Need query to find rows

hi
i need query to find out for one perticular row from parent table ,,which references how many rows in how many child tables , and child to child tables. in herirachy fashion.
if anybody know please help me

I guess it would be difficult to come up with a generic script to get answer that can help in all situations...you have to see the table definitions in your schema and try...like in case of emp and dept table,
select * from dept where deptno in
(select deptno,count(*) from emp group by deptno having count(*)>1)
(this gives rows from dept table where , for corresponding deptno there are more than one rows in emp table).
Thanks
Nirav

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    Hi All,
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    SAS - count is 1
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    Girish GWhenever you have a question, please post a little sample data (CREATE TABLE and INSERT statements) and the results you want from that data.
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       1 S-S-S-P-S-B-S-S-C-S-P     SC                                 1
       1 S-S-S-P-S-B-S-S-C-S-P     SP                                 2
       1 S-S-S-P-S-B-S-S-C-S-P     SS                                 3
       2 S-S-A-S-S-A-A-C-S-P-S-A   SAAC                               1
       2 S-S-A-S-S-A-A-C-S-P-S-A   SAS                                1
       2 S-S-A-S-S-A-A-C-S-P-S-A   SP                                 1
       2 S-S-A-S-S-A-A-C-S-P-S-A   SS                                 2(This corresponds to what you posted, except that there is only one output row for 'SS' when x_id=1.)
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         ,     REGEXP_COUNT (txt, 'S')     AS s_cnt
         FROM     table_x
    ,     cntr          AS
         SELECT     LEVEL     AS n
         FROM     dual
         CONNECT BY     LEVEL <= (
                                   SELECT  MAX (s_cnt)
                             FROM    got_s_cnt
    ,     got_pattern     AS
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         ,     s.txt
         ,     c.n
         ,     REGEXP_SUBSTR ( REPLACE ( SUBSTR ( txt
                                                 , INSTR (s.txt, 'S', 1, c.n)
                         , 'SA*[^A]'
                            )       AS pattern
         FROM    got_s_cnt  s
         JOIN     cntr        c  ON  c.n  <= s.s_cnt
    SELECT       x_id
    ,       txt
    ,       pattern
    ,       COUNT (*)     AS cnt
    FROM       got_pattern
    WHERE       pattern     IS NOT NULL
    GROUP BY  x_id
    ,            txt
    ,       pattern
    ORDER BY  x_id
    ,            pattern
    ;At the heart of this query is the call to REGEXP_SUBSTR:
    REGEXP_SUBSTR ( x
               , 'SA*[^A]'
               )which is looking for:
    S     = the letter S
    A*     = the letter A, repeated 0 or more times
    [^A]     = anything except the letter A

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    Thanks for the offer, unfortunately we would need something certified by Adobe.
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