のmysql - 二つのテーブル

Question

から行を数えるので、私はこのテーブル

log_inに

+---------+------------+---------+ 
| date_id | date_today | user_id | 
+---------+------------+---------+ 
| 1 | 2017/03/19 | 16 | 
| 2 | 2017/03/20 | 17 | 
| 3 | 2017/03/20 | 12 | 
| 4 | 2017/03/21 | 16 | 
| 5 | 2017/03/22 | 10 | 
| 6 | 2017/03/22 | 11 | 
+---------+------------+---------+ 

file_downloads

+---------+------------+---------+ 
| date_id | date_today | user_id | 
+---------+------------+---------+ 
| 1 | 2017/03/20 | 17 | 
| 2 | 2017/03/20 | 17 | 
| 3 | 2017/03/20 | 12 | 
| 4 | 2017/03/20 | 17 | 
| 5 | 2017/03/20 | 12 | 
| 6 | 2017/03/20 | 12 | 
| 7 | 2017/03/20 | 12 | 
| 8 | 2017/03/20 | 12 | 
| 9 | 2017/03/22 | 10 | 
| 10 | 2017/03/22 | 10 | 
| 11 | 2017/03/22 | 11 | 
+---------+------------+---------+ 

を持っており、これが私の望ましい結果である：

+------------+-----------+--------+ 
| date_today | login_num | dl_num | 
+------------+---------+----------+ 
| 2017/03/19 |  1  | 0 | 
| 2017/03/20 |  2  | 8 | 
| 2017/03/21 |  1  | 0 | 
| 2017/03/22 |  2  | 3 | 
+------------+-----------+--------+ 

I任意の助けをまだmysqlに新しいです非常に高く評価されるだろう。ありがとうございました！ :)

Answer 1

私は最も簡単な方法は、union allとgroup byだと思う：

select date_today, sum(login) as logins, sum(dl) as dls 
from ((select date_today, 1 as login, 0 as dl from log_in 
    ) union all 
     (select date_today, 0, 1 from file_downloads 
    ) 
    ) lfd 
group by date_today;