在将两个表与第三个表连接时计算记录的总和答案

【问题标题】：Calculate sum on the records while joining two tables with a third table在将两个表与第三个表连接时计算记录的总和
【发布时间】：2017-10-02 19:19:36
【问题描述】：

我有三张桌子：

mysql> select * from a;
+----+---------+
| ID | Name    |
+----+---------+
|  1 | John    |
|  2 | Alice   |
+----+---------+

mysql> select * from b;
+------+------------+----------+
| UID  | date       | received |
+------+------------+----------+
|    1 | 2017-10-02 |        5 |
|    1 | 2017-09-30 |        1 |
|    1 | 2017-09-29 |        4 |
+------+------------+----------+

mysql> select * from c;
+------+------------+------+
| UID  | date       | sent |
+------+------------+------+
|    1 | 2017-09-25 |    7 |
|    1 | 2017-09-30 |    2 |
|    1 | 2017-09-29 |    3 |
+------+------------+------+

如果我尝试计算为 John 发送的总数，则为 12。而对于已接收，则为 10。但是如果我尝试加入所有三个表，结果很奇怪。这是我连接三个表的查询：

mysql> select sum(sent), sum(received) from a
    -> join c on c.UID = a.ID
    -> join b on b.UID = a.ID
    -> where a.ID = 1;
+-----------+---------------+
| sum(sent) | sum(received) |
+-----------+---------------+
|        36 |            30 |
+-----------+---------------+

但我需要正确的数字（分别为 12 和 10）。我怎样才能得到正确的数字？

【问题讨论】：

标签： mysql sql query-optimization

【解决方案1】：

你应该加入聚合结果而不是原始表

select a.uid, t1.received, t2.sent
from a
inner join (
    select uid, sum(received) received
    from b 
    group by uid
    ) t1 on t1.uid = a.id
inner join (
    select uid, sum(sent) sent
    from c 
    group by uid
    ) t2 on t2.uid = a.id
 where a.id = 1

【讨论】：

抛出一个期望：“字段列表”中的未知列“已发送”。
有没有办法避免子查询？
为了得到正确的结果，不......如果你尝试直接加入表格，你会在表格行之间得到一个笛卡尔积（你得到 3 倍的总和，因为你有 3 行......尝试使用4 或 5 行）
@Amir - from 子句中的子查询很好，不要避免使用它们。
@scaisEdge - 查询错误 - sent 来自 c。

【解决方案2】：

你可以试试下面

select bx.id, recieved, sum(c.sent) sent from
(
SELECT a.id, sum(b.received) recieved
from a 
INNER JOIN b 
ON a.id=b.uid
group by a.id
) bx
INNER JOIN c 
ON c.uid=bx.id 
group by bx.id, bx.recieved;

>>>Demo<<<

【讨论】：

相同的结果，30 和 36。
@Amir 你现在可以试试。
谢谢，拉维。有没有办法避免子查询？
@Amir 实际上并不完全，但我尝试并避免了一个子查询。

【解决方案3】：

这摆脱了子查询，但引入了一些你可能不想要的东西：

( SELECT  uid, 'Received' AS direction, SUM(received) AS HowMany
      WHERE  uid = 1
      GROUP BY uid )
UNION ALL
( SELECT  uid, 'Sent' AS direction, SUM(sent) AS HowMany
      WHERE  uid = 1
      GROUP BY uid )

【讨论】：