Reg: Hierarchical query (using connection by)

Hi all
I got the result with the hierarchical query in the form:
* / qxxh *.
* / qxxh/jxobcbg *.
* / qxxh/jxobcbg/n00wcp4 *.
* / qxxh/jxobcbg/n00wcp4 / 000263 x *.
* / qxxh/jxobcbg/n00wcp4 / x 000263 / p0263 *.
* / qxxh/jxxocbg *.
* / qxxh/jxxocbg/n00voc1 *.
* / qxxh/jxxocbg/n00voc1 / x 000589 *.
* / qxxh/jxxocbg/n00voc1 / x 000589 / p0589 *.
* / qxxh/jxuwxxh *.
* / qxxh/jxuwxxh/n00xpxf *.
* / qxxh, jxuwxxh, n00xpxf, m00bxpl *.
* / qxxh/jxuwxxh/n00xpxf/m00bxpl / 000522 x *.
* / qxxh/jxuwxxh/n00xpxf/m00bxpl / 000522 x / p0522 *.

Here, I want to select only the maximum path. Here I used "SYS_CONNECT_BY_PATH.
Please let meknow how to do this?
Thanks in advance.

Published by: udeffcv on December 9, 2009 22:03

udeffcv wrote:
Hi all
I got the result with the hierarchical query in the form:
* / qxxh *.
* / qxxh/jxobcbg *.
* / qxxh/jxobcbg/n00wcp4 *.
* / qxxh/jxobcbg/n00wcp4 / 000263 x *.
* / qxxh/jxobcbg/n00wcp4 / x 000263 / p0263 *.
* / qxxh/jxxocbg *.
* / qxxh/jxxocbg/n00voc1 *.
* / qxxh/jxxocbg/n00voc1 / x 000589 *.
* / qxxh/jxxocbg/n00voc1 / x 000589 / p0589 *.
* / qxxh/jxuwxxh *.
* / qxxh/jxuwxxh/n00xpxf *.
* / qxxh, jxuwxxh, n00xpxf, m00bxpl *.
* / qxxh/jxuwxxh/n00xpxf/m00bxpl / 000522 x *.
* / qxxh/jxuwxxh/n00xpxf/m00bxpl / 000522 x / p0522 *.

Here, I want to select only the maximum path. Here I used "SYS_CONNECT_BY_PATH.
Please let meknow how to do this?
Thanks in advance.

Published by: udeffcv on December 9, 2009 22:03

What do you mean by maximum path? is this...
* / qxxh/jxobcbg/n00wcp4 / x 000263 / p0263 *.
* / qxxh/jxxocbg/n00voc1 / x 000589 / p0589 *.
* / qxxh/jxuwxxh/n00xpxf/m00bxpl / 000522 x / p0522 *.

is it child nodes?
so, you would like to see
Column nickname... CONNECT_BY_ISLEAF example, you can find it in the link below
http://download.Oracle.com/docs/CD/B14117_01/server.101/b10759/pseudocolumns001.htm#sthref670

Ravi Kumar

Tags: Database

Similar Questions

  • Help in hierarchical query

    Hello

    I have data like this.

    EMPID ENAME MANAGERID
    1A
    2 1
    1 OF 3
    4 2
    5 2
    6 3
    4 OF 7

    And I want to show the data in this way. SUPERMNAGERID is always 1, regardless of the fact that that which is his immediate superior.

    EMPID ENAME MANAGERID SUPERMNAGERID
    1A
    2-1-1
    3-1-1
    4-2-1
    5-2-1
    6-3-1
    7-4-1

    Your help on this would be really useful.

