Distinct unique value and sum others in mysql - mysql

Question

I am trying to get the order_payment_total of the unique od_grp_id once but while using sum it get added.
CREATE TABLE IF NOT EXISTS `subscription` (
`id` int(11) unsigned NOT NULL,
`od_grp_id` int(11) unsigned NULL,
`user_id` int(11) NOT NULL,
`order_discount` decimal(10, 2) null,
PRIMARY KEY (`id`)
) DEFAULT CHARSET = utf8;
INSERT INTO `subscription` (
`id`, `od_grp_id`, `user_id`, `order_discount`
)
VALUES
(123994, NULL, 115, null),
(124255, NULL, 115, null),
(124703, 1647692222, 115, null),
(125788, 1647692312, 115, '25.00'),
(125789, 1647692312, 115, '5.00');
CREATE TABLE IF NOT EXISTS `online_payment_against_subscription` (
`subscription_od_grp_id` int(11) unsigned NOT NULL,
`order_payment_total` decimal(10, 2) unsigned NOT NULL,
`user_id` int(11) NOT NULL
) DEFAULT CHARSET = utf8;
INSERT INTO `online_payment_against_subscription` (
`subscription_od_grp_id`, `order_payment_total`, `user_id`
)
VALUES
(1643695200, '45.00', 115),
(1647692312, '250.00', 115),
(1647692222, '30.00', 115);
SELECT
sum(y.order_payment_total),
sum(s.order_discount)
FROM
subscription s
LEFT JOIN(
SELECT
SUM(order_payment_total) as order_payment_total,
user_id,
subscription_od_grp_id
FROM
online_payment_against_subscription
GROUP BY
subscription_od_grp_id
) y ON y.subscription_od_grp_id = s.od_grp_id
WHERE
find_in_set(
s.id, '123994,124255,124703,125788,125789'
)
group by
s.user_id
Current Output:
| sum(y.order_payment_total) |sum(s.order_discount) |
|----------------------------|-----------------------|
| 530 | 30 |
Expected Ouput:
| sum(y.order_payment_total) |sum(s.order_discount) |
|----------------------------|-----------------------|
| 280 | 30 |
Sql Fiddle: http://sqlfiddle.com/#!9/5628f5/1

Answer 1

If I understand correctly, The problem is caused by some duplicate od_grp_id from subscription table, so you might remove the duplicate od_grp_id before JOIN, so we might do that in a subquery.
Query 1:
SELECT
SUM(order_payment_total),
SUM(order_discount)
FROM (
SELECT od_grp_id,SUM(order_discount) order_discount
FROM subscription
WHERE find_in_set(id, '123994,124255,124703,125788,125789')
GROUP BY od_grp_id
) s
LEFT JOIN online_payment_against_subscription y ON y.subscription_od_grp_id=s.od_grp_id
Results:
| SUM(order_payment_total) | SUM(order_discount) |
|--------------------------|---------------------|
| 280 | 30 |

Answer 2

I think you are getting this error because every subscription doesn't have an order payment that is you are getting NULL values.
You can try to remove them by using this -
SELECT y.order_payment_total
FROM subscription s
LEFT JOIN(SELECT SUM(order_payment_total) AS order_payment_total, user_id, subscription_od_grp_id
FROM online_payment_against_subscription
GROUP BY subscription_od_grp_id) y ON y.subscription_od_grp_id = s.od_grp_id
WHERE FIND_IN_SET(s.id, '11258,22547,18586')
AND y.order_payment_total IS NOT NULL;
Or you can make NULL values 0 if you required -
SELECT COALESCE(y.order_payment_total, 0) AS order_payment_total
FROM subscription s
LEFT JOIN(SELECT SUM(order_payment_total) AS order_payment_total, user_id, subscription_od_grp_id
FROM online_payment_against_subscription
GROUP BY subscription_od_grp_id) y ON y.subscription_od_grp_id = s.od_grp_id
WHERE FIND_IN_SET(s.id, '11258,22547,18586');