MYSQL calculate amount from two table? - mysql

Question

I have 2 mysql tables:
"Orders" table:
customer_id | money
3 120
5 80
3 45
3 70
6 20
"collecting" table:
customer_id | money
3 50
3 70
4 20
4 90
I want a result like:
"Total" table:
customer_id | Amount
3 115
4 110
5 80
6 20
"Total" table "customer_id" should be singular
Amount = (SUM(All customer orders.money) - SUM(All customer collecting.money))
"Money" can be NULL
"Orders" table can have customer_id and "Collecting" table may not have
Or
"Collecting" table can have customer_id and "Orders" table may not have
How can i write a single query for output "Total" table?

Answer 1

The following returns the result you expect.
SELECT
customer_id,
SUM(amount) as amount
FROM (
SELECT customer_id, SUM(money) as amount
FROM orders GROUP BY customer_id
UNION ALL
SELECT customer_id, SUM(money) * -1 as amount
FROM collecting GROUP BY customer_id
) as tb
GROUP BY customer_id;
customer_id = 4 returns -110, not 110, since it's only in the collecting table.
Example: http://www.sqlfiddle.com/#!2/3b922/5/0

Answer 2

The fastest way is to union your data with the money value being negative on the collecting table:
-- load test data
create table orders(customer_id int, money int);
insert into orders values
(3,120),
(5,80),
(3,45),
(3,70),
(6,20);
create table collecting(customer_id int,money int);
insert into collecting values
(3,50),
(3,70),
(4,20),
(4,90);
-- populate Total table
create table Total(customer_id int,Amount int);
insert into Total
select oc.customer_id,sum(oc.money) Amount
from (
select customer_id,coalesce(money,0) money from orders
union all
select customer_id,coalesce(-money,0) money from collecting
) oc
group by oc.customer_id;
-- return results
select * from Total;
SQL Fiddle: http://www.sqlfiddle.com/#!2/deebc

Answer 3

You need to do this by pre-aggregating the data. If I assume that the orders come first, you can use left outer join:
select o.customer_id, (o.money - coalesce(c.money)) as Amount
from (select o.customer_id, sum(o.money) as money
from orders o
group by o.customer_id
) o left outer join
(select c.customer_id, sum(c.money) as money
from collecting c
group by c.customer_id
) c
on o.customer_id = c.customer_id;

Answer 4

Something like this should work:
SELECT CASE
WHEN c.customer_id IS NULL
THEN o.customer_id
ELSE c.customer_id
END
,sum(o.MONEY) - sum(c.MONEY)
FROM orders o
OUTER JOIN collecting c ON c.customer_id = o.customer_id
GROUP BY 1

Answer 5

Try this:
SELECT customer_id, SUM(o_money - c_money) AS Amount
FROM (
SELECT customer_id, money AS o_money, 0 AS c_money FROM orders
UNION ALL
SELECT customer_id, 0 AS o_money, money AS c_money FROM collecting
) total
GROUP BY customer_id;
According to your description, you need a FULL [OUTER] JOIN, which is a combination of LEFT JOIN and RIGHT JOIN. Many databases don't support FULL JOIN, so you need to use UNION ALL to reach the same effect.