How to subtract the sum of value using left join? - mysql

Question

I have two tables orders and customers:
select * from orders;
+------+---------------------+-------------+--------+
| oid | date | customer_id | amount |
+------+---------------------+-------------+--------+
| 102 | 2009-10-08 00:00:00 | 4 | 300 |
| 100 | 2009-10-08 00:00:00 | 3 | 15000 |
| 101 | 2008-10-08 00:00:00 | 2 | 1300 |
| 105 | 2010-10-08 00:00:00 | 1 | 400 |
| 106 | 2014-12-23 00:00:00 | 3 | 300 |
+------+---------------------+-------------+--------+
select * from customers;
+------+--------+------+-----------+----------+
| id | name | age | address | salary |
+------+--------+------+-----------+----------+
| 1 | ramesh | 32 | Ahmedabad | 2000.00 |
| 2 | khilan | 25 | delhi | 1500.00 |
| 3 | muffy | 22 | bhopal | 8500.00 |
| 4 | suresh | 48 | mumbai | 24000.00 |
| 1 | ramesh | 32 | Ahmedabad | 300.00 |
| 5 | akil | 21 | madurai | 1000.00 |
| 6 | rajesh | 22 | delhi | 5000.00 |
+------+--------+------+-----------+----------+
What I'm trying to do is to do SUM(salary) from customers and subtract it with the SUM(amount) from orders table. I have tried with this query:
SELECT id ,NAME,SUM(salary),SUM(amount),SUM(salary)-SUM(amount)
FROM customers a LEFT JOIN orders b ON a.id=b.customer_id
GROUP BY NAME;
This will return the following result, which some of them return incorrect value:
+------+--------+-------------+-------------+-------------------------+
| id | name | SUM(salary) | SUM(amount) | SUM(salary)-SUM(amount) |
+------+--------+-------------+-------------+-------------------------+
| 5 | akil | 1000.00 | NULL | NULL |
| 2 | khilan | 1500.00 | 1300 | 200.00 |
| 3 | muffy | 17000.00 | 15300 | 1700.00 |
| 6 | rajesh | 5000.00 | NULL | NULL |
| 1 | ramesh | 2300.00 | 800 | 1500.00 |
| 4 | suresh | 24000.00 | 300 | 23700.00 |
+------+--------+-------------+-------------+-------------------------+
My expected output is as following:
+------+--------+-------------+-------------+-------------------------+
| id | name | SUM(salary) | SUM(amount) | SUM(salary)-SUM(amount) |
+------+--------+-------------+-------------+-------------------------+
| 5 | akil | 1000.00 | NULL | 1000 |
| 2 | khilan | 1500.00 | 1300 | 200.00 |
| 3 | muffy | 8500 | 15300 | -6800 |
| 6 | rajesh | 5000.00 | NULL | 5000 |
| 1 | ramesh | 2300.00 | 400 | 1900.00 |
| 4 | suresh | 24000.00 | 300 | 23700.00 |
+------+--------+-------------+-------------+-------------------------+

Answer 1

One way is to make calculation on orders into sub-query.
To cater for NULL value, you can use IFNULL(value,0).
SELECT id,NAME,SUM(salary),amt,SUM(salary)-IFNULL(amt,0)
FROM customers a LEFT JOIN
(SELECT customer_id, SUM(amount) amt FROM orders GROUP BY customer_id) b
ON a.id=b.customer_id
GROUP BY NAME;