly available tables (suc as scott.emp, who has a hierarchy of level 4), then you do not have ot post sample data, just the results and explanations.
    Always say what version of Oracle you are using (for example, 11.2.0.2.0).
    See the FAQ forum {message identifier: = 9360002}

    If your hierarchy consists only of 2 levels, then a self-join will be probably more effective, simpler code and easier to maintain.

    If you don't know how many levels in the hierarchy, then self-join is not an option. Use CONNECT BY or, if you have Oracle 11.2, a WITH recursive clause.

    If you have a fixed number of levels (or an upper limit) greater than 2, then CONNECT BY (or a WITH recursive clause) will probably be the best.

  • How to order a tree balanced with SQL hierarchical queries

    by searching the forum I found this

    Re: Generate tree balanced with SQL hierarchical queries

    is there a way I can order this tree in alphabetical order so that the result looks like

    LEVEL INDENTED_ENAME
    ---------- --------------------
    1 KING BED
    2 BLAKE
    3 ALLEN
    3 JAMES
    MARTIN 3
    3 TURNER
    WARD 3
    2 CLARK
    3 MILLER
    2 JONES
    3 FORD
    4 SMITH
    3 SCOTT
    4 ADAMS

    -the original query-

    SELECT THE LEVEL
    , LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
    FROM scott.emp
    START WITH mgr IS NULL
    CONNECT BY PRIOR empno = mgr
    ;

    LEVEL INDENTED_ENAME
    ---------- --------------------
    1 KING BED
    2 JONES
    3 SCOTT
    4 ADAMS
    3 FORD
    4 SMITH
    2 BLAKE
    3 ALLEN
    WARD 3
    MARTIN 3
    3 TURNER
    3 JAMES
    2 CLARK
    3 MILLER

    Hello

    Bodin wrote:
    Hi Frank, I can order it selectively depending on the level, which means that only siblings stopped at third level, but rest of the brothers and sisters remain Nations United ordered

    It's actually quite difficult. You can "ORDER of brothers and SŒURS BY CASE... ', like this:

    SELECT  LEVEL
    ,      LPAD (' ', 3 * LEVEL) || ename     AS indented_ename
    FROM      scott.emp
    START WITH        mgr     IS NULL
    CONNECT BY         mgr      = PRIOR empno
    ORDER SIBLINGS BY  CASE
                   WHEN  job = 'MANAGER'  THEN  ename
                                              ELSE  NULL
                 END
    ;
    

    In this case to get desired results in table scott.emp, as the lines are LEVEL = 2 if and only if use = "MANAGER".
    But if you reference LEVEL in the CASE expression (for example, if you replace ' job = 'MANAGER' ' with "2 LEVEL =" above "), then you will get the error" ORA-00976: LEVEL, PRIOR or ROWNUM not allowed here. "
    The best way I can think to do exactly what you asked is to do 2 CONNECT BY queries; one just to get the LEVEL and the other for the brothers and SŒURS ORDER BY:
    {code}
    WITH got_lvl AS
    (
    SELECT LEVEL AS lvl
    Bishop
    empno
    ename
    FROM scott.emp
    START WITH mgr IS NULL
    CONNECT BY PRIOR empno = mgr
    )
    SELECT lvl
    , LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
    OF got_lvl
    START WITH mgr IS NULL
    CONNECT BY PRIOR empno = mgr
    BROTHERS AND SŒURS OF ORDER OF CASES
    ONCE lvl = 2 THEN ename
    ANOTHER NULL
    END
    ;
    {code}
    Why you can't simply "Brothers and SŒURS of ORDER BY ename" at all levels? If all you care is the order of the items of LEVEL = 2, then this is probably the most effective and simplest way. It really hurt anything if nodes on levels 3, 4, 5,... are in order, too?

    Here's something you can do if you want to order by different unique things to different levels:
    {code}
    WITH got_sort_key AS
    (
    SELECT LEVEL AS lvl
    , LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
    empno
    SYS_CONNECT_BY_PATH (LPAD (CASE
    WHEN LEVEL = 2
    THEN ename
    Of OTHER TO_CHAR (empno)
    END
    10
    )
    , ','
    ) AS sort_key
    FROM scott.emp
    START WITH mgr IS NULL
    CONNECT BY PRIOR empno = mgr
    )
    SELECT lvl
    indented_ename
    empno
    OF got_sort_key
    ORDER BY sort_key
    ;
    {code}
    However, all possible values of the CASE expression must uniquely identify the node; otherwise, the output won't necessarily hierarchical order. You can assign arbitrary unique IDS to the lines (using the ROW_NUMBER analytic function, for example), but that requires another subquery and is also complex and perhaps as ineffective as the solution above with 2 garages CONNECT.

  • to solve the problems of performance (running slow SQL queries), which should be the systematic approach

    Hello

    I am new to oracle.  Need some advice on how we should approach to problems of performance especially for SQL queries running slow on the database.

    Thanks in advance.

    6e85b3c2-5f02-40CB-A8AD-d2279f63b4be wrote:

    Hello

    I am new to oracle.  Need some advice on how we should approach to problems of performance especially for SQL queries running slow on the database.

    Thanks in advance.

    Start here;  http://docs.Oracle.com/CD/E11882_01/server.112/e10822/TOC.htm

    Continue here: http://docs.oracle.com/cd/E11882_01/server.112/e41573/toc.htm

  • hierarchical queries: output: display all folder paths. entry: Folder1, folder2

    I have a table called 'File' with the data indicated below. I want the sql query to show the hierarchical output.
    Enter values for the query: 2-11 tree shows two points (see "Input2" and "Input11"). The query now takes the values 2 and 11 and outputs hierarchical tree from root (see "Root") to the final sheets ('2, '12'), while the entries are as points of 'touch' inside the full tree-path, so I have the necessary output should be like 5 rows:

    (0, null)-parent of the two inputs
    (2, 0) - input2
    (10,0) - parents of 'input11 '.
    (11: 10) - input11
    (12: 11) - leaf


    I don't know how to write this query. Can help you.
    Something like:

    "Select...".
    Connect prior ID = PARENT_ID.

    --

    Table "Folder" (ID, PARENT_ID).

    Column PARENT_ID reference ID to create the hierarchy.
    Example of data in the table:

    (0, null)-root
    (1, 0) - child '1 '.
    (2, 0) - Input2
    (3, 1)
    (10,0) - child "10".
    (11: 10) - chilc-child "11", Input11
    (12: 11) - chilc-child "12".

