Question

所以我正在优化SQL查询。我已经清理了很多，但仍有一些工作要做。请参阅以下查询：

比赛tbl： ID，fb_share_points，tw_share_points

comp_mapper tbl： id，referrer_user_id，user_subscription_id，has_fb_shared，has_tw_shared，competition_id

SELECT 
    IF (cm.has_fb_shared = 1, IFNULL(c.fb_share_points,0), 0) 
    + IF (cm.has_tw_shared = 1, IFNULL(c.tw_share_points,0), 0) 
    + (SELECT count(*) FROM comp_mapper as cm2
        WHERE cm2.comp_id = cm.comp_id 
        AND cm2.referrer_user_id = us.user_id)
    as shares
FROM competitions AS c
JOIN comp_mapper as cm ON cm.competition_id = c.id
JOIN user_subscription as us on cm.user_subscription_id = us.id
WHERE c.id = :id
ORDER BY shares DESC
LIMIT :limit

当引用用户时，引荐用户的ID被放入竞赛者的referrer_user_id列中。子查询计算有多少参赛者引用其他用户。我目前正试图在没有子查询的情况下找到另一种编写方法，因为查询本身需要很长时间才能运行。如果有人有任何建议或建议，请分享！

Answer 1

这应该返回相同的结果。这里的区别在于，我们有一个内联视图（MySQL称之为派生表）而不是相关子查询，其别名为rf。

SELECT IF(cm.has_fb_shared = 1, IFNULL(c.fb_share_points,0), 0) 
     + IF(cm.has_tw_shared = 1, IFNULL(c.tw_share_points,0), 0) 
     + IFNULL(rf.ref_count,0) AS shares
  FROM competitions c
  JOIN comp_mapper cm 
    ON cm.competition_id = c.id
  JOIN user_subscription us
    ON us.id = cm.user_subscription_id
  LEFT
  JOIN (  
         SELECT cm2.referrer_user_id 
              , COUNT(1) AS ref_count
           FROM comp_mapper cm2
          WHERE cm2.comp_id = :id
          GROUP BY cm2.referrer_user_id
       ) rf
    ON rf.referrer_user_id = us.user_id
 WHERE c.id = :id
 ORDER BY shares DESC
 LIMIT :limit

有点奇怪的是，查询返回的行集只有一列（shares），而没有其他列。如果意图是返回单个值，则需要在选择列表中有一个聚合函数（即SUM()）。

不使用子查询重写查询

1 个答案: