MySQL 连续计算 2 列并与另一个查询连接答案

【问题标题】：MySQL count 2 columns in a row and join with another queryMySQL 连续计算 2 列并与另一个查询连接
【发布时间】：2013-10-05 04:50:27
【问题描述】：

我有一张这样的桌子visits：

+--------------+-------------+------+-----+---------------------+----------------+
| Field        | Type        | Null | Key | Default             | Extra          |
+--------------+-------------+------+-----+---------------------+----------------+
| id           | int(11)     | NO   | PRI | NULL                | auto_increment |
| vis_id       | int(11)     | NO   |     | NULL                |                |
| unit         | int(11)     | NO   |     | NULL                |                |
| time_in      | timestamp   | NO   |     | CURRENT_TIMESTAMP   |                |
| time_out     | timestamp   | NO   |     | 0000-00-00 00:00:00 |                |
| in_username  | varchar(16) | NO   |     | NULL                |                |
| out_username | varchar(16) | NO   |     | NULL                |                |
+--------------+-------------+------+-----+---------------------+----------------+

还有一个像这样的表users：

+------------+-------------+------+-----+---------------------+----------------+
| Field      | Type        | Null | Key | Default             | Extra          |
+------------+-------------+------+-----+---------------------+----------------+
| id         | int(11)     | NO   | PRI | NULL                | auto_increment |
| fname      | varchar(32) | NO   |     | NULL                |                |
| lname      | varchar(32) | NO   |     | NULL                |                |
| date_added | timestamp   | NO   |     | CURRENT_TIMESTAMP   |                |
| username   | varchar(16) | NO   |     | NULL                |                |
| password   | varchar(40) | NO   |     | NULL                |                |
| auth_level | int(1)      | NO   |     | 1                   |                |
| last_login | timestamp   | NO   |     | 0000-00-00 00:00:00 |                |
+------------+-------------+------+-----+---------------------+----------------+

我希望能够计算每个用户在in_username 中的次数 AND out_username... 我之前使用的查询如下所示：

select count(*) as "count", u.fname as "fname" 
    from visits v
        inner join users as u on u.username = v.in_username
    group by u.username order by u.fname

但这只会返回有多少in_username...如果可能的话，我希望两者都在同一个查询中，所以我可以得到这样的结果：

+----------+-----------+----------+
| count_in | count_out | fname    |
+----------+-----------+----------+
|      118 |       224 | Bo       |
|       27 |        64 | James    |
|      147 |       138 | Jeremy   |
|       23 |        37 | Jim      |
|      182 |       172 | Robert   |
|      120 |       158 | Tom      |
+----------+-----------+----------+

其中count_in 是他们的用户名在visits.in_username 中出现的次数，count_out 是他们的用户名在visits.out_username 中出现的次数

我用UNION 尝试的所有操作似乎都将计数加在一起，或者出于某种原因删除了行。有什么想法吗？

【问题讨论】：

标签： mysql sql join count union

【解决方案1】：

执行子查询以获取每个总数，将它们与UNION 合并，然后将它们与SUM() 合并。

SELECT SUM(count_in) count_in, SUM(count_out) count_out, fname
FROM (SELECT COUNT(*) count_in, 0 count_out, in_username fname
      FROM visits v
      GROUP BY fname
      UNION
      SELECT 0 count_in, COUNT(*) count_out, out_username fname
      FROM visits v
      GROUP BY fname) combined

【讨论】：

好极了，这正是我所需要的……我以前没见过combined，那是MySQL的关键字吗？
子查询需要有别名，仅此而已。