Question

所以我有一张这样的表

id  | user_id | point | created_at
-------------------------------------------
1   | 1       | 10    | 2015-08-03 00:08:25 
2   | 3       | 20    | 2015-08-01 00:08:25 
3   | 4       | 30    | 2015-08-13 00:08:25 
4   | 3       | 10    | 2015-08-25 00:08:25 
5   | 2       | 20    | 2015-09-02 00:08:25 
6   | 1       | 10    | 2015-09-14 00:08:25 
7   | 4       | 50    | 2015-09-22 00:08:25 
8   | 2       | 80    | 2015-09-30 00:08:25 
9   | 1       | 30    | 2015-10-02 00:08:25 
10  | 5       | 90    | 2015-10-02 00:08:25

我可以生成一个查询来制作这样的结果吗？

start_date | end_date   | total_point_100_or_more | total_point_less_than_100
------------------------------------------------------------------------------
2015-08-01 | 2015-08-10 | 3 User                  | 7 User
2015-08-11 | 2015-08-20 | 8 User                  | 5 User
2015-08-21 | 2015-08-30 | 10 User                 | 5 User
2015-08-31 | 2015-09-09 | 4 User                  | 6 User

我现在没有任何代码，因为我完全不知道从哪里开始。

Answer 1

我还没有对此进行过测试，因为您的示例答案与您的示例数据不对应，但是非常类似以下内容应该有效。在这里，我假设你想知道用户如何获得高于或低于100的总分（在每个时期）。

SELECT  start_date,
        DATEADD(day, 9, start_date) as end_date,
        SUM(CASE WHEN period_points >= 100 THEN 1 ELSE 0 END) as total_point_100_or_more,
        SUM(CASE WHEN period_points < 100 THEN 1 ELSE 0 END) as total_point_less_than_100
FROM (
        SELECT  start_date,
                user_id,
                SUM(point) as period_points
        FROM (
                SELECT  DATEADD(
                                day,
                                DATEDIFF(day, '2015-08-01 00:00:00', created_at)/10*10,
                                '2015-08-01 00:00:00'
                        ) as start_date,
                        id,
                        user_id,
                        point
                FROM    points
        ) AS p
        GROUP BY start_date, user_id
) as q
GROUP BY start_date

请注意，我没有将整数计数转换为文本并附加＆＃34;用户＆＃34;如你的样本。

MySQL查询生成报告的时间为10天

1 个答案: