How can I do this faster than cursor? - oracle

Question

I have 2 tables:
first is : dept
second is : payment
I want to compare these tables and make one result table .
Tables include :
table debt:
customerid order amount
1 1 30
1 2 50
1 3 70
table payment:
customerid recorddate amount
1 20080101 10
1 20080102 20
1 20080103 180
And I want result table is :
customerid recorddate amount order
1 20080101 10 1
1 20080102 20 1
1 20080103 50 2
1 20080103 70 2
1 20080103 60 -
I make this result with 2 cursor. And I have 1 million record and It takes too long. How can I make this faster?
thanks in advance
EDIT
I did it with this code:
DECLARE
V_RECORDDATE DATE;
V_CUSTOMERID VARCHAR2(500CHAR);
V_PAYMENT NUMBER;
CURSOR TAH_HES IS
SELECT /*+ PARALLEL(16) */ * FROM
payment_table
WHERE customerid='1'
ORDER BY 3,1;
BEGIN
EXECUTE IMMEDIATE 'alter session force parallel query parallel 16';
EXECUTE IMMEDIATE 'alter session force parallel dml parallel 16';
OPEN TAH_HES;
LOOP
FETCH TAH_HES INTO V_RECORDDATE, V_CUSTOMERID,V_PAYMENT;
EXIT WHEN TAH_HES%NOTFOUND;
FOR CUR_X IN (
SELECT /*+ PARALLEL(16) */
COMPENENT,
AMOUNT
FROM
DEBT_TABLE
WHERE 1=1
AND CUSTOMERID=V_CUSTOMERID
ORDER BY "ORDER"
)
LOOP
IF(CUR_X.AMOUNT<=V_PAYMENT)
THEN
INSERT INTO
RESULT_TABLE
SELECT /*+ PARALLEL(16) */
V_CUSTOMERID,V_RECORDDATE,CUR_X.COMPENENT,CUR_X.AMOUNT
FROM
DUAL;
COMMIT;
DELETE FROM
DEBT_TABLE
WHERE CUSTOMERID=V_CUSTOMERID
AND COMPENENT=CUR_X.COMPENENT
AND AMOUNT=CUR_X.AMOUNT;
COMMIT;
UPDATE
PAYMENT_TABLE
SET PAYMENT=PAYMENT-CUR_X.AMOUNT
WHERE CUSTOMERID=V_CUSTOMERID
AND HISLEMTARIH=V_RECORDDATE;
COMMIT;
ELSE
INSERT INTO
RESULT_TABLE
SELECT /*+ PARALLEL(16) */
V_CUSTOMERID,V_RECORDDATE,CUR_X.COMPENENT,V_PAYMENT
FROM
DUAL;
COMMIT;
UPDATE
DEBT_TABLE
SET AMOUNT=AMOUNT-V_PAYMENT
WHERE CUSTOMERID=V_CUSTOMERID
AND COMPENENT=CUR_X.COMPENENT ;
COMMIT;
DELETE FROM
PAYMENT_TABLE
WHERE CUSTOMERID=V_CUSTOMERID
AND PAYMENT=V_PAYMENT
AND RECORDDATE=V_RECORDDATE;
COMMIT;
EXIT;
END IF;
END LOOP;
END LOOP;
END;
INSERT INTO
RESULT_TABLE
SELECT /*+ PARALLEL(16) */
CUSTOMERID,
RECORDDATE,
'-',
PAYMENT
FROM
PAYMENT_TABLE;
COMMIT;

Answer 1

If payments are used to clear off the debt then your result is a bit unexpected.
More logically
payments 10 and 20 will clear off order 1: 30 = 10 + 20
payment 180 will clear off order 2: 50 (180 - 50 = 130 remaining)
payment 180 will clear off order 3: 70 (130 - 70 = 60 remaining)
Or in pure SQL
with debt(customerid, ord, amount) as
(
select 1, 1, 30 from dual
union all select 1, 2, 50 from dual
union all select 1, 3, 70 from dual
),
payment(customerid, recorddate, amount) as
(
select 1, 20080101, 10 from dual
union all select 1, 20080102, 20 from dual
union all select 1, 20080103, 180 from dual
),
allocation as
(
select *
from
(select d.customerid, d.ord, d.amount, p.recorddate, p.amount as pay_amount
from debt d
join payment p on d.customerid = p.customerid)
model ignore nav
partition by (customerid)
dimension by (recorddate, ord)
measures(amount, pay_amount, 0 allocated)
rules
(
allocated[any, any] order by ord, recorddate =
least(pay_amount[cv(recorddate), cv(ord)] -
sum(allocated)[cv(recorddate), ord <= cv(ord)]
,amount[cv(recorddate), cv(ord)] -
sum(allocated)[recorddate <= cv(recorddate), cv(ord)])
)
)
select a.*, pay_amount -
sum(allocated) over (partition by recorddate order by ord) remaining
from allocation a
where allocated > 0
order by ord, recorddate;
CUSTOMERID RECORDDATE ORD AMOUNT PAY_AMOUNT ALLOCATED REMAINING
---------- ---------- ---------- ---------- ---------- ---------- ----------
1 20080101 1 30 10 10 0
1 20080102 1 30 20 20 0
1 20080103 2 50 180 50 130
1 20080103 3 70 180 70 60
PL/SQL solution would be more optimal for this task (but not your implementation).