我必须在sql server中跟随表:
date | status 2009-01-01 12:00:00 OK 2009-01-01 12:03:00 Failed 2009-01-01 12:04:00 OK 2009-01-01 12:06:20 OK 2009-01-01 12:07:35 Failed 2009-01-01 12:07:40 Failed 2009-01-01 12:20:40 Failed 2009-01-01 12:25:40 OK
我需要以下内容:从2009年1月1日12:00开始,每隔10分钟,我需要查看OK和Failed的数量.
就像是:
INTERVAL Failed OK 2009-01-01 12:00:00-2009-01-01 12:15:00 1 2 2009-01-01 12:15:01-2009-01-01 12:30:00 0 1
等等..
解决方法
好的,首先..
你提到10分钟,并提供一个15分钟的例子..另外你的样本数据应该返回不同于你发布的结果..
解决方案使用Pivot
Declare @datetimestart datetime Declare @interval int Set @datetimestart = '2009-01-01 12:00:00' Set @interval = 15 Select * From ( Select DateAdd( Minute,Floor(DateDiff(Minute,@datetimestart,[date])/@interval)*@interval,@datetimestart),DateAdd( Minute,@interval + Floor(DateDiff(Minute,status From dtest ) As W([from],[to],status) Pivot (Count(status) For status In ([ok],[Failed])) p
这将回来
From To Ok Failed 2009-01-01 12:00:00.000 2009-01-01 12:15:00.000 3 3 2009-01-01 12:15:00.000 2009-01-01 12:30:00.000 1 0
评论后更新
此版本将包含数据库中没有值的时间间隔.
我们需要动态创建一个临时表..
Declare @datetimestart datetime,@datetimeend datetime,@datetimecurrent datetime Declare @interval int Set @datetimestart = '2009-01-01 12:00:00' Set @interval = 10 Set @datetimeend = (Select max([date]) from dtest) SET @datetimecurrent = @datetimestart declare @temp as table ([from] datetime,[to] datetime) while @datetimecurrent < @datetimeend BEGIN insert into @temp select (@datetimecurrent),dateAdd( minute,@interval,@datetimecurrent) set @datetimecurrent = dateAdd( minute,@datetimecurrent) END Select * From ( Select [from],status From @temp t left join dtest d on d.[date] between t.[from] and t.[to] ) As W([from],status) Pivot (Count(status) For status In ([ok],[Failed])) p
现在使用10分钟的间隔,显示没有值的句点,返回..
From To Ok Failed 2009-01-01 12:00:00.000 2009-01-01 12:10:00.000 3 3 2009-01-01 12:10:00.000 2009-01-01 12:20:00.000 0 0 2009-01-01 12:20:00.000 2009-01-01 12:30:00.000 1 0
原文地址:https://www.jb51.cc/mssql/76850.html
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。