本文介绍了如何在SQL存储过程中获取if else中的多个输出参数值？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
I am writing stored procedure for login in which i want result through output parameters usertype and loginstatus values as 1 and 1 respectively.
Following is stored procedure code:
CREATE procedure [dbo].[usp_login]
(@userid nvarchar(50), @password nvarchar(50), @usertype int output, @loginstatus int output)
As
Begin
if exists(Select COUNT(User_Id) from User_Master where User_Id = @userid and Password = @password)
begin
select @usertype = User_Type from User_Master where User_Id = @userid and Password = @password;
set @loginstatus = 1;
end
else
begin
set @usertype = 0
set @loginstatus = 0;
end
End
I am executing the sp in ssms as follows:
declare @usrtype int;
declare @lgnstat int;
exec usp_login 'a', 'a', @usrtype output, @lgnstat output
print @usrtype
print @lgnstat
Now when i pass correct userid password i get result as 1,1 which is as expected.
But, when i pass wrong userid or password i get result as 1 which is expected as 0,0.
Please tell me where's the problem either in sp definition code or execution code ?
什么我试过了：
What I have tried:
CREATE procedure [dbo].[usp_login]
(@userid nvarchar(50), @password nvarchar(50), @usertype int output, @loginstatus int output)
As
Begin
if exists(Select COUNT(User_Id) from User_Master where User_Id = @userid and Password = @password)
begin
select @usertype = User_Type from User_Master where User_Id = @userid and Password = @password;
set @loginstatus = 1;
end
else
begin
set @usertype = 0
set @loginstatus = 0;
end
End
declare @usrtype int;
declare @lgnstat int;
exec usp_login 'z', 'z', @usrtype output, @lgnstat output -- passing wrong userid password
print @usrtype
print @lgnstat推荐答案
if exists(Select COUNT(User_Id) from User_Master where User_Id = @userid and Password = @password)请改为尝试：
if exists(Select 1 from User_Master where User_Id = @userid and Password = @password)
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