    Edited by: user10827825 may 6, 2012 12:36

    Most inefficient way. If you want to use a hierarchical query, use START WITH and to get the "supermanager" use CONNECT_BY_ROOT:

    WITH t AS
         (SELECT 1 EMPID, 'A' ENAME, NULL MANAGER_ID
            FROM DUAL
          UNION ALL
          SELECT 2 EMPID, 'B' ENAME, 1 MANAGER_ID
            FROM DUAL
          UNION ALL
          SELECT 3 EMPID, 'C' ENAME, 1 MANAGER_ID
            FROM DUAL
          UNION ALL
          SELECT 4 EMPID, 'D' ENAME, 2 MANAGER_ID
            FROM DUAL
            UNION ALL
          SELECT 5 EMPID, 'E' ENAME, 2 MANAGER_ID
            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
      START WITH MANAGER_ID IS NULL
      CONNECT BY MANAGER_ID = PRIOR EMPID
      ORDER BY EMPID;
    

    Compare the number of lines hierarchy is generated by bot queries:

    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    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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    I have a hierarchical query basically he brings an organization chart. We start with the Manager id, get everything that employees of the person. If the employee is also a Manager I want to get the employees of that person and turn them right after that person. I don't bother with the entire query but relevant part is:
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    Beth        Frank
    Alex        Beth
    Charles     Beth
    Ed          Beth
    Dean        Frank
    George      Frank
    Benny       George
    David       George
    Sam         George
    Dan         Sam
    Harry       Sam
    John        Sam
    Terry       George
    James       Frank
    Ken         Frank
    Mike        Ken
    Warren      Ken
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    Published by: kendenny on July 28, 2010 07:31

    Make use of the ORDER of Friars and SŒURS clause in the hierarchical queries to define the order of child columns.

    START WITH em.mgr_id = pi_mgr_id
          CONNECT BY nocycle PRIOR em.emp_id = em.mgr_id
            *order siblings by name1;*
    
  • Getting the line without the use of hierarchical query values

    Hi all


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    1,
    2, 1
    3, 2
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    10, 1

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    1
    2
    3
    4

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    Thanks in advance.


    Thank you
    PAL

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      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  ),
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     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
    

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    Published by: Frank Kulash, November 26, 2008 14:09

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    Hello

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

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    Cambrault                        148                                 100                 2

    Bates                              172                                 148                 3

    Bloom                             169                                  148                3

    Fox                                 170                                  148                3

    Kumar                             173                                  148                3

    Ozer                                168                                  148               3

    Smith                               171                                  148              3

    De Haan                           102                                   100             2

    Hunold                              103                                   102             3

    Austin                               105                                   103             4

    Ernst                                 104                                   103             4

    Lorentz                              107                                   103             4

    Pataballa                           106                                    103            4

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    Ande                                 166                                    147             3

    Banda                               167                                     147             3

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    Kind regards

    mebu

    Mebu wrote:

    I used a hierarchical query. But based on Login, I want to display only members under him.

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    Employees

    START WITH employee_id = 100

    CONNECT BY PRIOR employee_id = manager_id

    Brothers and SŒURS of ORDER BY last_name;

    See what/when/where? How to get the answers from the forum

    Describe the requirements clearly and completely, using the APEX, Oracle and terminology of web standard.

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    Select 2: the nurse, '123' Str of all union double

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    Select option 4: the nurse, 'john doe' double Str

    )

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    /

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    Dear all

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

    KING

    | __JONES

    |   | __SCOTT

    |   | __ADAMS

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    |      | __SMITH

    | __BLAKE

    |   | __ALLEN

    |   | __WARD

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    create table APPL_TAXONOMY_TL (module_id varchar2(30), description varchar2(100), user_module_name varchar2(30), language varchar2(5));
    create table APPL_TAXONOMY (MODULE_ID    varchar2(30),    MODULE_NAME    varchar2(30), MODULE_TYPE varchar2(10),    MODULE_KEY varchar2(30));
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    insert into APPL_TAXONOMY_TL values ('10', null, 'General Ledger', 'US' );
    
    insert into APPL_TAXONOMY_HIERARCHY values ('1', 'DDDDDDDD');
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    insert into APPL_TAXONOMY_HIERARCHY values ('20', '2');
    
    insert into APPL_TAXONOMY values ('1', 'Fusion', 'PROD', 'Fusion');
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    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
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                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
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    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
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    Thanks in advance for your suggestions!

    John

    Hello

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    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

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    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.

  • Hierarchical data, how to aggregate over levels in hierarchical query?

    Hello

    I hope someone can help me.

    I held in a data table ("" what part was built in what other part of when when? "')
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    1 NULL NULL NULL
    2/1 2010 2012
    3 2 2011 2013
    4 2 2013 NULL

    What are the parts is stored in a separate table.

    Now I want to know when when which part was built in the first level, in the example, I want to know that
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    3 was part of 1 between 2011 and 2012
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    What version of Oracle you are on?

    In 11.2.x, you can use the recursive subquery factoring. Something like

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    select 3, 2, 2011, 2013 from dual union all
    select 4, 2, 2013, null from dual
    )
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    select id, id, parent_id, 0, 9999
    from t
    where parent_id is null
    union all
    select root_id, t.id, t.parent_id
    , greatest(nvl(t.build_in,0), nvl(b.build_in,0))
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    from c1 b, t
    where b.id = t.parent_id
    )
    select * from c1
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    ROOT_ID ID    PARENT_ID  BUILD_IN  BUILD_OUT
    ------- ----- ---------- --------- ----------
    1       1                0         9999
    1       2     1          2010      2012
    1       3     2          2011      2012      
    

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    from t
    start with parent_id is null
    connect by parent_id = prior id
    )
    where max_in < min_out
    ;
    

    but it is not powerful enough. This version compares the dates between a current and previous levels, but the recursive subquery is to compare the dates in the current level for the winners of the comparisons to the previous level. Not sure if it's an important distinction for your needs, however.

    If you are on 11.2 I advise to use the recursive subquery factoring. If this isn't the case, you can try the link by version.

    Kind regards
    Bob

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    Thank you

    Sentinel

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    I have two tables:

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    Now, I need to run a query with the following recursive logic

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    } < /i >

    (I hope the pseudo-code is understandable - the query is executed as < i > calcQuota (originalFolderId, true) < /i >).

    Now, my question is if I can accomplish this with a single hierarchical query, or if I should implement as a recursive procedure stored.

    Thank you!

    P.S. I use Oracle XE (10g)

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    My problem is that my hierarchical query seems only to cut values that do not meet my criteria, but still includes their children. When my query hits a record that does not meet my criteria, I want it to stop there. I tried including the criteria in just the ' ' a where clause of the query and also put the criteria in clause "connect by" as well, but nothing has been set. Will you please keep in mind that I'm using Oracle 8i, so I can't use some of the "nice" statements for hierarchical queries that they introduced in 9. I'm stuck with "Start With... Connect in "."

    I have examples of tables/data I can post if someone needs to see to help me, but to start with, here my current query:
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              ,     c_bill.qty_per                          AS     c_qty_per_p
              ,     c_bill.qty_per_type                     AS     c_qty_per_type
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              ,     c_bill.comp_off_adj                     AS     rqd_offset
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              AND     c_bill.begn_eff_dt     <=     SYSDATE     
         ) a
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    Hello

    The criterion in the outer query

    c_part_type = 'M'
    

    applies only to the results; You can still get the descendants of these lines in the output. This seems to be the only thing that is not repeated in the CONNECT BY clause.

    If you don't want to repeat the criteria in the WHERE clause and the CONECT BY clause, you can apply them once in a view online.
    The following example works in Oracle 8.1:

    SELECT     LPAD ( ' '
              , 3 * (LEVEL - 1)
              ) || ename          AS iname
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    ,     mgr
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         SELECT     ename
         ,     empno
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    ;
    

    Not only is excluded BLAKE, but the descendants of JAMES BLAKE, MARTIN, etc. are excluded, too, without repeating the condition.

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