Question

SELECT 
  (SELECT COUNT(*) FROM votes WHERE votes.vote = 1) AS upvotes,
  (SELECT COUNT(*) FROM votes WHERE votes.vote = -1) AS downvotes,
FROM votes WHERE link = <linkid>

直截了当的问题;怎么可以优化？我想不出更好的方法，但我对MySQL不好。

感谢您的回复！

编辑：为了清楚起见：我希望它返回一行有两列; upvotes和downvotes

Answer 1

SELECT sum(case when vote = 1 then 1 end) as upvotes,
    sum(case when vote = -1 then 1 end) as downvotes
FROM votes
WHERE link = < linkid >

Answer 2

尝试使用SUM而不是COUNT：

SELECT 
  SUM(vote = 1) AS upvotes,
  SUM(vote = -1) AS downvotes
FROM votes WHERE link = <linkid>

请注意，不会为您提供与您发布的查询相同的结果。我认为您发布的查询错误！

Answer 3

我会用这个：

SELECT COUNT(*), vote AS numvotes FROM votes WHERE vote = -1 OR vote = 1 GROUP BY vote

Answer 4

SELECT 
  COUNT(case vote when  1 then 1 else null end) AS upvotes,
  COUNT(case vote when -1 then 1 else null end) AS downvotes
FROM votes WHERE link = <linkid>

学习时间：count仅计算非空值，因此每当您看到then 1也可能是then 'Yay!'时。空值不计算在内。

Answer 5

我猜所有评论员都认为您的查询应该是：

SELECT 
  (SELECT COUNT(*) FROM votes WHERE link = <linkid> AND vote = 1 ) AS upvotes,
  (SELECT COUNT(*) FROM votes WHERE link = <linkid> AND vote = -1) AS downvotes

到目前为止，您至少有4种其他方式与此方法相同（3种方式完全相同，1种方式稍有不同，给出相同的结果，但是在2行中）。使用您的数据进行测试并选择更快（或保留所有数据并稍后再次测试，当表变大时）。

(link,vote)上的复合索引对所有版本都有用。

Answer 6

恕我直言，你不应该计算苍蝇的总投票数。相反，您应该将总投票数与商品一起存储，并使用详细的投票日志作为确保结果完整性的工具，并让用户有权撤回投票。

优化多个SELECT

6 个答案: