Question

有一个记录用户登录的表。

列是：

UserID，Date，IsSuccessful（bit）

我需要找到那些有3个以上失败的考生的用户才能连续登录。

我试过了：

select  UserID, Date, IsSuccessful,
        count(*) over ( partition by UserID, IsSuccessful order by Date ) Cnt
from    UserLogins
order   by UserID, Date

Cnt列显示IsSuccessful值的序号，但在组更改时不会重新启动：

UserID, IsSuccessful, Cnt
555, 1, 1
555, 1, 2
555, 1, 3
555, 0, 1
555, 0, 2
555, 1, 4
555, 0, 3
555, 1, 5
555, 1, 6
555, 1, 7
555, 0, 4
555, 0, 5
555, 0, 6
555, 1, 8

我需要的是：

UserID, IsSuccessful, Cnt
555, 1, 1
555, 1, 2
555, 1, 3
555, 0, 1
555, 0, 2
555, 1, 1
555, 0, 1
555, 1, 1
555, 1, 2
555, 1, 3
555, 0, 1
555, 0, 2
555, 0, 3
555, 1, 1

Answer 1

您可以使用以下查询：

SELECT UserID, [Date], IsSuccessful,
       ROW_NUMBER() OVER (PARTITION BY UserID, IsSuccessful, grp 
                          ORDER BY [Date]) AS cnt
FROM (
  SELECT UserID, [Date], IsSuccessful,
         ROW_NUMBER() OVER (PARTITION BY UserID 
                            ORDER BY [Date]) -
         ROW_NUMBER() OVER (PARTITION BY UserID, IsSuccessful 
                            ORDER BY [Date]) AS grp
FROM UserLogins) AS t

Answer 2

另一种解决方案，不太通用，但解决了我的具体任务（连续检测到3次失败）：

select  UserID, [Date], IsSuccessful,
        SUM( 1 - CAST(IsSuccessful AS INT) ) OVER( PARTITION BY UserID ORDER BY DATE ROWS BETWEEN 2 PRECEDING AND CURRENT ROW ) AS NumberOfFailures
from    UserLogins
order   by UserID, [Date]

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2 个答案: