Question

SELECT SUM(case when p.status = 2 then p.value end) as 'val_accepted'
      FROM
        props AS p
            INNER JOIN (p_contents AS pc
              INNER JOIN contents AS c ON c.id = pc.library_id)
            ON p.id = pc.prop_id
      WHERE p.account_id = 3
      GROUP BY (pc.library_id)

所以，发生了什么：有两个p_contents与道具相关联。这两个p_contents具有相同的library_id，指向相应的内容因此，p.value的SUM是应该的两倍，因为有两个p_contents指向相同的内容

如何不将p.value加倍？

编辑：
我想出了如何使用DISTINCT，但我仍然需要访问内部列......

SELECT    c.name as 'library_name',
   SUM(case when p.status = 2 then p.value end) as 'val_accepted',

FROM
  props AS p
  INNER JOIN 
  (
    SELECT DISTINCT(pc.library_id), prop_id
    FROM prop_contents AS pc
    INNER JOIN 
    (
      SELECT name, visibility, id, updated_at 
      FROM contents AS c
    ) as c
      ON c.id = pc.library_id
  )as pc
  ON p.id = pc.prop_id
 WHERE p.account_id = 3
 GROUP BY (pc.library_id)

现在我收到了错误：

Unknown column 'c.name' in 'field list')

Answer 1

这是一个解决方案。首先将集合减少到派生表中的不同行，然后将GROUP BY应用于该结果：

SELECT SUM(case when d.status = 2 then d.value end) as 'val_accepted'
FROM (
  SELECT DISTINCT p.id, p.status, p.value, pc.library_id
  FROM props p
  INNER JOIN p_contents AS pc ON p.id = pc.prop_id
  INNER JOIN contents AS c ON c.id = pc.library_id
  WHERE p.account_id = 3) AS d
GROUP BY d.library_id

您在示例中使用DISTINCT(pc.library_id)，就像DISTINCT仅适用于括号内的列一样。这是一个普遍的误解。 DISTINCT适用于选择列表的所有列。 DISTINCT不是一个函数;它是一个查询修饰符。

MySQL当我使用带有多个INNER JOINS的SUM和COUNT时，如何不包含重复的行？

1 个答案: