带有子查询的两个表的总和答案

【问题标题】：Sum of two tables with subqueries带有子查询的两个表的总和
【发布时间】：2014-01-27 10:26:46
【问题描述】：

我有问题

SELECT c.user_id, 
   (SELECT COUNT(*) FROM jos_findme_bestof b WHERE b.user_id = c.user_id) as bCount,
   (SELECT COUNT(*) FROM jos_findme_pair   p WHERE p.user_id = c.user_id) as pCount,
   (SELECT COUNT(*) FROM jos_findme_style  s WHERE s.user_id = c.user_id) as sCount,

FROM   jos_findme c

如何使用：bCount + pCount + sCount 创建一列？

谢谢

【问题讨论】：

标签： mysql sql select sum left-join

【解决方案1】：

如果要在查询中添加列：

   SELECT *,(bCount+pCount+sCount) as sumCount FROM(
    SELECT c.user_id, 
       (SELECT COUNT(*) FROM jos_findme_bestof b WHERE b.user_id = c.user_id) as bCount,
       (SELECT COUNT(*) FROM jos_findme_pair   p WHERE p.user_id = c.user_id) as pCount,
       (SELECT COUNT(*) FROM jos_findme_style  s WHERE s.user_id = c.user_id) as sCount,

    FROM   jos_findme c)

【讨论】：

【解决方案2】：

试试这个

SELECT SUM(t.bCount+t.pCount+t.sCount) As TotalCount , t.user_id
FROM (
SELECT c.user_id as user_id, 
   (SELECT COUNT(*) FROM jos_findme_bestof b WHERE b.user_id = c.user_id) as bCount,
   (SELECT COUNT(*) FROM jos_findme_pair   p WHERE p.user_id = c.user_id) as pCount,
   (SELECT COUNT(*) FROM jos_findme_style  s WHERE s.user_id = c.user_id) as sCount
FROM jos_findme c ) t ;

【讨论】：

【解决方案3】：

试试这个：

SELECT c.user_id, 
      (IFNULL(b.bCount, 0) + IFNULL(p.pCount, 0) + IFNULL(s.sCount, 0)) AS totalCount
FROM jos_findme c
LEFT JOIN (SELECT b.user_id, COUNT(1) bCount 
           FROM jos_findme_bestof b 
           GROUP BY b.user_id 
         ) AS b ON b.user_id = c.user_id 
LEFT JOIN (SELECT p.user_id, COUNT(1) pCount 
           FROM jos_findme_pair p 
           GROUP BY p.user_id 
         ) AS p ON p.user_id = c.user_id 
LEFT JOIN (SELECT s.user_id, COUNT(1) sCount 
           FROM jos_findme_style s 
           GROUP BY s.user_id
         ) AS s ON s.user_id = c.user_id

【讨论】：

【解决方案4】：

你可以试试这个

SELECT c.user_id, sum(b.count(*)+p.count(*)+s.count(*)) as totalcount 
    from jos_findme c 
    join  jos_findme_bestof b 
        on b.user_id = c.user_id 
    join jos_findme_pair p 
        on p.user_id = c.user_id 
    join jos_findme_style  s 
        on s.user_id = c.user_id

【讨论】：