    Hello

    If you want to ignore a few lines in the file and only pay attention to the parameters, their ancestors and their descendants. Is this fair?
    If so, do CONNECT it BY query on the results of a subquery which includes only the desired lines. In the example below, this query Tahina is be a UNION of two other CONNECT BY queries: one to get the ancestors of the parameters and the other for their descndents.

    VARIABLE     input_a     NUMBER
    VARIABLE     input_b     NUMBER
    EXEC  :input_a := 2;
    EXEC  :input_b := 11;
    
    WITH     universe     AS
    (
         -- Descendents of parameters
         SELECT     id,     parent_id
         FROM     folder
         START WITH     parent_id     IN (:input_a, :input_b)
         CONNECT BY     parent_id     = PRIOR id
         --
        UNION
         -- Ancestors of parameters
         SELECT  id,     parent_id
         FROM     folder
         START WITH     id     IN (:input_a, :input_b)
         CONNECT BY     id     = PRIOR parent_id
    )
    SELECT     SYS_CONNECT_BY_PATH (id, '/')     AS path
    FROM     universe
    START WITH     parent_id     IS NULL
    CONNECT BY     parent_id     = PRIOR id
    ;
    

    Output of your sample data:

    PATH
    ------------------------------
    /0
    /0/2
    /0/10
    /0/10/11
    /0/10/11/12
    
  • Hierarchical queries for subnets

    I need to write a hierarchical query for quite a number of (contiguous groups of IP addresses) IP subnets. What I'm hoping to do, is let the database do the bulk of the work here and I will not need to write code to do this. I'll do that if I have to but I would really rather not.

    General information may be in order for those who do not know subnets. IP subnets have unique characteristics, based on network management standards (caveat: I'm not a guru/subnet on the network, but I know enough to be dangerous). The first thing to know is that a subnet is a contiguous block of IP addresses that help define the IP network protocol. Another thing is that an IP address can be converted to a binary or decimal/numeric value. This is useful because it can take us out of the sphere of analysis of strings and comparing a group of text values, which will eventually be converted to numbers in order to make meaningful comparisons anyway.

    The result is that subnets can be defined by digital upper and lower limits. And the trick here is that some of these blocks of IP addresses live in the other blocks. Look at the data below and I'll explain if you need. In addition, the network address is the lower limit of the subnet and dissemination is the upper limit. This is a create table and examples of data from my dataset where you can see the IP addresses and name of each subnet and numeric limits for each subnet.

    CREATE TABLE "SUBNET_DECIMAL_VALS" 
       (    
        "SUBNET" VARCHAR2(4000 BYTE), 
        "CIDR_BLOCK" VARCHAR2(30 BYTE), 
        "NETWORK" VARCHAR2(4000 BYTE), 
        "BROADCAST" VARCHAR2(4000 BYTE), 
        "NETWORK_DEC" NUMBER, 
        "BROADCAST_DEC" NUMBER
       ) ;
    

    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.0.0/20','/20','128.110.0.0','128.110.15.255',2154692608,2154696703);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.1.0/24','/24','128.110.1.0','128.110.1.255',2154692864,2154693119);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.4.0/22','/22','128.110.4.0','128.110.7.255',2154693632,2154694655);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.8.0/22','/22','128.110.8.0','128.110.11.255',2154694656,2154695679);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.12.0/22','/22','128.110.12.0','128.110.15.255',2154695680,2154696703);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.16.0/21','/21','128.110.16.0','128.110.23.255',2154696704,2154698751);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.16.0/22','/22','128.110.16.0','128.110.19.255',2154696704,2154697727);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.20.0/22','/22','128.110.20.0','128.110.23.255',2154697728,2154698751);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.24.0/22','/22','128.110.24.0','128.110.27.255',2154698752,2154699775);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.28.0/22','/22','128.110.28.0','128.110.31.255',2154699776,2154700799);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.28.0/24','/24','128.110.28.0','128.110.28.255',2154699776,2154700031);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.29.0/24','/24','128.110.29.0','128.110.29.255',2154700032,2154700287);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.30.0/24','/24','128.110.30.0','128.110.30.255',2154700288,2154700543);
    Insert into SUBNET_DECIMAL_VALS (SUBNET,CIDR_BLOCK,NETWORK,BROADCAST,NETWORK_DEC,BROADCAST_DEC) values ('128.110.31.0/24','/24','128.110.31.0','128.110.31.255',2154700544,2154700799);
    

    SELECT * FROM SUBNET_DECIMAL_VALS
    subnet            cidr   network          broadcast        network_dec   broadcast_dec
    128.110.0.0/20    /20    128.110.0.0      128.110.15.255   2154692608    2154696703
    128.110.1.0/24    /24    128.110.1.0      128.110.1.255    2154692864    2154693119
    128.110.4.0/22    /22    128.110.4.0      128.110.7.255    2154693632    2154694655
    128.110.8.0/22    /22    128.110.8.0      128.110.11.255   2154694656    2154695679
    128.110.12.0/22   /22    128.110.12.0     128.110.15.255   2154695680    2154696703
    128.110.16.0/21   /21    128.110.16.0     128.110.23.255   2154696704    2154698751
    128.110.16.0/22   /22    128.110.16.0     128.110.19.255   2154696704    2154697727
    128.110.20.0/22   /22    128.110.20.0     128.110.23.255   2154697728    2154698751
    128.110.24.0/22   /22    128.110.24.0     128.110.27.255   2154698752    2154699775
    128.110.28.0/22   /22    128.110.28.0     128.110.31.255   2154699776    2154700799
    128.110.28.0/24   /24    128.110.28.0     128.110.28.255   2154699776    2154700031
    128.110.29.0/24   /24    128.110.29.0     128.110.29.255   2154700032    2154700287
    128.110.30.0/24   /24    128.110.30.0     128.110.30.255   2154700288    2154700543
    128.110.31.0/24   /24    128.110.31.0     128.110.31.255   2154700544    2154700799
    

    What I tried to do is a hierarchical query traditional and used more than and less than the comparisons in the where clause. I'm trying to find the limits of subnet (top and bottom numbers) that match among other numerical limits. This is an example, and it gives me the results I need.

