Question

在这里我需要得到消息。如果我在应用程序中输入电子邮件ID，我应该收到电子邮件已经存在的消息？

IF not EXISTS(SELECT * FROM Resource WHERE Email= @Email)
if(@status = 'Insert')
BEGIN
     Insert into resource(FirstName,LastName,Password,Role,Email,Phone,OrganisationID) 
     values (@FirstName,@LastName,@Password,@Role,@Email,@Phone,@OrganisationID)
END

Answer 1

添加ELSE块：

IF not EXISTS(SELECT * FROM Resource WHERE Email= @Email)
BEGIN
    if(@status = 'Insert')
    BEGIN

      Insert into resource(FirstName, LastName, Password, 
                           Role, Email, Phone, OrganisationID) 
      values (@FirstName,@LastName,@Password,
              @Role,@Email,@Phone,@OrganisationID)
    end
END
ELSE  
BEGIN
    RETURN 0;
END;

或：您可以将条件移至IF，如下所示：

IF not EXISTS(SELECT * FROM Resource WHERE Email= @Email)
   AND @status = 'Insert'
BEGIN
      Insert into resource(FirstName, LastName, Password, 
                           Role, Email, Phone, OrganisationID) 
      values (@FirstName,@LastName,@Password,
              @Role,@Email,@Phone,@OrganisationID);
END
ELSE  
BEGIN
    RETURN 0; -- It depends on the return type 
END;

Answer 2

if not Exists (select Email from Resource where Email= @Email)
begin
 set @status='Insert'
Insert into resource(FirstName,LastName,Password,Role,Email,Phone,OrganisationID) 
values (@FirstName,@LastName,@Password,@Role,@Email,@Phone,@OrganisationID)
end
else begin  set @status='EmailExists='+@Email end

这是我的存储过程，用于检查是否应该有相同的电子邮件ID

2 个答案: