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Question and Answer

Chris Saxon

Thanks for the question, Radha Sri Seshu.

Asked: July 25, 2017 - 11:16 am UTC

Last updated: July 25, 2017 - 1:24 pm UTC

Version: 11.2

Viewed 1000+ times

You Asked

SQL> CREATE TABLE A(NO1 NUMBER(2) PRIMARY KEY,NO2 NUMBER(2));

Table created.

SQL> CREATE TABLE B(NO1 NUMBER(2) PRIMARY KEY,NO2 NUMBER(2));

Table created.

SQL> ALTER TABLE A ADD CONSTRAINT AA FOREIGN KEY(NO2) REFERENCES B(NO1);

Table altered.

SQL> ALTER TABLE B ADD CONSTRAINT BB FOREIGN KEY(NO2) REFERENCES B(NO1);

Table altered.

SQL> INSERT INTO A VALUES(10,20);
INSERT INTO A VALUES(10,20)
*
ERROR at line 1:
ORA-02291: integrity constraint (SUBK.AA) violated - parent key not found


SQL> INSERT INTO B VALUES(10,20);
INSERT INTO B VALUES(10,20)
*
ERROR at line 1:
ORA-02291: integrity constraint (SUBK.BB) violated - parent key not found

and Chris said...

If you have foreign keys that form a circle you need to declare at least one of them as deferrable (initially deferred):

CREATE TABLE A(NO1 NUMBER(2) PRIMARY KEY,NO2 NUMBER(2));
CREATE TABLE B(NO1 NUMBER(2) PRIMARY KEY,NO2 NUMBER(2));

ALTER TABLE A ADD CONSTRAINT AA FOREIGN KEY(NO2) REFERENCES B(NO1);
ALTER TABLE B ADD CONSTRAINT BB FOREIGN KEY(NO2) REFERENCES B(NO1)
  deferrable initially deferred;
  
INSERT INTO B VALUES(10,20);
INSERT INTO A VALUES(20,10);

select * from a;

NO1  NO2  
20   10 

select * from b;

NO1  NO2  
10   20


This delays validating the constraint until you commit. So if you add rows which never get a parent you'll see and error when the transaction ends:

INSERT INTO B VALUES(11,30);

commit;

ORA-02091: transaction rolled back
ORA-02291: integrity constraint (CHRIS.BB) violated - parent key not found

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Comments

Excellent

Radha Sri Seshu Kolla, July 27, 2017 - 9:23 am UTC

Thank you very much. Theoretically i know this concept, but i could not able to apply to the situation.

Thanks once again.
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