Question

我有以下选项从日志中选择并分组到分钟（不包括秒和milisec）：

SELECT DATEPART(YEAR, [Date]) AS YEAR, DATEPART(MONTH, [Date]) AS MONTH,
       DATEPART(DAY, [Date]) AS DAY, DATEPART(HOUR, [Date]) AS HOUR,
       DATEPART(MINUTE, [Date]) AS MIN, COUNT(*) AS COUNT

FROM [database].[dbo].[errorlog]

GROUP BY DATEPART(YEAR, [Date]), DATEPART(MONTH, [Date]), DATEPART(DAY, [Date]),
         DATEPART(HOUR, [Date]), DATEPART(MINUTE, [Date])

ORDER BY DATEPART(YEAR, [Date]) DESC, DATEPART(MONTH, [Date]) DESC,
         DATEPART(DAY, [Date]) DESC, DATEPART(HOUR, [Date]) DESC,
         DATEPART(MINUTE, [Date]) DESC;

但是正如你可以看到的那样只是为了获得一个计数，所以我想知道是否有更好的方法对它进行分组，所以我按年，月，日和小时分组到几分钟？ / p>

Answer 1

这应该有效：

select CAST([Date] AS smalldatetime) as time_stamp, count(*) as count
FROM [database].[dbo].[errorlog]
group by CAST([Date] AS smalldatetime)
order by CAST([Date] AS smalldatetime) desc;

对此答案的评论后更新：

select dateadd(second,-datepart(ss,[Date]),[Date]) as time_stamp, count(*) as count
FROM [database].[dbo].[errorlog]
group by dateadd(second,-datepart(ss,[Date]),[Date]) 
order by dateadd(second,-datepart(ss,[Date]),[Date]) desc ;

第一个解决方案将时间戳向上舍入到最近的分钟。我意识到这不是OP想要的。因此，第二个解决方案只是从时间戳中减去seconds部分，并将时间戳保留为秒zero（假设[日期]没有小数秒）

Answer 2

DATEADD(minute,DATEDIFF(minute,'20010101',[Date]),'20010101')

应将所有Date列值舍入到最接近的分钟。所以：

SELECT DATEADD(minute,DATEDIFF(minute,'20010101',[Date]),'20010101'),
     COUNT(*) AS COUNT
FROM [database].[dbo].[errorlog]
GROUP BY DATEADD(minute,DATEDIFF(minute,'20010101',[Date]),'20010101')
ORDER BY DATEADD(minute,DATEDIFF(minute,'20010101',[Date]),'20010101') DESC;

（如果您想进一步减少重复，可以将此表达式移动到子查询中）

Answer 3

你可以做这样的事情来获得

declare @now datetime
set @now = GETDATE()

select dateadd(minute, mm, @now) as date, c from (
  select DATEDIFF(minute, @now, [Date]) as mm, COUNT(1) as c
  from [database].[dbo].[errorlog]
  group by DATEDIFF(minute, @now, [Date])
) t

是否有更好的方法可以在几分钟内对日志进行分组？

3 个答案: