FOR ALL WITH RETURNING

Question

Hi All,
My requirement is to return a set of values using forall returning clause.For that I am using
begin
FORALL J IN 1..M_T_TABLE1.COUNT SAVE EXCEPTIONS
INSERT INTO TABLE2 VALUES M_T_TABLE1(J)
RETURNING CUST_CODE BULK COLLECT INTO P_T_CUST_CODE;
exception
end;
This code is working fine and It is returning CUST_CODE values using P_T_CUST_CODE.
But there is a situation that i've to run this code in loop like..
begin
FOR I IN 1..NO_OF_ITERATIONS LOOP
FORALL J IN 1..M_T_TABLE1.COUNT SAVE EXCEPTIONS
INSERT INTO TABLE2 VALUES M_T_TABLE1(J)
RETURNING CUST_CODE BULK COLLECT INTO P_T_CUST_CODE;
END LOOP;
exception
end;
NOW IT IS RETURNING THE VALUES INSERTED IN LAST ITERATION ONLY .BUT I WANT THE VALUES INSERTED IN ALL ITERATIONS.
ANY IDEA HOW TO ACHIVE THAT?
THANKS IN ADVANCE.
NM

Answer 1

Hi,
You need to declare table of table (two dimensional table in pl/sql).
  1   declare
  2     type
  3       t1 is table of table2%rowtype index by binary_integer;
  4     type
  5       t2 is table of t1 index by binary_integer;
  6     no_of_iterations number := 3;
  7     m_t_table1 t2;
  8     p_t_cust_code t2;
  9   begin
10     m_t_table1(1)(1).cust_code := 1;
11     m_t_table1(1)(2).cust_code := 2;
12     m_t_table1(2)(1).cust_code := 3;
13     m_t_table1(2)(2).cust_code := 4;
14     m_t_table1(3)(1).cust_code := 5;
15     m_t_table1(3)(2).cust_code := 6;
16     delete from table2;
17     --your code
18     begin
19       FOR I IN 1..NO_OF_ITERATIONS LOOP
20         FORALL J IN 1..M_T_TABLE1(I).COUNT SAVE EXCEPTIONS
21           INSERT INTO TABLE2 VALUES M_T_TABLE1(I)(J)
22           RETURNING CUST_CODE BULK COLLECT INTO P_T_CUST_CODE(I);
23       END LOOP;
24     end;
25     --your code
26     for I in 1..NO_OF_ITERATIONS LOOP
27       for J in M_T_TABLE1(I).first..M_T_TABLE1(I).last loop
28         dbms_output.put_line('P_T_CUST_CODE('||I||', '||J||')='||to_char(P_T_CUST_CODE(I)(J).cust_code));
29       end loop;
30     end loop;
31*  end;
32  /
P_T_CUST_CODE(1, 1)=1
P_T_CUST_CODE(1, 2)=2
P_T_CUST_CODE(2, 1)=3
P_T_CUST_CODE(2, 2)=4
P_T_CUST_CODE(3, 1)=5
P_T_CUST_CODE(3, 2)=6
PL/SQL procedure successfully completed.Bartek
Edited by: brtk on 2009-06-18 08:25 edited line 28.
Edited by: brtk on 2009-06-18 08:26