从3个表中求和的SQL查询

Question

I have following 3 tables

ITEM table
===========
id   desc
===========


IN table
=======================
id    id_item    number
=======================


OUT table
=========================
id     id_item    number
=========================

and this data on

item: 1 - GECO

in: 1 - 1 - 40 2 - 1 - 2

out: 1 - 1 - 3 2 - 1 - 2 3 - 1 - 3 4 - 1 - 2

This is my query:

SELECT item.id,
       SUM(in.number) AS Sum_IN,
       SUM(out.number) AS Sum_OUT,
       (SUM(in.number) - SUM(out.number)) AS Dif
FROM item
LEFT OUTER JOIN IN ON item.id = IN.id_item
LEFT OUTER JOIN OUT ON item.id = OUT.id_item
GROUP BY item.id

And tis is the result

id - Sum_IN - Sum_OUT

1 - 168 - 20

But i want

id - Sum_IN - Sum_OUT

1 - 42 - 10

Where is the problem in my query?

Answer 1

Try with INNER JOIN like

SELECT `item`.`id`,
       SUM(`in`.`number`) AS Sum_IN,
       SUM(`out`.`number`) AS Sum_OUT
FROM item
INNER JOIN `IN` ON item.id = `IN`.id_item
INNER JOIN `OUT` ON item.id = `OUT`.id_item
GROUP BY item.id

Answer 2

SELECT item.id, A.Sum_IN, B.Sum_OUT, (A.Sum_IN - B.Sum_OUT) AS Dif 
from item 
left outer join(
select  id_item as id, SUM(number) AS Sum_IN
FROM in   GROUP BY id_item) as A ON item.id = A.id
 left outer join(
select  id_item as id, SUM(number) AS Sum_OUT
FROM out  GROUP BY id_item) as B ON item.id = B.id

Answer 3

You have two IN records and 4 OUT records. You join them, so you get 8 records in total. You get each OUT record twice and each IN record four-fold in this intermediate result. Thus your sums are multiplied when you aggregate. You can divide by the number of joined records to get back to the original sums:

SUM(in.number) / COUNT(DISTINCT OUT.id) AS Sum_IN,
SUM(out.number) / COUNT(DISTINCT IN.id) AS Sum_OUT,

This works, but is rather strange to read. The question is: Why do you join all those single records, when you are only interested in their sums? Use derived tables instead that hold the already aggregated data , ie the sums:

select 
  item.id,
  in_sum.value as sum_in,
  out_sum.value as sum_out,
  in_sum.value - out_sum.value as dif
from item
left outer join (select id_item, sum(number) as value from in group by id_item) in_sum
  on item.id = in_sum.id_item
left outer join (select id_item, sum(number) as value from out group by id_item) out_sum
  on item.id = out_sum.id_item
group by item.id