sql server表峰值时间

我有一个应用程序，有1000个用户在不同的时间出于不同的目的登录。 现在的任务是以某种方式计算出“高峰时间的用户数” 我们在sql中记录的是userLoginTime，timespunt。 这里的问题是

如何实际计算应用程序的峰值时间。 以及如何计算高峰时间的用户数

---

在Sql中是否可能？我已经做了一个游戏-我正在使用记录的开始和结束日期时间2值的会话，但希望您可以调整当前数据以符合以下要求：

样本数据（如果我的答案错了，也许您可以采用此方法，将其添加到您的问题中，并添加更多样本和预期输出）：

以及查询：

--Logically, the highest number of simultaneous users was reached at some point when a session started
;with StartTimes as (
    select distinct SessionStart as Instant from #Sessions
), Overlaps as (
    select
        st.Instant,COUNT(*) as Cnt,MIN(s.SessionEnd) as SessionEnd
    from
        StartTimes st
            inner join
        #Sessions s
            on
                st.Instant >= s.SessionStart and
                st.Instant < s.SessionEnd
    group by
        st.Instant
), RankedOverlaps as (
    select Instant as SessionStart,Cnt,SessionEnd,RANK() OVER (ORDER BY Cnt desc) as rnk
    from Overlaps
)
select * from RankedOverlaps where rnk = 1

drop table #Sessions

---

另一种方法仍然使用上述方法，但如果您还想分析“不太峰值”值，则如下所示：

--An alternate approach - arrange all of the distinct time values from Sessions into order
;with Instants as (
    select SessionStart as Instant from #Sessions
    union --We want distinct here
    select SessionEnd from #Sessions
), OrderedInstants as (
    select Instant,ROW_NUMBER() OVER (ORDER BY Instant) as rn
    from Instants
), Intervals as (
    select oi1.Instant as StartTime,oi2.Instant as EndTime
    from
        OrderedInstants oi1
            inner join
        OrderedInstants oi2
            on
                oi1.rn = oi2.rn - 1
), IntervalOverlaps as (
    select
        StartTime,
        EndTime,
        COUNT(*) as Cnt
    from
        Intervals i
            inner join
        #Sessions s
            on
                i.StartTime < s.SessionEnd and
                s.SessionStart < i.EndTime
    group by
        StartTime,
        EndTime
)
select * from IntervalOverlaps order by Cnt desc,StartTime

---

这里解释了一个解决方案：

userLoginTime

和

timespunt

的数据类型是什么？您正在尝试查找整个集合中登录用户最多的时间间隔？（我说的是间隔，以防出现平局）（时间间隔很可能比任何覆盖时间段的单个用户都小）数据类型分别是DateTime和Integer。记录的时间是sql中的当前日期时间，因此可能是15:24 12:34

SessionStart           Cnt         SessionEnd             rnk
---------------------- ----------- ---------------------- --------------------
2012-01-03 00:00:00.00 2           2012-01-05 00:00:00.00 1
2012-01-07 00:00:00.00 2           2012-01-08 00:00:00.00 1

--An alternate approach - arrange all of the distinct time values from Sessions into order
;with Instants as (
    select SessionStart as Instant from #Sessions
    union --We want distinct here
    select SessionEnd from #Sessions
), OrderedInstants as (
    select Instant,ROW_NUMBER() OVER (ORDER BY Instant) as rn
    from Instants
), Intervals as (
    select oi1.Instant as StartTime,oi2.Instant as EndTime
    from
        OrderedInstants oi1
            inner join
        OrderedInstants oi2
            on
                oi1.rn = oi2.rn - 1
), IntervalOverlaps as (
    select
        StartTime,
        EndTime,
        COUNT(*) as Cnt
    from
        Intervals i
            inner join
        #Sessions s
            on
                i.StartTime < s.SessionEnd and
                s.SessionStart < i.EndTime
    group by
        StartTime,
        EndTime
)
select * from IntervalOverlaps order by Cnt desc,StartTime

StartTime              EndTime                Cnt
---------------------- ---------------------- -----------
2012-01-03 00:00:00.00 2012-01-05 00:00:00.00 2
2012-01-07 00:00:00.00 2012-01-08 00:00:00.00 2
2012-01-01 00:00:00.00 2012-01-03 00:00:00.00 1
2012-01-05 00:00:00.00 2012-01-07 00:00:00.00 1
2012-01-08 00:00:00.00 2012-01-09 00:00:00.00 1