    select level, subnet, cidr_block cidr, network, broadcast, network_dec, broadcast_dec
    from poc_subnet_decimal_vals
    connect by nocycle prior network_dec > network_dec and broadcast_dec < broadcast_dec
    

    Do I have to create from this, it is something like this:

    Level    Subnet           CIDR    Network        Broadcast         Network_Dec   Broacast_Dec
    1        128.110.0.0/20    /20    128.110.0.0    128.110.15.255    2154692608    2154696703
    2        128.110.1.0/24    /24    128.110.1.0    128.110.1.255     2154692864    2154693119
    2        128.110.4.0/22    /22    128.110.4.0    128.110.7.255     2154693632    2154694655
    2        128.110.8.0/22    /22    128.110.8.0    128.110.11.255    2154694656    2154695679
    2        128.110.12.0/22   /22    128.110.12.0   128.110.15.255    2154695680    2154696703
    1        128.110.16.0/21   /21    128.110.16.0   128.110.23.255    2154696704    2154698751
    2        128.110.16.0/22   /22    128.110.16.0   128.110.19.255    2154696704    2154697727
    2        128.110.20.0/22   /22    128.110.20.0   128.110.23.255    2154697728    2154698751
    2        128.110.24.0/22   /22    128.110.24.0   128.110.27.255    2154698752    2154699775
    2        128.110.28.0/22   /22    128.110.28.0   128.110.31.255    2154699776    2154700799
    3        128.110.28.0/24   /24    128.110.28.0   128.110.28.255    2154699776    2154700031
    3        128.110.29.0/24   /24    128.110.29.0   128.110.29.255    2154700032    2154700287
    3        128.110.30.0/24   /24    128.110.30.0   128.110.30.255    2154700288    2154700543
    3        128.110.31.0/24   /24    128.110.31.0   128.110.31.255    2154700544    2154700799
    

    Once, I get a hierarchical query of good work I can do the rest of the formatting with the values of the path and CYCLE but I'm not able to get a basic querying work. Does anyone have an idea of how to go on this subject using SQL only? I can write a PL/SQL procedure to add an ID of parent to each row, and then I know that I can create a hierarchical query from that but I was trying to avoid this method if possible.

    Hello

    Earl Lewis wrote:

    I need to write a hierarchical query for quite a number of (contiguous groups of IP addresses) IP subnets. What I'm hoping to do, is let the database do the bulk of the work here and I will not need to write code to do this. I'll do that if I have to but I would really rather not.

    General information may be in order for those who do not know subnets. IP subnets have unique characteristics, based on network management standards (caveat: I'm not a guru/subnet on the network, but I know enough to be dangerous). The first thing to know is that a subnet is a contiguous block of IP addresses that help define the IP network protocol. Another thing is that an IP address can be converted to a binary or decimal/numeric value. This is useful because it can take us out of the sphere of analysis of strings and comparing a group of text values, which will eventually be converted to numbers in order to make meaningful comparisons anyway.

    The result is that subnets can be defined by digital upper and lower limits. And the trick here is that some of these blocks of IP addresses live in the other blocks. Look at the data below and I'll explain if you need. In addition, the network address is the lower limit of the subnet and dissemination is the upper limit. This is a create table and examples of data from my dataset where you can see the IP addresses and name of each subnet and numeric limits for each subnet.

    1. CREATE TABLE 'SUBNET_DECIMAL_VALS '.
    2. (
    3. VARCHAR2 (4000 BYTE) "SUBNET."
    4. VARCHAR2 (30 BYTE) "CIDR_BLOCK."
    5. VARCHAR2 (4000 BYTE) 'NETWORK ',.
    6. "BROADCASTING" VARCHAR2 (4000 BYTE),
    7. NUMBER OF "NETWORK_DEC."
    8. NUMBER OF 'BROADCAST_DEC '.
    9. ) ;
    1. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.0.0/20','/20','128.110.0.0','128.110.15.255',2154692608,2154696703);
    2. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.1.0/24','/24','128.110.1.0','128.110.1.255',2154692864,2154693119);
    3. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.4.0/22','/22','128.110.4.0','128.110.7.255',2154693632,2154694655);
    4. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.8.0/22','/22','128.110.8.0','128.110.11.255',2154694656,2154695679);
    5. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.12.0/22','/22','128.110.12.0','128.110.15.255',2154695680,2154696703);
    6. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.16.0/21','/21','128.110.16.0','128.110.23.255',2154696704,2154698751);
    7. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.16.0/22','/22','128.110.16.0','128.110.19.255',2154696704,2154697727);
    8. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.20.0/22','/22','128.110.20.0','128.110.23.255',2154697728,2154698751);
    9. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.24.0/22','/22','128.110.24.0','128.110.27.255',2154698752,2154699775);
    10. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.28.0/22','/22','128.110.28.0','128.110.31.255',2154699776,2154700799);
    11. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.28.0/24','/24','128.110.28.0','128.110.28.255',2154699776,2154700031);
    12. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.29.0/24','/24','128.110.29.0','128.110.29.255',2154700032,2154700287);
    13. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.30.0/24','/24','128.110.30.0','128.110.30.255',2154700288,2154700543);
    14. Insert into SUBNET_DECIMAL_VALS (subnet, CIDR_BLOCK, NETWORK, BROADCAST, NETWORK_DEC, BROADCAST_DEC) values ('128.110.31.0/24','/24','128.110.31.0','128.110.31.255',2154700544,2154700799);
    1. SELECT * FROM SUBNET_DECIMAL_VALS
    2. subnet cidr network network_dec broadcast_dec broadcast
    3. 128.110.0.0/20 20 128.110.0.0 128.110.15.255 2154692608 2154696703
    4. 128.110.1.0/24 24 128.110.1.0 128.110.1.255 2154692864 2154693119
    5. 22 128.110.4.0/22 128.110.4.0 128.110.7.255 2154693632 2154694655
    6. 22 128.110.8.0/22 128.110.8.0 128.110.11.255 2154694656 2154695679
    7. 22 128.110.12.0/22 128.110.12.0 128.110.15.255 2154695680 2154696703
    8. 128.110.16.0/21 21 128.110.16.0 128.110.23.255 2154696704 2154698751
    9. 22 128.110.16.0/22 128.110.16.0 128.110.19.255 2154696704 2154697727
    10. 22 128.110.20.0/22 128.110.20.0 128.110.23.255 2154697728 2154698751
    11. 22 128.110.24.0/22 128.110.24.0 128.110.27.255 2154698752 2154699775
    12. 22 128.110.28.0/22 128.110.28.0 128.110.31.255 2154699776 2154700799
    13. 128.110.28.0/24 24 128.110.28.0 128.110.28.255 2154699776 2154700031
    14. 128.110.29.0/24 24 128.110.29.0 128.110.29.255 2154700032 2154700287
    15. 128.110.30.0/24 24 128.110.30.0 128.110.30.255 2154700288 2154700543
    16. 128.110.31.0/24 24 128.110.31.0 128.110.31.255 2154700544 2154700799

    What I tried to do is a hierarchical query traditional and used more than and less than the comparisons in the where clause

    . I'm trying to find the limits of subnet (top and bottom numbers) that match among other numerical limits. This is an example, and it gives me the results I need.

