시도 :
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"
출력 :
UserID DaysBetweenSessions
----------- -------------------
1 8
1 5
1 103
5 4
5 117
(5 row(s) affected)
당신은 일치하는 다음 행 아웃합니다 (WHERE 주석)을 가진 사람을 포함하고,이 결과 세트를 얻을하지 않으려면 :
를
UserID DaysBetweenSessions
----------- -------------------
1 8
1 5
5 4
(3 row(s) affected)
너무 나쁜 SQL Server에는 LEAD/LAG가 없습니다. ... ( –