如何写这个查询MYSQL

Question

我有这个数据库：

| id | name  | email           | control_number | created    |   |   |   |   |          |
|:--:|-------|-----------------|----------------|------------|---|---|---|---|---|
| 1  | john  | john@gmail.com  | 1              | 14/09/2016 |   |   |   |   |   |
| 2  | carl  | carl@gmail.com  | 1              | 13/08/2016 |   |   |   |   |   |
| 3  | frank | frank@gmail.com | 2              | 12/08/2016 |   |   |   |   |   |

我想通过control_number获取过去12个月中的COUNT。

基本上是一个COUNT，其中control_number = 1，但按月计。

因此，如果查询在今天（9月）完成，则应从9月开始到2015年10月，并显示每个月的记录数。

结果应为：

09/2016 = 50
08/2016 = 35
07/2016 = 20
06/2016 = 50
05/2016 = 21
04/2016 = 33
03/2016 = 60
02/2016 = 36
01/2016 = 11
12/2015 = 0
11/2015 = 0
10/2015 = 0

Answer 1

嗯。 获取0值可能很棘手。 假设您每个月都有一些数据（即使不是“ 1”），则可以执行以下操作：

select extract(year_month from created) as yyyymm,
       sum(control_number = 1)
from t
where created >= date_sub(curdate(), interval 12 month)
group by extract(year_month from created)
order by yyyymm;

如果您每个月都没有至少一条记录，那么您将需要一个left join和一个每月有一行的表。

Answer 2

尝试这个：

select CONCAT(SUBSTRING(ym, 5, 2), '/', SUBSTRING(ym, 1, 4)) Month, Count from ( 
  select EXTRACT(YEAR_MONTH FROM created) ym, count(*) Count
  from mytable
  where EXTRACT(YEAR_MONTH FROM created) > (EXTRACT(YEAR_MONTH FROM SUBDATE(NOW(), INTERVAL 1 YEAR))
  group by 1
  order by 1 desc) x

Answer 3

尝试：

select concat(month(created),'/',year(created)) as period, count(*) as cnt
from mytable
where control_number=1 and TIMESTAMPDIFF(year, created, now())=0
group by (month(created));