    1. Select the level, subnet, cidr_block cidr, broadcast network, network_dec, broadcast_dec
    2. of poc_subnet_decimal_vals
    3. nocycle to connect before network_dec > network_dec and broadcast_dec<>

    Do I have to create from this, it is something like this:

    1. Level of subnet CIDR network broadcast Network_Dec Broacast_Dec
    2. 1 20 128.110.0.0/20 128.110.0.0 128.110.15.255 2154692608 2154696703
    3. 2 128.110.1.0/24 24 128.110.1.0 128.110.1.255 2154692864 2154693119
    4. 2 128.110.4.0/22 22 128.110.4.0 128.110.7.255 2154693632 2154694655
    5. 2 128.110.8.0/22 22 128.110.8.0 128.110.11.255 2154694656 2154695679
    6. 2 128.110.12.0/22 22 128.110.12.0 128.110.15.255 2154695680 2154696703
    7. 1 128.110.16.0/21 21 128.110.16.0 128.110.23.255 2154696704 2154698751
    8. 2 128.110.16.0/22 22 128.110.16.0 128.110.19.255 2154696704 2154697727
    9. 2 128.110.20.0/22 22 128.110.20.0 128.110.23.255 2154697728 2154698751
    10. 2 128.110.24.0/22 22 128.110.24.0 128.110.27.255 2154698752 2154699775
    11. 2 128.110.28.0/22 22 128.110.28.0 128.110.31.255 2154699776 2154700799
    12. 3 128.110.28.0/24 24 128.110.28.0 128.110.28.255 2154699776 2154700031
    13. 3 128.110.29.0/24 24 128.110.29.0 128.110.29.255 2154700032 2154700287
    14. 3 128.110.30.0/24 24 128.110.30.0 128.110.30.255 2154700288 2154700543
    15. 3 128.110.31.0/24 24 128.110.31.0 128.110.31.255 2154700544 2154700799

    Once, I get a hierarchical query of good work I can do the rest of the formatting with the values of the path and CYCLE but I'm not able to get a basic querying work. Does anyone have an idea of how to go on this subject using SQL only? I can write a PL/SQL procedure to add an ID of parent to each row, and then I know that I can create a hierarchical query from that but I was trying to avoid this method if possible.

    This is the right forum. I can see why there is only a single forum for SQL and PL/SQL, but I do not see why they suggest that there are separate forums.

    I'm not sure that you understand the problem.  In the ouptu (the line whose number 11) you have

    1. 2 128.110.28.0/22 22 128.110.28.0 128.110.31.255 2154699776 2154700799

    I guess this level = 2 means that the row is a child of a few rows, but which?  It was numbered line 7:

    1. 1 128.110.16.0/21 21 128.110.16.0 128.110.23.255 2154696704 2154698751

    ?  This would mean that the child has a higher than its parent network_dec.

    I think you want something like this:

    WITH got_parent AS

    (

    SELECT s. *- or whatever the desired columns

    ,       (

    SELECT MIN (ss.subnet) DUNGEON (DENSE_RANK FIRST ORDER BY ss.broadcast_dec - ss.network_dec)

    OF subnet_decimal_vals ss

    WHERE ss.network_dec<=>

    AND ss.broadcast_dec > = s.broadcast_dec

    AND ss.subnet <> s.subnet

    ) As a parent

    OF s subnet_decimal_vals

    )

    SELECT LEVEL AS lvl

    subnet, cidr_block, network, broadcast, network_dec, broadcast_dec

    OF got_parent

    START WITH parent IS NULL

    Parent = subnet PRIOR CONNECTION

    ;

    The result is:

    LVL SUBNET NETWORK BROADCAST NETWORK_DEC BROADCAST_DEC CID

    --- ---------------- --- ---------------- ---------------- ----------- -------------

    1 20 128.110.0.0/20 128.110.0.0 128.110.15.255 2154692608 2154696703

    2 128.110.1.0/24 24 128.110.1.0 128.110.1.255 2154692864 2154693119

    2 128.110.4.0/22 22 128.110.4.0 128.110.7.255 2154693632 2154694655

    2 128.110.8.0/22 22 128.110.8.0 128.110.11.255 2154694656 2154695679

    2 128.110.12.0/22 22 128.110.12.0 128.110.15.255 2154695680 2154696703

    1 128.110.16.0/21 21 128.110.16.0 128.110.23.255 2154696704 2154698751

    2 128.110.16.0/22 22 128.110.16.0 128.110.19.255 2154696704 2154697727

    2 128.110.20.0/22 22 128.110.20.0 128.110.23.255 2154697728 2154698751

    1 22 128.110.24.0/22 128.110.24.0 128.110.27.255 2154698752 2154699775

    1 22 128.110.28.0/22 128.110.28.0 128.110.31.255 2154699776 2154700799

    2 128.110.28.0/24 24 128.110.28.0 128.110.28.255 2154699776 2154700031

    2 128.110.29.0/24 24 128.110.29.0 128.110.29.255 2154700032 2154700287

    2 128.110.30.0/24 24 128.110.30.0 128.110.30.255 2154700288 2154700543

    2 128.110.31.0/24 24 128.110.31.0 128.110.31.255 2154700544 2154700799

    Close enough to what you have requested, but not exact.  If this isn't what you want, explain how you can tell if a line is one ancestor of the other in the tree.

    I guess that this subnet is unique.  If this is not the case, the above query will have to be changed a little, but only a little.

  • Slow running queries on a table partitioned by date

    I have a table partitioned by COLDATE, when I write a query to get the data is slow, even if I try to only get 1 day data. How can I correct this query runs faster?

    Oracle Version: 10g

    Number of lines:

    See the below query

    select COL1, COL2 RANK() OVER (PARTITION BY   COL1, COLDATE ORDER BY OTHER_DATE DESC) SHOW_RANK
    from 
    SAMPLETABLE
    where COLDATE BETWEEN '01-JAN-2015' AND '01-JAN-2015'
    
    
    create table SAMPLETABLE
    (
      COL1   NUMBER(10) not null,
      COL2        NUMBER(10) not null,
      COLDATE       DATE not null,
      OTHER_DATE DATE
    )
    partition by range (COLDATE)
    (
      partition PARTITION_01_2015 values less than (TO_DATE(' 2015-02-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
        tablespace MYTABLESPACE_2015
        pctfree 20
        initrans 1
        maxtrans 255
        storage
        (
          initial 64K
          minextents 1
          maxextents unlimited
        ),
      partition PARTITION_02_2015 values less than (TO_DATE(' 2015-03-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
        tablespace MYTABLESPACE_2015
        pctfree 20
        initrans 1
        maxtrans 255
        storage
        (
          initial 64K
          minextents 1
          maxextents unlimited
        ),
      .
      .
      .
      .
      .
      .
      partition PARTITION_12_2016 values less than (TO_DATE(' 2017-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
        tablespace MYTABLESPACE_2016
        pctfree 20
        initrans 1
        maxtrans 255
        storage
        (
          initial 64K
          minextents 1
          maxextents unlimited
        )
    

    Force the session in parallel mode and check with the parallel hint option.

    SQL > alter session force parallel dml.

    SQL > select / * + parallel (table_name, none of the servers) * / colA, colB from table_name;

  • Legacy and hierarchical queries

    Hi all

    Using the RDBMS 11.2.0.3, I would like to know if there was a way to "inherit" a parent row values when you use a hierarchical query.

    For example, with these values:

    CREATE TABLE foohh (
     id NUMBER,
     parent NUMBER,
     name VARCHAR2(64),
     continent VARCHAR2(32)
    );
    
    INSERT INTO foohh VALUES ( 1, null, 'United States', 'AMERICA' ) ;
    INSERT INTO foohh VALUES ( 2, 1, 'California', null ) ;
    INSERT INTO foohh VALUES ( 3, 2, 'San Francisco', null ) ;
    INSERT INTO foohh VALUES ( 4, 3, 'Golden Gate', null ) ;
    INSERT INTO foohh VALUES ( 5, null, 'China', 'ASIA' ) ;
    INSERT INTO foohh VALUES ( 6, 5, 'Beijing', null ) ;
    INSERT INTO foohh VALUES ( 7, 6, 'Great Wall', null ) ;
    
    

    I would like to be able to view the continent of all lines.

    The following query returns the leaves, but I don't see their continent:

    SELECT
     id
     ,connect_by_root(parent) parent
     ,sys_connect_by_path(name, '-->') path
     ,nvl( continent, 'unknown' ) continent
    FROM foohh
    WHERE
     connect_by_root(parent) is null
     and connect_by_isleaf=1
    CONNECT BY PRIOR id=parent
    ;
    
    

    CONTINENT OF PARENT ID PATH
    ---------- ---------- ------------------------------------------------------------ ------------------------------
    4-> United States-> California-> San Francisco-> Golden Gate unknown
    7-> China-> Beijing-> great unknown wall

    Is there a clever way to do this?

    Thank you

    Anthony

    Include below in him SELECT query.

    Continent CONNECT_BY_ROOT

  • Hierarchical queries with Rollup sum (CONNECTED BY GROUP BY ROLLUP)

    Hi all

    Imagine the following scenario: I have an ACCOUNT table that contains the accounts and their hierarchy (currently 5 levels) and a BALANCE table that holds the record for the balance of the accounts. Only CHILD accounts (level 5) have records in the table for BALANCE. Simple example:
    CREATE TABLE accounts (account_code VARCHAR2(30), parent_account VARCHAR2(30), account_desc VARCHAR2(400));
    CREATE TABLE balances (account_code VARCHAR2(30), balance_amount NUMBER(18,2));
    INSERT INTO ACCOUNTS VALUES ('TOT',NULL,'Total');
    INSERT INTO ACCOUNTS VALUES ('ANA1','TOT','General Expenses');
    INSERT INTO ACCOUNTS VALUES ('4801001','ANA1','Small Expenses');
    INSERT INTO ACCOUNTS VALUES ('4801002','ANA1','Transportation');
    INSERT INTO ACCOUNTS VALUES ('ANA2','TOT','Health Expenses');
    INSERT INTO ACCOUNTS VALUES ('4802001','ANA2','Healthcare');
    INSERT INTO ACCOUNTS VALUES ('4802002','ANA2','Facilities');
    
    
    INSERT INTO BALANCES VALUES ('4801001', 2000);
    INSERT INTO BALANCES VALUES ('4801002', 1000);
    INSERT INTO BALANCES VALUES ('4802001', 3000);
    INSERT INTO BALANCES VALUES ('4802002', 4000);
    What I need in this scenario is to run a hierarchical query, where each node, I calculate the sum of all its children (in TERMINAL nodes that are child accounts, this amount is the value of the balances itself). End result would be:
    TOT -> 10000
      ANA1 -> 3000
        4801001 -> 2000
        4801001 -> 1000
      ANA2 -> 7000
        4802001 -> 3000
        4802002 -> 4000
    I tried many ways and found a solution that works for a fixed amount of levels, basically he built the hierarchy and calculates the SYS_CONNECT_BY_PATH, then divides it as a regular expression and using GROUP BY ROLLUP to calculate the highest levels. Then I assemble again, now with the calculated values. Here's the example query:
    select level
        , NVL (vfinal.child_account,'TOTAL') ||' - '||
                            ( SELECT account_desc
                                FROM accounts 
                               WHERE account_code = vfinal.child_acct ) account_name
    
         , to_char(sum_bal, 'fm999g999g999g990') as rolled_up_balance
      from 
    (
    select coalesce( princ.lvl3, princ.lvl2, princ.lvl1 ) child_acct
         , DECODE ( princ.lvl2 , NULL 
                                     , NULL 
                                     , DECODE ( princ.conta_lvl3, NULL
                                     , princ.conta_lvl1,princ.conta_lvl2 ) ) parent_acct
         , sum(princ.balance_amount) sum_bal
    from (
    select hier.lvl1
         , hier.lvl2
         , hier.lvl3
         , hier.parent_account
         , hier.account_code child_acc
         , bal.balance_amount
      from ( select level  
                  , sys_connect_by_path( account_code, '/' ) hierarchy_acct
                  , REGEXP_SUBSTR(sys_connect_by_path( account_code, '/' ),'[^/]+',1,3) lvl3
                  , REGEXP_SUBSTR(sys_connect_by_path( account_code, '/' ),'[^/]+',1,2) lvl2
                  , REGEXP_SUBSTR(sys_connect_by_path( account_code, '/' ),'[^/]+',1,1) lvl1
                  , account_code
                  , parent_account  
               from accounts acc
               where level <= 3
               start with parent_account is null
               connect by nocycle prior account = parent_account
               order siblings by parent_account
               ) hier
          , balances  bal
      where bal.cod_conta  = hier.account_code
    ) princ
    where princ.lvl1 is not null
    group by rollup ( princ.lvl1
                    , princ.lvl2
                    , princ.lvl3 )
    
    order by princ.conta_lvl1
           , princ.conta_lvl2
           , princ.conta_lvl3
    ) vfinal
    where child_acct is not null
    start with parent_acct is null
    connect by nocycle prior child_acct = parent_acct
    All is said and done, what I need is to do the same thing for infinite levels, because this query has 3 fixed levels. Do you know how can I structure a new query where, regardless of the number of levels, amounts of parent are all wound like that?

    Thank you very much in advance! Best regards!
    Thiago

    Published by: Thiago Sep 6, 2011 11:31

    Published by: Thiago Sep 6, 2011 13:01
    select  account_code,
            (
             select  sum(balance_amount)
               from  accounts a2,
                     balances b
               where b.account_code(+) = a2.account_code
               start with a2.account_code = a1.account_code
               connect by a2.parent_account = prior a2.account_code
            ) balance_amount
      from  accounts a1
    /
    
    ACCOUNT_CODE    BALANCE_AMOUNT
    --------------- --------------
    TOT                      10000
    ANA1                      3000
    4801001                   3000
    4801002
    ANA2                      7000
    4802001                   7000
    4802002
    
    7 rows selected.
    
    SQL> 
    

    SY.

  • Building the tree balanced with SQL hierarchical queries

    Hi all

    I have the following hierarchical data with different levels of subtree:

    A0
    -A001
    -A00101
    A1
    -A101
    A2
    -A201
    -A20101
    -A201010001

    A0 subtree has 3 levels, A1 subtree has 2 levels and subtree of the A3 is level 4. I want to generate a tree balanced on the data with all levels of the subtree equal to the maximum number of levels available in the whole tree which, in this particular case, is 4.

    I don't know that it is possible with SQL. Script to generate the above mentioned are as below:

    CREATE TABLE codes_tree
    (node_id VARCHAR2 (10))
    parent_node_id VARCHAR2 (10)
    );
    INSERT INTO codes_tree VALUES ('A0', NULL);
    INSERT INTO codes_tree VALUES ('A001', 'A0');
    INSERT INTO codes_tree VALUES ('A00101', 'A001');
    ---
    INSERT INTO codes_tree VALUES ('A1', NULL);
    INSERT INTO codes_tree VALUES ('A101', 'A1');
    ---
    INSERT INTO codes_tree VALUES ('A2', NULL);
    INSERT INTO codes_tree VALUES ('A201', 'A2');
    INSERT INTO codes_tree VALUES ('A20101', 'A201');
    INSERT INTO codes_tree VALUES ('A201010001', 'A20101');

    Any help will be much appreciated.

    Thank you... Best regards

    Published by: naive2Oracle on May 12, 2011 19:40

    Published by: naive2Oracle on May 12, 2011 19:41

    Hello

    Of course, you can do it in SQL.
    One way is to take the normal output of hierarchical and manipulate the result set so that the leaves are repeated as often as necessary to make all branches of the same length. I have Oracle 10.2 available right now, so here's a solution that will work in Oracle 10 (and more):

    WITH     original_hierarchy     AS
    (
         SELECT     node_id
         ,     LEVEL               AS lvl
         ,     CONNECT_BY_ISLEAF     AS isleaf
         ,     ROWNUM               AS rnum
         FROM     codes_tree
         START WITH     parent_node_id     IS NULL
         CONNECT BY     parent_node_id     = PRIOR node_id
    )
    ,     got_max_lvl     AS
    (
         SELECT     o.*
         ,     MAX (lvl) OVER ()     AS max_lvl
         FROM     original_hierarchy     o
    )
    SELECT       LPAD ( ' '
                , 3 * ( ( d.lvl
                     + NVL (c.rnum, 1)
                     - 1
                     )
                   - 1
                   )
                ) || CASE
                   WHEN c.rnum > 1
                   THEN '*' || d.node_id || '*'
                   ELSE        d.node_id
                  END          AS display_id
    FROM            got_max_lvl     d
    LEFT OUTER JOIN       got_max_lvl     c  ON     d.isleaf     = 1
                           AND     c.rnum          <= 1 + d.max_lvl - d.lvl
    ORDER BY  d.rnum
    ,       c.rnum
    ;
    

    With the help of Oracle 11.2, it would be preferable to generate original_hierarchy as above, but to manipulate using a WITH recursive clause.
    Analytical functions often interfere with CONNECT BY, so I used a separate subquery to get max_lvl, do CONNECT BY in a sub-querry and analytic function in a separate subquery. I don't know what is needed on all versions.

