Question

我的数据包含秒[时间]值。我想根据每秒的记录数计算我的数据。

这样的工作原理如下：

SELECT [Time], count(*) 

FROM [dbo].[Times]
GROUP BY
[Time] ORDER By [Time]

这给了我：

+----------+----+
| 00:00:00 | 4  |
+----------+----+
| 00:00:01 | 2  |
+----------+----+
| 00:00:02 | 1  |
+----------+----+
| 00:00:08 | 2  |
+----------+----+
...

我希望以5秒的间隔对它们进行分组和计数，结果如下：

+----------+----+
| 00:00:05 | 7  |   <--   # of records between 0:00:00 and 0:00:05
+----------+----+
| 00:00:10 | 2  |   <--   # of records between 0:00:05 and 0:00:10
+----------+----+
...

你有什么好办法在MS SQL中做到这一点吗？

Answer 1

这是一种方法：

select cast(dateadd(second, (datediff(second, 0, [time]) / 5) * 5, 0) as time(0)) as [time],
       count(*)
from [dbo].[Times] t
group by dateadd(second, (datediff(second, 0, [time]) / 5) * 5, 0)
order by min([time]);

编辑：

正如评论中所指出的，你可以写：

select v.[Time], count(*)
from [dbo].[Times] t cross apply
     (values (cast(dateadd(second, (datediff(second, 0, [time]) / 5) * 5, 0) as time(0)) )
     ) as v([Time])
group by v.[Time]
order by v.[Time];

这使复杂的计算只在一个地方 - 防止我在编辑原始答案时所犯的错误类型。

Answer 2

如果您尝试使用BigObject数据库，它会提供time_seg()功能来轻松完成此操作。如下所示的查询将返回您想要的结果：

mysql> select time_seg(datetime,5s) as ti, COUNT(*) from t group by ts order by ti;
+---------------------+----------+
| ti                  | COUNT(*) |
+---------------------+----------+
| 2018-05-09 00:00:00 |        3 |
| 2018-05-09 00:00:05 |        1 |
+---------------------+----------+

以5秒的间隔分组[时间]

2 个答案: