Hierarchical query help

Version: Oracle Database 11 g Enterprise Edition Release 11.2.0.2.0 - 64 bit Production

Hello

I can't just writing this hierarchical query.

CREATE TABLE resources
(
    resource_name   VARCHAR2( 20 )
   ,resource_id     NUMBER
   ,resource_mgr_id NUMBER
);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Phupinder',29269,29298);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Terry',29298,NULL);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Matt',4876,32942);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Bryan',29429,29269);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Cathy',32942,4922);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Scott',4257,29429);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Joe',44026,4876);

INSERT INTO resources(resource_name,resource_id,resource_mgr_id)
VALUES ('Greg',4922,4257);

I also tried this:

  1. SELECT DISTINCT R.resource_name
  2. r.resource_id
  3. From r resources
  4. WHERE r.resource_id IN (SELECT DISTINCT r2.resource_mgr_id
  5. RESOURCE r2
  6. WHERE r.resource_id = r2.resource_mgr_id OR
  7. R2.resource_mgr_id IS NULL)
  8. ORDER BY 1;

Which gives correct results, but isn't that a hierarchical query?

It was the (almost) good solution to your problem.

The only thing is:

-you don't need to do a correlated subquery to solve the problem

-you don't need to separate your subqurey

-you don't need to separate your mainquery

SELECT resource_name, id_ressource

RESOURCES

where id_ressource in (select r2.resource_mgr_id from r2 resources);

would be sufficient.

HTH

Tags: Database

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    Published by: user13276471 on December 31, 2012 03:21

    Hello

    Welcome to the forum!

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        FROM DUAL
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      UNION ALL
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        UNION ALL
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        FROM DUAL
      UNION ALL
      SELECT 6 EMPID, 'F' ENAME, 3 MANAGER_ID
        FROM DUAL
      UNION ALL
      SELECT 7 EMPID, 'G' ENAME, 4 MANAGER_ID
        FROM DUAL
        )
SELECT EMPID, ENAME, MANAGER_ID, CONNECT_BY_ROOT EMPID
  FROM t
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  CONNECT BY MANAGER_ID = PRIOR EMPID
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  2       (SELECT 1 EMPID, 'A' ENAME, NULL MANAGER_ID
  3          FROM DUAL
  4        UNION ALL
  5        SELECT 2 EMPID, 'B' ENAME, 1 MANAGER_ID
  6          FROM DUAL
  7        UNION ALL
  8        SELECT 3 EMPID, 'C' ENAME, 1 MANAGER_ID
  9          FROM DUAL
 10        UNION ALL
 11        SELECT 4 EMPID, 'D' ENAME, 2 MANAGER_ID
 12          FROM DUAL
 13          UNION ALL
 14        SELECT 5 EMPID, 'E' ENAME, 2 MANAGER_ID
 15          FROM DUAL
 16        UNION ALL
 17        SELECT 6 EMPID, 'F' ENAME, 3 MANAGER_ID
 18          FROM DUAL
 19        UNION ALL
 20        SELECT 7 EMPID, 'G' ENAME, 4 MANAGER_ID
 21          FROM DUAL
 22          )
 23  SELECT count(*)
 24    FROM t
 25    CONNECT BY PRIOR  MANAGER_ID = EMPID
 26    ORDER BY EMPID;

  COUNT(*)
----------
        18

SQL> WITH t AS
  2       (SELECT 1 EMPID, 'A' ENAME, NULL MANAGER_ID
  3          FROM DUAL
  4        UNION ALL
  5        SELECT 2 EMPID, 'B' ENAME, 1 MANAGER_ID
  6          FROM DUAL
  7        UNION ALL
  8        SELECT 3 EMPID, 'C' ENAME, 1 MANAGER_ID
  9          FROM DUAL
 10        UNION ALL
 11        SELECT 4 EMPID, 'D' ENAME, 2 MANAGER_ID
 12          FROM DUAL
 13          UNION ALL
 14        SELECT 5 EMPID, 'E' ENAME, 2 MANAGER_ID
 15          FROM DUAL
 16        UNION ALL
 17        SELECT 6 EMPID, 'F' ENAME, 3 MANAGER_ID
 18          FROM DUAL
 19        UNION ALL
 20        SELECT 7 EMPID, 'G' ENAME, 4 MANAGER_ID
 21          FROM DUAL
 22          )
 23  SELECT count(*)
 24    FROM t
 25    START WITH MANAGER_ID IS NULL
 26    CONNECT BY MANAGER_ID = PRIOR EMPID
 27    ORDER BY EMPID;

  COUNT(*)
----------
         7

SQL>

    Now imagine the table real a couple thousand lines.

    SY.

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,     mgr
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     ,     mgr
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)
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select id, id, parent_id, 0, 9999
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    Dear all

    I use Oracle Database 10g and I need a hierarchical query giving following release

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where H.source_module_id = b.module_id
and b.module_id = vl.module_id
and vl.module_type not in ('PAGE', 'LBA')
UNION ALL
select distinct table_name, table_name,  regexp_substr(table_name,'[^_]+',1,1) 
from TABLES --needed as a link between TABLES and APPL_TAXONOMY
union all
select module_key as source_module_id, module_name as user_module_name, module_id as target_module_id  from appl_taxonomy VL where VL.module_type = 'APP')
SELECT  case when connect_by_isleaf = 1 then 0
            when level = 1             then 1
            else                           -1
       end as status,
LEVEL,
user_module_name as title,
null as icon,
ltrim(user_module_name, ' ') as value,
null as tooltip,
null as link
      FROM MODULES
      START WITH source_module_id = '1'
      CONNECT BY PRIOR source_module_id = target_module_id
ORDER SIBLINGS BY user_module_name;

    In Oracle APEX, this gives a tree with the appropriate data, you can see here:

    https://Apex.Oracle.com/pls/Apex/f?p=32581:29 (username: guest, pw: app_1000);

    The SQL of the query results are:

    STATUSTITLE LEVELVALUE

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

    11 oracle FusionOracle Fusion
    -12 financial tablesFinancials
    -13 accounts payableAccounts payable
    04 APAP
    -1General Accounting 3General Accounting
    -14 GLGL
    05 GL_JE_CATEGORIESGL_JE_CATEGORIES
    05 GL_JE_SOURCES_TLGL_JE_SOURCES_TL
    02 human resourcesHuman resources

    The lowest level is the name of the table to level 5. HR is not any level under level 2, in the same way, "AP" (level 4) has nothing below, i.e. no level 5 and that's why I don't want to show these nodes. Is this possible with the above query?

    Thanks in advance for your suggestions!

    John

    Hello

    The following query will include only the nodes of level = 5 and their ancestors (or descendants):

    WITH modules LIKE

    (

    SELECT h.source_module_id

    b.user_module_name AS the title

    h.target_module_id

    To appl_taxonomy_hierarchy:

    appl_taxonomy_tl b

    appl_taxonomy vl

    WHERE h.source_module_id = b.module_id

    AND b.module_id = vl.module_id

    AND vl.module_type NOT IN ('PAGE', "LBA")

    UNION ALL

    SELECT DISTINCT

    table-name

    table_name

    , REGEXP_SUBSTR (table_name, ' [^ _] +')

    From the tables - required as a link between the TABLES and APPL_TAXONOMY

    UNION ALL

    SELECT module_key AS source_module_id

    AS user_module_name module_name

    module_id AS target_module_id

    Of appl_taxonomy vl

    WHERE vl.module_type = 'APP '.

    )

    connect_by_results AS

    (

    SELECT THE CHECK BOX

    WHEN CONNECT_BY_ISLEAF = 1 THEN 0

    WHEN LEVEL = 1 THEN 1

    OF ANOTHER-1

    The END as status

    LEVEL AS lvl

    title

    -, NULL AS icon

    , LTRIM (title, "") AS the value

    -, NULL as ToolTip

    -, Link AS NULL

    source_module_id

    SYS_CONNECT_BY_PATH (source_module_id - or something unique

    , ' ~' - or anything else that may occur in the unique key

    ) || ' ~' AS the path

    ROWNUM AS sort_key

    Modules

    START WITH source_module_id = '1'

    CONNECT BY PRIOR Source_module_id = target_module_id

    Brothers and SŒURS of ORDER BY title

    )

    SELECT the status, lvl, title, value

    -, icon, tooltip, link

    OF connect_by_results m

    WHEN THERE IS)

    SELECT 1

    OF connect_by_results

    WHERE the lvl = 5

    AND the path AS ' % ~' | m.source_module_id

    || '~%'

    )

    ORDER BY sort_key

    ;

    You may notice that subqueries modules and the connect_by_results are essentially what you've posted originally.  What was the main request is now called connect_by_results, and it has a couple of additional columns that are necessary in the new main request or the EXISTS subquery.

    However, I am suspicious of the 'magic number' 5.  Could you have a situation where the sheets you are interested in can be a different levels (for example, some level = 5 and then some, into another branch of the tree, at the LEVEL = 6, or 7 or 4)?  If so, post an example.  You have need of a Query of Yo-Yo, where you do a bottom-up CONNECT BY query to get the universe of interest, and then make a descendant CONNECT BY query on this set of results.

  • Support for hierarchical query

    Hi all

    I must be tired and can't think clearly, so I am a little confused the following query.

    The environment is Oracle 9i:

    Oracle9i Enterprise Edition Release 9.2.0.8.0 - 64 bit Production

    PL/SQL Release 9.2.0.8.0 - Production

    CORE Production 9.2.0.8.0

    AMT for HP - UX: 9.2.0.8.0 - Production Version

    NLSRTL Version 9.2.0.8.0 - Production

    Suppose I have the following data:

    with mydata as

    (

    Select the code 1, code_high, null, 'John' cname 'Smith' csurname, 'X' union resp. double all the

    Select 2 code, 1 code_high, cname 'Bill', 'White' csurname, RESP null in union double all the

    Select 3 code, code_high 2, 'Fred' cname 'Reed' csurname, 'X' union resp. double all the

    Select 4 code, code_high, null, 'Tim' cname 'Hackman' csurname, 'X' union resp. double all the

    Select code 5, code_high 4, 'John', 'Reed' cname csurname resp null in union double all the

    Select 6 code, code_high 5, cname 'Bill', 'Hakcman' csurname, 'X' union resp. double all the

    Select the code 7, code_high 6, cname 'Fred' csurname 'White', null union resp. double all the

    Select code 8, code_high 7, 'Bill' cname 'Smith' csurname, resp. union null double all the

    Select 9 code, code_high 8, cname "Tom", "Reed" csurname, null double RESP

    )

    Select *.

    of mydata;

    CODE CODE_HIGH CNAME CSURNAME RESP

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

    John Smith 1 X

    2 1 bill White

    3 2 Fred Reed X

    4 Tim Hackman X

    5 4 John Reed

    6 5 bill Hakcman X

    7 6 Fred white

    8 7 bill Smith

    9 8 Tom Reed

    It is a hierarchical query where code_high represents the father.

    I need to find in the hierarchy of higher level responsible for each code.

    Suppose I want to find in the hierarchy, one with resp = 'X '.

    Run the following query I find for the code = 9

    Select phone, cname, csurname code

    of mydata

    When resp = 'X '.

    and rownum = 1

    Connect prior code_high = code

    start with code = 9;

    CODE CNAME CSURNAME

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

    Bill 6 Hakcman

    Is there a way to get the full list with the loaded correspondents.

    The expected results are:

    CODE CODE_HIGH CNAME CSURNAME RESP. RESP_CODE RESP_NAME RESP_SURNAME

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

    1 John Smith John Smith 1 X

    2 1 bill White 1 John Smith

    3 2 Fred Reed X 3 Fred Reed

    Tim Hackman 4 X 4 Tim Hackman

    5 4 John Smith 4 Tim Hackman

    6 5 bill Hakcman Bill Hakcman 6 X

    7 6 Fred White 6 Bill Hakcman

    8 7 bill Smith 6 Bill Hakcman

    9 8 Tom Reed 6 Bill Hakcman

    Kind regards.

    Alberto

    Hi, Alberto.

    I know that you are using Oracle 9; That's why I mentioned that you would have to use a substitute for CONNECT_BY_ROOT.  Before I could show how, I saw the solution of the Padders, which is probably simpler and more efficient for this work.  Padders used REGEXP_SUBSTR, which is not available in Oracle 9, but you can use SUBSTR and INSTR instead.

    Here is the solution of the Padders for Orcle 9:

    WITH got_resp_path AS

    (

    SELECT m.*

    RTRIM (SYS_CONNECT_BY_PATH (CASE

    WHEN resp = 'X '.

    THEN the code

    END

    , ' '

    )

    ) AS resp_path

    OF mydata m

    START WITH code_high IS NULL

    CONNECT BY code_high = code PRIOR

    )

    C. SELECT

    r.code AS resp_code

    r.cname AS resp_name

    r.csurname AS resp_surname

    OF got_resp_path c

    JOIN mydata r ON r.code = TO_NUMBER (SUBSTR (c.resp_path

    INSTR (c.resp_path

    , ' '

    -1

    )

    )

    )

    ORDER BY c.code

    ;

    I agree that what you posted in your last post is not very satisfactory.  Rather than make a CONNECT a separate query for each column of resp_ you want to view, you can modify it to get only the unique code and then use it in a join, as Padders, to get all the other columns you need.

  • In a hierarchical query, is it possible for a line having more than one immediate ancestor?

    Hello

    Question:

    In a hierarchical query, is it possible for a line having more than one immediate ancestor?

    Answer:



    Isn't it?  Surely this is Yes?


    Thank you

    Jason

    Line may have ancestors as much as you want. It goes the same for successors.

    SY.

  • Getting the line without the use of hierarchical query values

    Hi all


    I want to know the hierarchical values without using a hierarchical query. I have two tables EMP, DEPT

    EMP table is to have two columns (empid, mgrid)
    DEPT table is to have two columns (deptid, empid)

    Data of the EMP

    1,
    2, 1
    3, 2
    4, 3

    Data DEPT

    10, 1

    Each time, I gave deptid = 10, I need to know the this deptid empid then who is this empid (child levels as well). In this case, the output should be
    1
    2
    3
    4

    I don't want to use hierarchical query.

    Thanks in advance.


    Thank you
    PAL

    You can use the RECURSIVE subquery, if you are 11 GR 2, like this

    SQL> with EMP (empid,mgrid) as
  2  (
  3  select 1,null from dual union all
  4  select 2, 1 from dual union all
  5  select 3, 2 from dual union all
  6  select 4, 3 from dual
  7  ),
  8  dept(deptid,empid) as
  9  (
 10  select 10, 1  from dual
 11  ),
 12  t(empid,mgrid) as
 13  (
 14  select empid,mgrid
 15  from emp e
 16  where empid in (
 17      select d.empid
 18      from dept d
 19      where deptid = 10
 20                  )
 21  union all
 22  select e.empid,e.mgrid
 23  from emp e, t
 24  where e.mgrid = t.empid
 25  )
 26  select *
 27  from t;

     EMPID      MGRID
---------- ----------
         1
         2          1
         3          2
         4          3

    Yet, the question is valid - why can't use you a hierarchical query?

    Published by: JAC on May 29, 2013 12:37

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