Question

我正在尝试创建一个存储过程来验证用户（登录），除了用户输入未注册的电子邮件之外，一切正常。您看到我已经处理了该异常，其中如果表中的电子邮件地址计数为0，则应该给出：

  Select 0 as AccountLocked,0 as Authenticated,0 as RetryAttempts,0 as Registered

如果它不是0，它将执行下一步。问题是当输入一个未注册的电子邮件时，它会返回两组结果，如下所示：

  Select 0 as AccountLocked,0 as Authenticated,0 as RetryAttempts,0 as Registered

和

 Select 1 as AccountLocked,0 as Authenticated,0 as RetryAttempts,1 as Registered

完整的过程可供参考。我在这里缺少什么？为什么它给了我第二个结果，我不想要？

Alter proc spValidateUser
    @EmailAdd nvarchar(30),
    @Password nvarchar(20)
as
begin
    Set Nocount on;
    Declare @UserId nvarchar(10),@LastLogin datetime,@RoleId int,@AccountLocked bit,@RetryCount int,@Count int

    Select @Count=Count(EmailAdd) from tblAllUsers
    where EmailAdd=@EmailAdd
    if(@Count = 0) begin
        Select 0 as AccountLocked,0 as Authenticated,0 as RetryAttempts,0 as Registered
    end else
        Select @AccountLocked=IsLocked from tblAllUsers where EmailAdd=@EmailAdd

    ----if account is already locked------
    if(@AccountLocked = 1) begin
       Select 1 as AccountLocked,0 as Authenticated,0 as RetryAttempts,1 as Registered
    end else begin
        -----check if username and password match-----
        Select @UserId = UserId, @LastLogin=LastLogin, @RoleId=RoleId
        from tblAllUsers where EmailAdd=@EmailAdd and Password=@Password
          ----if match found--------
        If @UserId is not null Begin
            Update tblAllUsers
            SET LastLogin= GETDATE(),RetryAttempts=0 WHERE UserId=@UserId

            Select @UserId [UserId],
            (Select Role from tblRoles where RoleId=@RoleId) [Roles],0 as AccountLocked,1 as Authenticated,0 as RetryAttempts,1 as Registered
        End Else Begin
            ------if match not found--------
            Select @RetryCount=ISNULL(RetryAttempts,0) from tblAllUsers where EmailAdd=@EmailAdd

            Set @RetryCount=@RetryCount+1
            if(@RetryCount<=3) Begin
                ----if retry attempts are not completed------
                Update tblAllUsers Set RetryAttempts=@RetryCount where EmailAdd=@EmailAdd

                Select 0 as AccountLocked,0 as Authenticated,@RetryCount as RetryAttempts,1 as Registered
            End Else Begin
                ------if retry attempts are completed--------
                Update tblAllUsers Set RetryAttempts=@RetryCount,IsLocked=1,LockedDateTime=GETDATE()
                where EmailAdd=@EmailAdd

                Select 1 as AccountLocked,0 as Authenticated,0 as RetryAttempts,1 as Registered
            End
        End
    End
End

编辑：看起来它也在执行以下代码：

 Begin
      ------if retry attempts are completed--------
      Update tblAllUsers Set RetryAttempts=@RetryCount,IsLocked=1,LockedDateTime=GETDATE()
      where EmailAdd=@EmailAdd
      Select 1 as AccountLocked,0 as Authenticated,0 as RetryAttempts,1 as Registered
      End

但为什么在电子邮件不匹配时执行上述操作？

Answer 1

我会改变你的程序逻辑，如下所示

        if exists(Select 1 from tblAllUsers where EmailAdd=@EmailAdd)
           Select 0 as AccountLocked,0 as 
           Authenticated,0 as RetryAttempts,0 as Registered
        else
           Select 1 as AccountLocked,0 as Authenticated,
           0 as RetryAttempts,1 as Registered

Answer 2

你遇到的问题是：

Select @Count=Count(EmailAdd) from tblAllUsers
where EmailAdd=@EmailAdd

现在@count将为零，如果出现以下情况，您将在以下方式中返回0作为AccountLocked：

if(@Count = 0) begin
    Select 0 as AccountLocked,0 as Authenticated,0 as Retry...
end else
    Select @AccountLocked=IsLocked from tblAllUsers where EmailAdd=@EmailAdd

这就是其他，这里没有开始+结束块。

在这里，您将检查帐户是否已锁定，但由于未找到@AccountLocked仍然为空

----if account is already locked------
if(@AccountLocked = 1) begin
   Select 1 as AccountLocked,0 as Authenticated,0 as Retry...
end else begin

现在你在这里检查密码，即使用户甚至找不到，所以这永远找不到任何东西：

    -----check if username and password match-----
    Select @UserId = UserId, @LastLogin=LastLogin, @RoleId=RoleId
    from tblAllUsers where EmailAdd=@EmailAdd and Password=@Password
      ----if match found--------
    If @UserId is not null Begin
   ....

然后第二个结果集来自这里：

    End Else Begin
        ------if match not found--------

存储过程提供了两组不需要的数据

2 个答案: