这是我的表格,显示用户名和他们得分的时间戳:
id user date
1 Aaron 23/02/2012 22:44
2 Betty 23/02/2012 22:47
3 Carlos 24/02/2012 16:01
4 David 28/02/2012 11:40
5 David 28/02/2012 12:32
6 David 28/02/2012 16:59
7 Aaron 2/03/2012 13:46
8 Aaron 30/03/2012 18:37
9 Betty 30/03/2012 19:58
10 Emma 9/04/2012 6:49
11 Emma 9/04/2012 13:19
12 Emma 9/04/2012 18:20
13 Emma 9/04/2012 20:46
14 Aaron 10/04/2012 15:47
15 Betty 10/04/2012 19:15
16 Betty 10/04/2012 20:40
17 Carlos 11/04/2012 9:44
18 Carlos 11/04/2012 20:01
19 David 11/04/2012 23:17
20 David 12/04/2012 17:09
以下是我尝试实现的结果表,即显示月 - 年的x轴,以及显示达到特定点阈值的用户数量的y轴月:
date 1 point First time? 2 points First time? 3 points First time? 4 points First time? Total
Feb-12 A,B,C A,B,C D D 4
Mar-12 B A A 3
Apr-12 A,B,C B,C,D B,C,D E E 4
我只计算了一个月内总点数和不同得分手总数:
SELECT DISTINCT CONCAT (MONTHNAME(date), ' ', YEAR(date)) as 'date', COUNT(id) as total_points, COUNT(distinct referrer_id) as number_of_scorers
from points
group by CONCAT (MONTH(date), ' ', YEAR(date))
order by YEAR(date), MONTH(date)
只给了我:
date total_points number_of_scorers
Feb-12 6 4
Mar-12 3 3
etc.
所以我的问题是:
由于
答案 0 :(得分:0)
您需要的基本查询是:
select date_format(date, '%Y-%m') as yyyymm, user, count(*) as points
from t
group by date_format(date, '%Y-%m') as yyyymm, user;
这可以获得一个月内每个用户的积分数。
其余的只是聚合,连接和条件:
select ymu.yyyymm,
group_concat(case when ymu.points = 1 then user end) as Points1_Users,
group_concat(case when ymu.points = 1 and ymu.yyyymm = u.min_yyyymm then user end) as Points1_Users_First,
group_concat(case when ymu.points = 2 then user end) as Points2_Users,
group_concat(case when ymu.points = 2 and ymu.yyyymm = u.min_yyyymm then user end) as Points2_Users_First
from (select date_format(date, '%Y-%m') as yyyymm, user, count(*) as points
from t
group by date_format(date, '%Y-%m') as yyyymm, user
) ymu join
(select user, min(yyyymm) as min_yyyymm
from (select date_format(date, '%Y-%m') as yyyymm, user, count(*) as points
from t
group by date_format(date, '%Y-%m') as yyyymm, user
) t
group by user
) u
on ymu.user = u.user
group by yyyymm
order by yyyymm;