2 COUNT 在带有 LEFT JOIN 的 SELECT 内答案

【问题标题】：2 COUNT inside a SELECT with LEFT JOIN2 COUNT 在带有 LEFT JOIN 的 SELECT 内
【发布时间】：2016-03-09 20:28:39
【问题描述】：

我有 2 张桌子：

用户：

ID | NAME 
1  | caio 
2  | mike
3  | peter

发送：

ID | ID_SEND | ID_RECEIVE
1  |    1    |     2
2  |    2    |     3
3  |    3    |     2
4  |    2    |     1

每次用户向另一位用户发送卡片（通过电子邮件发送）时，都会为餐桌提供信息。

问题：我想写一个SELECT 来了解用户在ID_SEND 中的次数以及用户在ID_RECEIVE 中的次数。我尝试过这样的事情：

SELECT 
    user.email, 
    COUNT(sends.id_receive) AS numberReceive, 
    COUNT(sends.id_send) AS numberSend
FROM
    user
LEFT JOIN 
    sends ON user.id = sends.id_send OR user.id = sends.id_receive
GROUP BY 
    user.email

问题是ID_SEND 和ID_RECEIVE 都返回相同的值，两者的总和......我哪里出错了？

【问题讨论】：

标签： mysql select sum left-join

【解决方案1】：

LEFT JOIN 两次，一次发送，一次接收

SELECT u.email,
       COUNT(r.id_receive) AS numberReceive,
       COUNT(s.id_send) AS numberSend
FROM user u
LEFT JOIN sends r ON u.id = r.id_receive
LEFT JOIN sends s ON u.id = s.id_send
GROUP BY u.email

或者，单个LEFT JOIN，使用 case 表达式进行条件计数：

SELECT u.email,
       SUM(case when u.id = s.id_receive then 1 else 0 end) AS numberReceive,
       SUM(case when u.id = s.id_send then 1 else 0 end) AS numberSend
FROM user u
LEFT JOIN sends s ON u.id IN (s.id_receive, s.id_send) 
GROUP BY u.email

【讨论】：

with 2 LEFT JOIN 我有一个错误：mysql_fetch_array() 期望参数 1 是资源
这很奇怪......应该可以正常执行。我将添加一个替代解决方案。 5. 回来查看。
是否可以将 numberReceive 和 numberSend 相加来进行 ORDER BY？
你可以做ORDER BY numberReceive + numberSend。

【解决方案2】：

试试这个。

SELECT S.*, 
   R.received 
FROM   (SELECT u.NAME, 
           Count(snd.id) sends 
    FROM   USER u 
           JOIN sends snd 
             ON snd.id_send = u.id 
    GROUP  BY u.NAME) S 
   LEFT JOIN (SELECT u.NAME, 
                     Count(rec.id) received 
              FROM   USER u 
                     JOIN sends rec 
                       ON rec.id_receive = u.id 
              GROUP  BY u.NAME) R 
          ON S.NAME = R.NAME

【讨论】：