Question

我有以下3个表

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


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


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

和

上的这些数据

项： 1 - GECO

在： 1 - 1 - 40 2 - 1 - 2

出： 1 - 1 - 3 2 - 1 - 2 3 - 1 - 3 4 - 1 - 2

这是我的疑问：

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

而结果是

id - Sum_IN - Sum_OUT

1 - 168 - 20

但我想要

id - Sum_IN - Sum_OUT

1 - 42 - 10

我的查询中的问题在哪里？

Answer 1

尝试使用INNER JOIN

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

您有两条IN记录和4条OUT记录。你加入他们，所以你总共得到8条记录。在这个中间结果中，每次OUT记录两次，每次IN记录四次。因此，当您汇总时，您的总和会成倍增加。您可以除以连接记录的数量以返回原始总和：

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

这很有效，但读起来很奇怪。问题是：当你只对他们的金额感兴趣时，为什么要加入所有这些单一记录？使用派生表来保存已经聚合的数据，即总和：

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

从3个表中求和的SQL查询

3 个答案: