Question

我有这张桌子

UserID   SessionID   SessionStart   SessionEnd
-----------------------------------------------
1        abc1        2010-1-1       2010-1-2
5        def3        2010-1-5       2010-1-9
1        llk0        2010-1-10      2010-1-11
5        spo8        2010-1-13      2010-1-15
1        pie7        2010-1-16      2010-1-29

我希望能够找到每个特定用户在一个会话结束到下一个会话开始之间的日子。

所以我希望得到像

这样的东西

UserID   DaysBetweenSessions
-----------------------------
1        8
1        5
5        4

谢谢！

Answer 1

尝试：

DECLARE @YourTable table (UserID int, SessionId char(5), SessionStart datetime, SessionEnd datetime)
INSERT @YourTable VALUES (1,'abc1','2010-1-1 ','2010-1-2')
INSERT @YourTable VALUES (5,'def3','2010-1-5 ','2010-1-9')
INSERT @YourTable VALUES (1,'llk0','2010-1-10','2010-1-11')
INSERT @YourTable VALUES (5,'spo8','2010-1-13','2010-1-15')
INSERT @YourTable VALUES (1,'pie7','2010-1-16','2010-1-29')

;WITH AllStarts AS
(SELECT
     UserID, SessionEnd,row_number() over (partition by UserID order by SessionStart) as EndRank
     FROM @YourTable
)
, AllEnds AS 
(SELECT 
     UserID, SessionStart, row_number() over (partition by UserID order by SessionEnd) as StartRank
     FROM @YourTable
)
SELECT 
    s.UserID, DATEDIFF(day,s.SessionEnd,ISNULL(e.SessionStart,GETDATE())) AS DaysBetweenSessions
    FROM AllStarts              s
        LEFT OUTER JOIN AllEnds e on s.UserID = e.UserID and e.StartRank=s.EndRank+1
    --WHERE e.UserID is not NULL  --include to remove "ones in progress"

输出：

UserID      DaysBetweenSessions
----------- -------------------
1           8
1           5
1           103
5           4
5           117

(5 row(s) affected)

如果您不想包含没有匹配的下一行的那些（取消注释WHERE）并获得此结果集：

UserID      DaysBetweenSessions
----------- -------------------
1           8
1           5
5           4

(3 row(s) affected)

SQL Server 2008：获取用户会话之间的持续时间

1 个答案: