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需求:有一个用户登陆日志表,记录用户每次登陆时间,然后想查找用户按天连续登陆的情况,找出每次连续登陆的最早时间和最后时间以及连续登陆天数。
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由于长久未写此类SQL,有点手生,本着走一步算一步的精神,慢慢来。
首先查看日志表
SELECT [Uid]
,[loginDate]
FROM [dbo].[Member_LoginLog]
WHERE [UID]=268
由于按天计算连续登陆,表中时间精确到毫秒,很难肉眼看出数据是否连续,于是考虑转换数据
而又由于我们只关心最早登陆时间和最后登陆时间,因此我们可以先按照天来统计用户最早登陆时间和最后登陆时间,并将时间转换成对应天数
--==============================================
--统计出用户每天最早登陆时间和最后登陆时间
SELECT T1.[UID]
,DATEDIFF(DAY,'2014-01-01',LoginDate) AS DiffDays
,MAX(LoginDate) AS MaxLoginDate
,MIN(LoginDate) AS MinLoginDate
INTO [dbo].[Member_LoginLog_Status1]
FROM [dbo].[Member_LoginLog] T1
GROUP BY T1.[UID],DATEDIFF(DAY,'2014-01-01',LoginDate)
--======================================
--查看效果
SELECT [UID]
,[DiffDays]
,[MaxLoginDate]
,[MinLoginDate]
FROM [dbo].[Member_LoginLog_Status1]
WHERE UID=268
从上图很容易看出第二天没连续登陆,是不是很容易看啊
接下来就是查找联系的天数了,如果我们按照UID分组,然后对DiffDays来排序求出排名来,依据DiffDays的增长量和RID量便可以判断出天数是否连续
SELECT
ROW_NUMBER()OVER(PARTITION BY UID ORDER BY [DiffDays] ASC) AS RID,
T1.*
FROM [dbo].[Member_LoginLog_Status1] T1
WHERE [UID]=268
这样我们便可以使用表的自连接来查找连续的登录,由于需要按照用户和天数来算出排名,因此我们可以先建立索引
CREATE CLUSTERED INDEX CIX_UID_Days ON
[dbo].[Member_LoginLog_Status1]
(
[UID],[DiffDays]
)
然后再求连续区间:
--==========================================
--查找连续的登录
;WITH Tem AS(
SELECT
ROW_NUMBER()OVER(PARTITION BY UID ORDER BY [DiffDays] ASC) AS RID,
T1.*
FROM [dbo].[Member_LoginLog_Status1] T1
)
,Tem1 AS(
SELECT ROW_NUMBER()OVER(
PARTITION BY T1.[UID],T1.[DiffDays]
ORDER BY T2.[diffdays]-T1.[diffdays] DESC) AS RID,
T1.[UID],
T1.MinLoginDate,
T2.MaxLoginDate,
T1.[diffdays] AS MinDiffDays,
T2.[diffdays] AS MAXDiffDays
FROM Tem AS T1
INNER JOIN Tem AS T2
ON T1.UID=T2.UID
AND T1.[diffdays]<=T2.[diffdays]
AND T2.[diffdays]-T1.[diffdays]= T2.RID-T1.RID
)
SELECT
[UID],
MinLoginDate,
MaxLoginDate,
MinDiffDays,
MAXDiffDays
INTO [dbo].[Member_LoginLog_Status2]
FROM Tem1 AS T1
WHERE T1.RID=1
--=========================================
--检查结果
SELECT [UID]
,[MinLoginDate]
,[MaxLoginDate]
,[MinDiffDays]
,[MAXDiffDays]
FROM [dbo].[Member_LoginLog_Status2]
WHERE [UID]=268
找出连续的区间后,我们会发现有很多区间不是最大连续区间,如第5天到第17天连续,但是比之更大的区间还有第3天到第17天,对于这种问题,解决办法就是依据maxDiffDays分组,求出最小的minDiffDays
由于此时要按照用户和maxDiffDays分组,然后按照MinDiffDays排序求最小值,因此先建立索引
CREATE CLUSTERED INDEX CIX_UID_MAXDiffDays
ON [AccMain_101].[dbo].[Member_LoginLog_Status2]
([UID],MAXDiffDays,MinDiffDays ASC)
然后再查询:
--====================================
--求出最大连续区间
;WITH CTE1 AS(
SELECT
ROW_NUMBER()OVER(PARTITION BY [UID],MAXDiffDays ORDER BY MinDiffDays ASC) AS RID,
[UID],
MinLoginDate,
MaxLoginDate,
MinDiffDays,
MAXDiffDays
FROM [AccMain_101].[dbo].[Member_LoginLog_Status2] AS T1
)
INSERT INTO [dbo].[Member_LoginLog_Status3]
([Uid]
,[firstLoginDate]
,[lastLoginDate]
,[loginNumber])
SELECT [UID],
MinLoginDate,
MaxLoginDate,
T1.MAXDiffDays-MinDiffDays AS ContinueDays
FROM CTE1 T1
WHERE T1.RID=1
--==================================
--查看结果
SELECT [Uid]
,[firstLoginDate]
,[lastLoginDate]
,[loginNumber]
FROM [dbo].[Member_LoginLog_Status3]
WHERE [UID]=268
查询结果:
结果正是我们想要的,因此打完收工,回家吃饭。
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总结:其实查找连续或查找孤岛这类原理,都是利用自连接然后看增长是否连续,多折腾几遍就好。
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在wwwwgou的回复中,指出一条更快捷的计算方式,同样使用排名来计算,但不使用关联,而是计算排名与登陆天数的差值,如果登陆天数连续增长,则排名也连续增长,两者的差值保持不变;如果登陆天数不连续,则登陆天数增长的值就会比排名增长的值高,这时两者的差值就会变大。
如下图:
随着天数不连续的次数增加,[天数-排名]的值会不断增大,因此可以使用[天数-排名]来分组,便可以定位到连续区间。
PS: 不会出现两个不同连续区间的[天数-排名]值一样的情况
查找代码:
--========================================
--感谢wwwwgou提供,
--此代码已略做修改
SELECT
[Uid],
mindt = MIN(mindt),
maxdt = MAX(maxdt),
logdays = COUNT(*)
FROM
(
SELECT
[Uid],
RowNo = ROW_NUMBER()
OVER(PARTITION BY [Uid]
ORDER BY DATEDIFF(DAY,'2014-01-01', loginDate)),
DiffDay = DATEDIFF(DAY,'2014-01-01', loginDate),
mindt = MIN(loginDate),
maxdt = MAX(loginDate)
FROM dbo.Member_LoginLog
GROUP BY [Uid], DATEDIFF(DAY,'2014-01-01', loginDate)
) T
GROUP BY [Uid], [RowNo] - DiffDay
ORDER BY [Uid], minDt
对wwwwgou筒子再次表示婶婶地感谢。
--===============================================
请原谅我苍白的讲解,让您们只能看代码。
妹子骚猴就上,不要着急。