Question

我有一个看起来像这样的mysql表：

id      level      time
1         1     2014-02-19 04:33:04
2         1     2014-03-19 04:33:04
3         1     2014-03-20 04:33:04
4         2     2014-03-21 04:53:04
5         1     2014-07-19 04:33:04
6         2     2014-07-19 04:33:04
7         1     2014-07-19 04:33:04
8         1     2014-08-19 04:33:04

我想得到第1级的结果：

level1count    year    month
  0            2014      1
  1            2014      2
  2            2014      3
  0            2014      4
  0            2014      5
  0            2014      6
  2            2014      7
  1            2014      8
  0            2014      9
  0            2014      10
  0            2014      11
  0            2014      12

我尝试了这个查询，但没有每月给出结果

SELECT YEAR(time) AS year, MONTH(time) AS month, COUNT(DISTINCT id) AS count FROM users where level = '1' GROUP BY year, month

Answer 1

您正在寻找的是将所有月份的结果存在或不存在于数据库中，对于结果集，您需要在所有月份进行查询，例如union，然后将基于年份和月份的条件与您的表联系起来

select coalesce(sum(`level` = 1),0) level1count
 coalesce(sum(`level` = 2),0) level2count ,y,m
from
(select 1 as m,2014 as y
union
.
.
.
union
select 12 as m ,2014 as y
) months
left join t on(months.m = month(t.time) and months.y = year(t.time))
group by months.m, year(t.time)

Demo

比联盟更好的方法是拥有一个包含所有月份年份表的表格，然后将其与表格一起加入

Answer 2

select sum(level = 1) as level1count, 
       year(`time`) as year, 
       month(`time`) as month
from your_table
group by year(`time`), month(`time`)

Mysql每个月计算一次

2 个答案:

Demo