Firing trigger to insert id in a table

Question

I am not able to insert the ID in CR_ID details when i press the create button in Apex Master detail form & Tabular form.
CREATE OR REPLACE TRIGGER COSTMAN."CM_RECIPE_T1"
BEFORE INSERT ON "CM_RECIPE"
FOR EACH ROW
DECLARE
L_ID NUMBER;
BEGIN
SELECT CM_RECIPE_SEQ.NEXTVAL INTO L_ID FROM DUAL;
IF :NEW.CR_ID IS NULL THEN
:NEW.CR_ID := L_ID;
END IF;
END;
Do i have to write a process in Apex form. This is a first time, i've come across a situation where i want the column to get populated using a trigger when pressing a Create button in an Apex form.
Please suggest..
Yogesh

Answer 1

Hi
If the trigger is on the table, then it will be fired when an INSERT takes place, you don't need to code for it specially.
Do lovely SQL> CREATE TABLE insert_test
(col1 NUMBER,
col2 VARCHAR(2));
Table created.
Elapsed: 00:00:00.24
Do lovely SQL> CREATE SEQUENCE insert_seq
START WITH 1
MAXVALUE 999999999999999999999999999
MINVALUE 1; 
Sequence created.
Elapsed: 00:00:00.00
Do lovely SQL> CREATE TRIGGER insert_trig
BEFORE INSERT ON insert_test
FOR EACH ROW
DECLARE
L_ID NUMBER;
BEGIN
SELECT insert_seq.NEXTVAL INTO L_ID FROM DUAL;
IF :NEW.col1 IS NULL THEN
:NEW.col1 := L_ID;
END IF;
END;
Trigger created.
Elapsed: 00:00:00.23
Do lovely SQL> INSERT INTO insert_test (col2) VALUES('TT');
1 row created.
Elapsed: 00:00:00.03
Do lovely SQL> COMMIT;
Commit complete.
Elapsed: 00:00:00.00
Do lovely SQL> SELECT * FROM insert_test;
      COL1 CO
         1 TT
Elapsed: 00:00:00.04See - you just need to do the insert.
Cheers
Ben