Question

我想要实现的是根据LoginAttempts值和LastLoginDate计算用户的登录尝试次数。例如，我需要在30天内使用2次登录尝试查询LastLoginDate。

结果应该是：

我拥有的是这个..我创建了临时表来提取信息，并且它似乎没有正确计数。这就是我被困住的地方。任何帮助都会受到赞赏!!

Answer 1

您的GROUP BY不正确。它包括LastLoginDateUTC。因此，您需要计算每个日期的登录次数，而不是每30天登录一次。

从GROUP BY中删除LastLoginDateUTC，并将SELECT子句更改为使用plt.imshow(acov,interpolation='nearest', cmap='Reds') plt.colorbar() plt.show()。这应该给你你想要的。

Answer 2

我不确定我是否完全理解您的请求，但这里的内容已经不在了。祝你好运！：

   select FirstName, StudentID, UserName, LastLoginDate, LoginAttempts,
    RowNumber = row_number() over(partition by StudentID order by StudentID, LoginAttempts)
    into #temp1
    from user
    where cast(LastLoginDate as date) >= getdate()- 30
    --should return all rows of data for past 30 days. 
    --should also rank each loginAttempt by student 

    select * from #temp1
    where RowNumber >= 3

Answer 3

这不是答案......更多建议。

我认为你有学生，你需要审核登录尝试这对我来说是一对多的关系所以我会保留学生表，从任何登录相关数据中删除这样可以简化您的Student表，减少行数，节省存储空间，不会反复使用相同的数据（名称，用户名等）。
该数据将在StudentLoginAttempts中，例如：

    Create Table StudentLoginAttempts (
        Id int not null identity(1,1),
        StudentId int not null,
        LoginDate datetime not null,
        Successful bit not null,

        Constraint PK_StudentLoginAttempts Primary Key Clustered (Id),
        Constraint FK_StudentLoginAttempts_Student Foreign Key (StudentId) References Student(StudentId)
    )
    go

    Create Index IX_StudentLoginAttempts_StudentId On StudentLoginAttempts(StudentId)
    go

    Create Index IX_StudentLoginAttempts_LoginDate On StudentLoginAttempts(LoginDate)
    go

所以事情可能会更清楚，你可以获得更多信息想想下面的例子：

Create Table #Student (
    StudentId int not null identity(1,1),
    Username varchar(50) not null,
    FirstName varchar(50) not null
)

Create Table #StudentLoginAttempts (
    Id int not null identity(1,1),
    StudentId int not null,
    LoginDate datetime not null,
    Successful bit not null
)

insert into #Student values
( 'Student001', 'JON' ),
( 'Student002', 'STEVE' )

insert into #StudentLoginAttempts values
( 1, '2016-01-01 09:12', 0 ),
( 1, '2016-02-01 09:12', 0 ),
( 1, '2016-03-01 09:12', 1 ),
( 2, '2016-03-02 10:12', 0 ),
( 2, '2016-04-02 10:12', 1 ),
( 2, '2016-05-02 10:12', 0 )

;with TotalAttemptsCte as (
    select StudentId, TotalLoginAttempts = count(*) from #StudentLoginAttempts group by StudentId
),
FailedCte as (
    select StudentId, FailedLogins = count(*) from #StudentLoginAttempts where ( Successful = 0 ) group by StudentId
),
SuccessfulCte as (
    select StudentId, SuccessfulLogins = count(*) from #StudentLoginAttempts where ( Successful = 1 ) group by StudentId
),
LastSuccessFulDateCte as (
    select StudentId, max(LoginDate) as LastSuccessfulLoginDate
    from
        #StudentLoginAttempts
    where
        ( Successful = 1 ) 
    group by StudentId
)
select
    a.*, b.TotalLoginAttempts, c.FailedLogins, d.SuccessfulLogins, e.LastSuccessfulLoginDate
from
    #Student a
    left join TotalAttemptsCte b       on ( a.StudentId = b.StudentId )
    left join FailedCte c              on ( a.StudentId = c.StudentId )
    left join SuccessfulCte d          on ( a.StudentId = d.StudentId )
    left join LastSuccessFulDateCte e  on ( a.StudentId = e.StudentId )

Drop Table #StudentLoginAttempts
Drop Table #Student

您还可以根据查询创建视图，以便更轻松地访问。

我提醒你，这只是一个建议，我会怎么做。

如何根据另一个日期列计算日期列？

3 个答案: