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我需要从 UPDATE 命令中的查询中引用 cartProduct 派生表。有人可以解释一下我做错了什么以及如何解决它。

UPDATE memberships SET points = (points + COALESCE( 
    (SELECT pts 
    FROM (
        SELECT pid0, SUM (S) as Pts
        FROM ( 
            (SELECT pid0, SUM (pts1) as S
            FROM 
                (SELECT DISTINCT pid0, pid0,pts1
                FROM cartProduct
                WHERE pid0 = i1) as pointsForPath 
            GROUP BY pid0)) as allPtsTbl
        GROUP BY pid0) as temp 
    WHERE memberships.cid = 57010 AND memberships.pid = temp.pid0 ),
0))
FROM (
    SELECT t0.pid as pid0,t1.pid as pid1,t1.invitedby as i1,t1.points as pts1, t2.pid as pid2,t2.invitedby as i2,t2.points as pts2,t3.pid as pid3,t3.invitedby as i3, t3.points as pts3
    FROM memberships t0, memberships  t1, memberships  t2, memberships  t3
    WHERE t0.cid = t1.cid AND t1.cid = t2.cid AND t2.cid = t3.cid) as cartProduct;

这个查询必须做什么并不重要，因为它是我想要优化的一小部分复杂查询。我唯一的问题是如何从 UPDATE 子查询中引用 cartProduct 。我要优化的查询是由 C 代码生成的......如您所见，我必须多次计算相同的东西。这个查询只有 3 个派生表，但它也可以是 100+，然后我必须等待 10 多个小时才能计算它。

UPDATE memberships SET points = points + COALESCE((
SELECT pts 
FROM (
    SELECT pid0, SUM (S) as Pts
    FROM ( 
        SELECT pid0, SUM (pts1) as S
        FROM 
            (SELECT DISTINCT pid0, pid1,pts1
            FROM (
                SELECT t0.pid as pid0, t1.pid as pid1, t1.invitedby as i1, t1.points as pts1 
                FROM memberships t0, memberships t1
                WHERE t0.cid = 57010 AND t1.cid = 57010) as paths
            WHERE pid0 = i1) as pointsForPath
        GROUP BY pid0
        UNION ALL
        SELECT pid0, SUM (pts2) as S
        FROM (
            SELECT DISTINCT pid0, pid2,pts2
            FROM (
                SELECT t0.pid as pid0, t1.pid as pid1, t1.invitedby as i1, t1.points as pts1 , t2.pid as pid2, t2.invitedby as i2, t2.points as pts2 
                FROM memberships t0, memberships t1, memberships t2
                WHERE t0.cid = 57010 AND t1.cid = 57010 AND t2.cid = 57010) as paths
            WHERE pid0 = i1 AND pid1 = i2) as pointsForPath
        GROUP BY pid0
        UNION ALL
        SELECT pid0, SUM (pts3) as S
        FROM (
            SELECT DISTINCT pid0, pid3,pts3
            FROM (
                SELECT t0.pid as pid0, t1.pid as pid1, t1.invitedby as i1, t1.points as pts1 , t2.pid as pid2, t2.invitedby as i2, t2.points as pts2 , t3.pid as pid3, t3.invitedby as i3, t3.points as pts3 
                FROM memberships t0, memberships t1, memberships t2, memberships t3
                WHERE t0.cid = 57010 AND t1.cid = 57010 AND t2.cid = 57010 AND t3.cid = 57010) as paths
            WHERE pid0 = i1 AND pid1 = i2 AND pid2 = i3) as pointsForPath
        GROUP BY pid0 ) as allPtsTbl
GROUP BY pid0) as temp WHERE memberships.cid = 57010 AND memberships.pid = temp.pid0 ),0)

对不起我的英语不好，非常感谢。

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1 回答 1

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如果性能是您的问题，您可以尝试：

update memberships
    set points = points + coalesce(s.points, 0)
    from (nasty union all'ed subquery) s
    where memberships.id = s.id

（这假设您将 pid0 字段重命名为 id。）

也就是说，您应该考虑以下几点：

  1. 学习正确的连接语法。在“where”子句中进行连接变得过时了，就像在英语中使用“hath”或“thee”一样。通过选择特定值（您正在执行的操作）隐式执行连接是非常危险的，因为您不知道从每个表中获取了多少条记录，并且对查询的微小更改可能会产生不利影响（例如选择两个用户而不是一个）。
  2. 将讨厌的子查询放入临时表中。通常，我不提倡使用临时表，但这可能是优化器混淆的情况（这在混合选择、删除和更新的查询中尤其可能）。
  3. 你为什么要在group by前做一个distinct by。首先，您可以执行“sum(distinct)”，尽管这几乎没有必要。这表明您的联接存在问题。而且，错误的连接会导致查询无法完成。
于 2012-05-27T14:03:10.313 回答