sql服務器表高峯時間

Question

我有一個應用程序，有1000個用戶在不同時間登錄到各種目的。 現在的任務是想出「高峯時用戶數」 我們在sql中記錄的是userLoginTime，timespent。 這裏的問題是

如何實際計算應用程序的高峯時間。 以及如何計算高峯時間的用戶數量。

是否有可能在SQL

Answer 1

我有一個發揮作用的 - 我與一個記錄會話開始工作，並最終datetime2值，但希望你能適應當前的數據符合本：

的樣本數據（如果我得到了答案錯了，也許你可以通過這一點，把它添加到你的問題，並添加更多的樣本和預期產出）：

create table #Sessions (
    --We'll treat this as a semi-open interval - the session was "live" at SessionStart, and "dead" at SessionEnd 
    SessionStart datetime2 not null, 
    SessionEnd datetime2 null 
) 
insert into #Sessions (SessionStart,SessionEnd) values 
('20120101','20120105'), 
('20120103','20120109'), 
('20120107','20120108') 

和查詢：

--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 

其中，與我樣本數據得出：

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 

---

的另一種方法，仍使用上述，但如果你也想分析「不太峯」值也，如下：

--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 

這一次，我輸出時間PE的所有 riods，隨着在時刻（順序從最高到最低）同時使用的用戶號碼一起：

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