如何从今天起每 3 小时对 MySQL 中的数据进行分组？

Question

我想为我的一个图表检索数据，并且想为它想出一个简单的解决方案，而不是执行 8 个子查询。

users 表如下所示：

user_id
1
2
3
4
5

pts 表如下所示：

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（有几个 cmets，我不知道该放什么）：

    "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
"

所以基本上查询需要在当天每 3 小时从数据库中收集一次数据（所以如果我使用上述选项，它将是 8 个子查询）。此外，如果间隔内尚未达到当前时间，则查询应返回 0 / null。如果当前时间在其中一个区间内，则它应该返回从区间开始到区间结束的数据。

我想要达到的结果：

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

这可以用 MySQL 实现吗？

Answer 1

你可以只对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。您的查询对数据进行了透视，但您想要的结果并未进行透视。

编辑：

如果你真的需要所有组的时间，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;