    Output of your sample data:

    DISPLAY_ID
    -------------------------------
    A0
       A001
          A00101
             *A00101*
    A1
       A101
          *A101*
             *A101*
    A2
       A201
          A20101
             A201010001
    

    Below is a generic version of the same query, which I used to test this on scott.emp:

    DEFINE     table_name     = scott.emp
    DEFINE     id_col          = empno
    DEFINE     parent_id_col     = mgr
    DEFINE     display_col     = ename
    
    WITH     original_hierarchy     AS
    (
         SELECT     &display_col          AS display_txt
         ,     LEVEL               AS lvl
         ,     CONNECT_BY_ISLEAF     AS isleaf
         ,     ROWNUM               AS rnum
         FROM     &table_name
         START WITH     &parent_id_col     IS NULL
         CONNECT BY     &parent_id_col     = PRIOR &id_col
    )
    ,     got_max_lvl     AS
    (
         SELECT     o.*
         ,     MAX (lvl) OVER ()     AS max_lvl
         FROM     original_hierarchy     o
    )
    SELECT       LPAD ( ' '
                , 3 * ( ( d.lvl
                     + NVL (c.rnum, 1)
                     - 1
                     )
                   - 1
                   )
                ) || CASE
                   WHEN c.rnum > 1
                   THEN '*' || d.display_txt || '*'
                   ELSE        d.display_txt
                  END          AS display_id
    FROM            got_max_lvl     d
    LEFT OUTER JOIN       got_max_lvl     c  ON     d.isleaf     = 1
                           AND     c.rnum          <= 1 + d.max_lvl - d.lvl
    ORDER BY  d.rnum
    ,       c.rnum
    ;
    

    Output:

    DISPLAY_ID
    -----------------------------
    KING
       JONES
          SCOTT
             ADAMS
          FORD
             SMITH
       BLAKE
          ALLEN
             *ALLEN*
          WARD
             *WARD*
          MARTIN
             *MARTIN*
          TURNER
             *TURNER*
          JAMES
             *JAMES*
       CLARK
          MILLER
             *MILLER*
    

    Published by: Frank Kulash, May 13, 2011 06:38
    Adding the generic version

  • hierarchical queries - error: ORA-30929: clause ORDERBY brothers and SŒURS unauthorized

    Hello

    I have a hierarchical query in which, I automatically generate unordered list. I needed to get the children in the same level, so I used the brothers and sisters of the order by clause, but the procedure did not compile and threw an error Ora-30929 - brothers and sisters Order By Clause not allowed here.

    This is the query.

    SELECT CASE WHEN LAG(LEVEL,1,0) (ORDER OF ROWNUM) > = LEVEL THEN "< li >".
    (Level) lead CASE of OTHER COURSES (ORDER OF ROWNUM) WHEN LEVEL THEN
    CASE WHEN ROWNUM = 1 THEN ' < ul id = "sidebarmenu1" 'ELSE' < ul' END | ' > < li > '
    On the OTHER CASE WHEN ROWNUM = 1 THEN ' < ul id = "sidebarmenu1" 'ELSE' < ul ' END: ' > < li > ' END END |
    "< span >' | daevmt.short_menu_item |' </span >' |
    LEVEL of CASE - LEAD(LEVEL,1,1) (ROWNUM ORDINANCE) WHEN - 1 THEN NULL WHEN 0 THEN "< /li >".
    ELSE REPLACE (LPAD ('* ', LEVEL-LEAD(LEVEL,1,1) (ORDER OF ROWNUM),'* '), ' * ',' < /li > < /ul > < /li > ')
    END | CASE WHEN AHEAD (LEVEL, 1, 0) (ORDER OF ROWNUM) = 0 THEN '< /ul >"END of another null unordered_list, daevmt.menu_item, daevmt.link_url, daevmt.menu_id,.
    daevmt.above_menu_id
    OF dae_vs_my_tasks daevmt
    START WITH daevmt.above_menu_id = "TOPMENU".
    CONNECT BY PRIOR = daevmt.above_menu_id Daevmt.menu_id
    Brothers and SŒURS of ORDER BY daevmt.display_order;

    Can someone please help what is wrong in my query?

    Thanks in advance,

    Natarajan

    Published by: Nikita on March 7, 2011 23:26

    Published by: Nikita on March 7, 2011 23:28

    Published by: Nikita on March 7, 2011 23:28

    According to the Oracle documentation, this error is caused by 'ORDER brothers' and SŒURS clause of having in a query that is not a "CONNECT BY" clause that is clearly not the case here. I suspect it's because you use analytical functions 'LEAD' and 'LAG '. I took your statement, replaced the names of table and column with some in a table, I also has a structure of hierarchical and got the same error. After removing all the functions of 'LEAD' and 'LAG' it run OK.
    I don't think that helps you all.
    You may create another table with the contents of this table by adding 4 columns for rownum, level (level) lead and lag (level). Then use this table to generate your unsorted list html code.

  • How do to cancel/return HTTP request (because of slow SQL queries)

    Hi everyone - I have a flex application that requires a connection first. The connection queries my database to check if it's a name of user and password valid. 95% of the time the query to check if it's a name of user and password valid takes almost no time (less than one second) and the user connects very, very quickly. But on occasion, for some reason, the query takes a long time or something where the valid/invalid result is not returned for about 20 to 40 seconds. If I hit the button submit once again, it will return the SQL query and very quickly (in general) work very well. It is a simple query to find the name of user and password, so I think that something is maybe going on why it it takes so long.

    Two questions:

    1. no idea why this could happen that some of the time, and most of the time it works fine? Any suggestions or fix for this?

    2. an idea to solve this problem I came up with, I could just cancel and resend the request. Anyone know how I can cancel and return the most recent httprequest object? Should I have it checked to see if I do a query, and the result is not returned within 2 seconds, then cancel the current request and return? This does not solve the real problem, but it provide a workaround. If I was a user of connect, and a login page lasted 30 seconds, 5% of the time I connect, I get bored and leave or think that it does not work.

    Note: I may queries on the site, what happens the same question, but the connection of the user request is the most visible and the most important. I would like to find a solution to avoid other problems too.

    Help, please! Thank you!!

    As a side note, if you want to see the site/application I'm talking, it's www.yahtzo.com

    The site is fully functional, just try to fix some bugs. If you have the time, play some games!

    Thank you!

    HTTPServcie has a requestTimeout property, I don't remember exactly if there is a default value. Together, a custom value and if you get an answer for this amount of time the HTTPService will trigger the ResultEvent. Inspect the event you could get some extra info to know that there was a timeout.

