Simulating constraint ON UPDATE option by TRIGGER - oracle

Question

I'm trying to simulate ON UPDATE option in my foreign keys by making trigger that will update child records after update.
Trigger looks like this:
CREATE OR REPLACE TRIGGER SOMETABLE_AU_ID_TRG
AFTER UPDATE OF ID ON SOMETABLE
FOR EACH ROW
BEGIN
UPDATE SOMECHILDTABLE SET SOMETABLE_ID = :NEW.ID WHERE ID = :OLD.ID;
END;
And I'm making the child table constraint deferrable like this:
set constraint
SOMECHILDTABLE_FK_SOMETABLE deferred;
But it still gives me ORA-02292 (child record found) on commit. How can I avoid this?

Answer 1

Kudos to #BrianCamire for spotting the bug:
Let's start by creating your trigger...
SQL> CREATE OR REPLACE TRIGGER SOMETABLE_AU_ID_TRG
AFTER UPDATE OF ID ON SOMETABLE
FOR EACH ROW
BEGIN
UPDATE SOMECHILDTABLE SET SOMETABLE_ID = :NEW.ID WHERE ID = :OLD.ID;
END; 2 3 4 5 6
7 /
Trigger created.
SQL> alter table somechildtable add constraint some_fk foreign key (sometable_id)
2 references sometable (id) deferrable initially deferred;
Table altered.
SQL>
Here is the data ...
SQL> select * from somechildtable;
ID SOMETABLE_ID
---------- ------------
11 1
12 2
13 6
14 4
15 5
SQL>
So, let's mess with the parent data....
SQL> update sometable set id = 3 where id = 6
2 /
1 row updated.
SQL> commit;
commit
*
ERROR at line 1:
ORA-02091: transaction rolled back
ORA-02292: integrity constraint (APC.SOME_FK) violated - child record found
SQL>
Aha! So, as Brian says we need to change the trigger code:
SQL> CREATE OR REPLACE TRIGGER SOMETABLE_AU_ID_TRG
AFTER UPDATE OF ID ON SOMETABLE
FOR EACH ROW
BEGIN
UPDATE SOMECHILDTABLE SET SOMETABLE_ID = :NEW.ID WHERE SOMETABLE_ID = :OLD.ID;
END;
/
2 3 4 5 6 7
Trigger created.
SQL> SQL> update sometable set id = 3 where id = 6;
1 row updated.
SQL> commit;
Commit complete.
SQL> select * from somechildtable
2 /
ID SOMETABLE_ID
---------- ------------
11 1
12 2
13 3
14 4
15 5
SQL>