如何从今天起每隔3小时对MySQL中的数据进行分组？

Question

我想检索一个图表的数据,我想为它提出一个简单的解决方案,而不是做8个子查询.

这就是users表格的样子:

user_id
1
2
3
4
5

这就是pts表格的样子:

id  user_id  status  points                 time
1         1       0      100  2014-08-23 03:34:54
2         4       0      100  2014-08-23 04:04:44
3         1       1      300  2014-08-23 08:34:21
4         1       0      300  2014-08-23 15:25:11
5         1       0      200  2014-08-23 22:23:12

查询看起来像这样(带有一些注释,我不知道要放什么):

    "SELECT  *,
        (
            SELECT  SUM(points)
            FROM    pts
            WHERE   status = 0 
            AND     user_id = 1
            AND     time >= "THIS DAY'S 00:00" AND <= "THIS DAY'S 03:00" 
        ),
        (
            SELECT  SUM(points)
            FROM    pts
            WHERE   status = 0 
            AND     user_id = 1
            AND     time >= "THIS DAY'S 03:00" AND <= "THIS DAY'S 06:00"
        ),
        (
            SELECT  SUM(points)
            FROM    pts
            WHERE   status = 0
            AND     user_id = 1 
            AND     time >= "THIS DAY'S 06:00" AND <= "THIS DAY'S 09:00"
        ),
        (
            SELECT  SUM(points)
            FROM    pts
            WHERE   status = 0 
            AND     user_id = 1
            AND     time >= "THIS DAY'S 12:00" AND <= "THIS DAY'S 15:00"
        ),
"ETC......"
FROM    pts
WHERE   user_id = 1
AND     status  = 1
"

因此,基本上查询需要在当天每隔3小时从数据库收集数据(因此,如果我选择上述选项,那么将是8个子查询).如果该时间间隔尚未达到当前时间,则查询应返回0/null.如果当前时间处于其中一个间隔中,则它应该从间隔的开始直到间隔的结束返回数据.

我希望实现的结果:

id  user_id  points                                       time
1         1       0  2014-08-23 00:00:00 - 2014-08-23 03:00:00
2         1     200  2014-08-23 03:00:00 - 2014-08-23 06:00:00
3         1     300  2014-08-23 06:00:00 - 2014-08-23 09:00:00
4         1       0  2014-08-23 09:00:00 - 2014-08-23 12:00:00
5         1       0  2014-08-23 12:00:00 - 2014-08-23 15:00:00
6         1     300  2014-08-23 15:00:00 - 2014-08-23 18:00:00
7         1       0  2014-08-23 18:00:00 - 2014-08-23 21:00:00
8         1     200  2014-08-23 21:00:00 - 2014-08-23 00:00:00

这有可能与MySQL？

Answer 1

您可以在小时使用聚合和算术:

select floor(hour(time) / 3) as hourgroup, sum(points)
from pts
where status = 0 and user_id = 1 and
      date(time) = "THIS DAY"
group by floor(hour(time) / 3);

对于您的特定输出:

select (hour(time) div 3) as id, user_id, sum(points),
       date('2014-08-23') + interval 3 * (hour(time) div 3) hour as starttime,
       date('2014-08-23') + interval 3 * ((hour(time) div 3) + 1) hour as endtime
from pts
where status = 0 and user_id = 1 and
      date(time) = date('2014-08-23')
group by floor(hour(time) / 3);

计数为零的值存在问题.我不确定这些有多重要.您的查询会对数据进行调整,但不会旋转您想要的结果.

编辑:

如果你真的需要所有团队的时间,你可以这样做:

select n.n as id, user_id, sum(points),
       date('2014-08-23') + interval 3 * n.n hour as starttime,
       date('2014-08-23') + interval 3 * (n.n + 1) hour as endtime
from (select 0 as n union all select 1 union all select 2 union all select 3 union all
      select 4 union all select 5 union all select 6 union all select 7
     ) n left join
     pts
     on n.n = hour(time) div 3
where status = 0 and user_id = 1 and
      date(time) = date('2014-08-23')
group by n.n;