by comparing the number of records in the table child

I have 3 tables below:

Table 1 - Parent Table
Table table2 child - who has the number of records for each record in Table1 (row_id in Table1 corresponds to par_row_id in Table2)
Child table table 3 - has that number of records for each record in Table1 (row_id in Table1 corresponds to par_row_id in table 3)

I would like to run an Oracle SQL query to identify all the separate row_id on Table1, for which the number of child records in Table2 does not match the number of child records in Table3

I ask for help to identify the best way to identify these records in the parent table. If possible an example too, please.

Thank you

Published by: cseshan on February 3, 2013 01:35

All by adding the simplest example (with the sample data that you were supposed to provide) below

create table table1(row_id number primary key);

create table table2(par_row_id number references table1);

create table table3(par_row_id number references table1);

insert into table1 values(1);
insert into table2 values(1);
insert into table3 values(1);

insert into table1 values(2);
insert into table2 values(2);
insert into table2 values(2);

insert into table1 values(3);
insert into table3 values(3);
insert into table3 values(3);

insert into table1 values(4);
insert into table2 values(4);
insert into table2 values(4);
insert into table2 values(4);
insert into table3 values(4);
insert into table3 values(4);

commit;

with child1_cnt as
(
  select par_row_id,count(*) cnt
  from table2
  group by par_row_id
),
child2_cnt as
(
  select par_row_id,count(*) cnt
  from table3
  group by par_row_id
),
all_counts as
(
  select p.row_id,nvl(c1.cnt,0) c1_cnt,nvl(c2.cnt,0) c2_cnt
  from table1 p
   left outer join child1_cnt c1
     on ( p.row_id = c1.par_row_id )
   left outer join child2_cnt c2
     on ( p.row_id = c2.par_row_id )
)
select row_id,c1_cnt,c2_cnt
from all_counts
where c1_cnt != c2_cnt;

ROW_ID C1_CNT C2_CNT
------ ------ ------
     2      2      0
     4      3      2
     3      0      2

Tags: Database

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