    C

  • Hierarchical queries - difficult

    Hi all

    Thanks in advance,

    Please answer as soon as possible
    I would like to know some suggestions for repatriated PRDCT_CUSTM_HIER_SK not duplicated using hierarchical values to oracle 10 g

    suggest on how to modify the query without changing the status of CONNECT BY PRIOR

    Only if the query should never return values in a double PRDCT_CUSTM_HIER_SK.
    If you view the example we have 4 duplicate values

    Consider the table as below

    PRDCT_CUSTM_HIER_SK - ITEM_ID - PRNT_ID
    27945 - 74-0
    28977 - 75-74
    28100 - 85-75
    28538 - 86-85
    29557 - 115-74
    29477 - 116-115
    28117 - 360-86
    30894 - 360-85
    29039 - 390-115
    30876 - 431-360
    30839 - 432-360
    30883 - 433-360
    30863 - 434-360



    I EXECUTED THE GIVEN SCRIPT

    SELECT
    LEVEL CUSTM_HIER_LVL_ID
    CUSTM_HIER_ITEM_ID ITEM_ID,
    PRNT_ID CUSTM_HIER_PRNT_ID
    lpad (PRDCT_CUSTM_HIER_SK, level + length (PRDCT_CUSTM_HIER_SK) * 10-10,'-') PRDCT_CUSTM_HIER_SK
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 2,'/ ')) CUSTM_HIER_SK_LVL1
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 3,'/ ')) CUSTM_HIER_SK_LVL2
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 4,'/ ')) CUSTM_HIER_SK_LVL3
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 5,'/ ')) CUSTM_HIER_SK_LVL4
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 6,'/ ')) CUSTM_HIER_SK_LVL5
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 7,'/ ')) CUSTM_HIER_SK_LVL6
    , TO_NUMBER (GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_SK,'/'), 8,'/ ')) CUSTM_HIER_SK_LVL7
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 2,'/ ') CUSTM_HIER_NM_LVL1
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 3,'/ ') CUSTM_HIER_NM_LVL2
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 4,'/ ') CUSTM_HIER_NM_LVL3
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 5,'/ ') CUSTM_HIER_NM_LVL4
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 6,'/ ') CUSTM_HIER_NM_LVL5
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 7,'/ ') CUSTM_HIER_NM_LVL6
    , GET_TOKEN (SYS_CONNECT_BY_PATH(PRDCT_CUSTM_HIER_LVL_1_NM,'/'), 8,'/ ') CUSTM_HIER_NM_LVL7
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 2,'/ ') CUSTM_HIER_ITEM_ID_LVL1
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 3,'/ ') CUSTM_HIER_ITEM_ID_LVL2
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 4,'/ ') CUSTM_HIER_ITEM_ID_LVL3
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 5,'/ ') CUSTM_HIER_ITEM_ID_LVL4
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 6,'/ ') CUSTM_HIER_ITEM_ID_LVL5
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 7,'/ ') CUSTM_HIER_ITEM_ID_LVL6
    , GET_TOKEN (SYS_CONNECT_BY_PATH(ITEM_ID,'/'), 8,'/ ') CUSTM_HIER_ITEM_ID_LVL7
    CONNECT_BY_ROOT PRDCT_CUSTM_HIER_SK PRDCT_CUSTM_ROOT_SK
    PRDCT_CUSTM_PRNT_SK PRDCT_CUSTM_HIER_SK PRIOR
    PRDCT_CUSTM_HIER_LVL_1_NM PRDCT_CUSTM_HIER_NM
    PRDCT_CUSTM_HIER_LVL_1_NM BEFORE CUSTM_HIER_PRNT_NM
    CONNECT_BY_ROOT PRDCT_CUSTM_HIER_LVL_1_NM CUSTM_HIER_ROOT_NM
    , SYS_CONNECT_BY_PATH (PRDCT_CUSTM_HIER_LVL_1_NM, ' / ') CUSTM_HIER_NM_PATH
    PRDCT_CUSTM_HIER_SS_CD PRDCT_CUSTM_HIER_SS_CD
    PRDCT_BRND_NM PRDCT_BRND_NM
    PRDCT_BRND_ID PRDCT_BRND_ID
    PRDCT_BRND_SK PRDCT_BRND_SK
    Of
    OBJECTIVE TMP_PRDCT_CUSTM_HIER
    START BY ITEM_ID = (SELECT MIN (ITEM_ID) OF BI. TMP_PRDCT_CUSTM_HIER)
    CONNECT PRIOR ITEM_ID = PRNT_ID
    /


    THE OUTPUT IS AS FOLLOWS

    1-74 - 0 - 27945
    2-115-74-29557
    3-390-115-29039
    3-116-115-29477
    2-75-74-28977
    3-85-75-28100
    4-360-85-30894
    *5 -- 431 -- 360 -- ----------------------------------------30876*
    *5 -- 432 -- 360 -- ----------------------------------------30839*
    *5 -- 433 -- 360 -- ----------------------------------------30883*
    *5 -- 434 -- 360 -- ----------------------------------------30863*
    4 -- 86 -- 85 -- ------------------------------28538
    5 -- 360 -- 86 -- ----------------------------------------28117
    *6 -- 431 -- 360 -- --------------------------------------------------30876*
    *6 -- 432 -- 360 -- --------------------------------------------------30839*
    *6 -- 433 -- 360 -- --------------------------------------------------30883*
    *6 -- 434 -- 360 -- --------------------------------------------------30863*

Maybe you are looking for

  • How do go back (or forward) several steps without having to navigate in the history tree?

    Internet Explorer has a down arrow button (drop) next to the rear and the next button which allows to skip several steps back (or forward) in two clicks. How do I do that in Firefox? Navigation in the history for this simple task tree is much heavier

  • many runtime errors and slow responses

    Originla title: what is WiseFixer and is a legitimate program? I get a lot of errors of execution and slow responses.  Ran the scan in the wise Fixer but was reluctant to unlock the registry to run the repair.

  • Windows XP must be activated by connecting to the laptop

    I have a problem with my laptop when I try to log on I put my password, then a message appears and says. This copy of windows must be activated with microsoft you can connect. You want to activate windows now? Yes or no. I click on Yes, then a window

  • Help with slow internet with new HP Envy

    I have a new HP Envy and my web sites load as dial-up, video breaks frequently.   With the help of Motorola modem/router combo.  Desktop HP loads and diffuses very quickly, HP laptop cheaper new husband load and distributes very quickly.   I'm almost

  • AccountKey ActiveSync/Accountid?

    Hello is it possible to get the accountkey (accountid) of Microsoft Exchange ActiveSync? I get the message "private" accounts, but I would like to send an e-mail via the account of buisness? Thank you.