创建查询时出错 [英] Error when Creating Query

问题描述

嘿伙计我需要你的帮助，我在创建这个查询时遇到错误。这个查询的作用很简单。它需要3个参数firstname，lastname和guid（唯一数字组合），然后检查该人是否已经存在。如果该人这样做，则输出该人的id，如果没有，则插入该人然后输出新的id。然而，它收到此错误：



消息102，级别15，状态1，过程Check_IF_Person_Exist，第4行

'@附近的语法不正确intPersonID'。

消息137，级别15，状态1，过程Check_IF_Person_Exist，第12行

必须声明标量变量@intPersonID。

消息137，级别15，状态1，过程Check_IF_Person_Exist，第19行

必须声明标量变量@intPersonID。



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Hey guys i need your help, i am getting errors when creating this query. What this query does is preety simple. It take 3 parameter firstname, lastname and guid(unique number combination), then it check if that person already exists. If the person does, then output the person's id and if not, insert the person then out the new id. However, it get this error:

Msg 102, Level 15, State 1, Procedure Check_IF_Person_Exist, Line 4
Incorrect syntax near '@intPersonID'.
Msg 137, Level 15, State 1, Procedure Check_IF_Person_Exist, Line 12
Must declare the scalar variable "@intPersonID".
Msg 137, Level 15, State 1, Procedure Check_IF_Person_Exist, Line 19
Must declare the scalar variable "@intPersonID".

Create Procedure Check_IF_Person_Exist(@FirstName nvarchar(50),@LastName  nvarchar (50),@Guid  int(11))


@intPersonID  int OUTPUT


AS
BEGIN
	IF EXISTS( SELECT ID FROM ITTESI_AssetTracker_Person WHERE Guid = @Guid)
		BEGIN
		    -- Assign existing PersonID to output parameter @intPersonID
			SET @intPersonID = @@IDENTITY
		END
	ELSE
		BEGIN
			INSERT INTO ITTESI_AssetTracker_Person VALUES (@Guid,@FirstName,@LastName)
			
			-- Assign newly generated PersonID to output parameter @intPersonID
			SET @intPersonID = @@IDENTITY
		END
END

感谢您的帮助。

Thanks for your help.

推荐答案

您需要添加：



DECLARE @intPersonID int

You need to add this:

DECLARE @intPersonID int

Create Procedure [dbo].[Check_IF_Person_Exist]

@FirstName nvarchar(50),
@LastName  nvarchar (50),
@Guid  int(11)
 

AS

DECLARE @intPersonID int


BEGIN
	IF EXISTS( SELECT ID FROM ITTESI_AssetTracker_Person WHERE Guid = @Guid)
		BEGIN
		    -- Assign existing PersonID to output parameter @intPersonID
			SET @intPersonID = @@IDENTITY
		END
	ELSE
		BEGIN
			INSERT INTO ITTESI_AssetTracker_Person VALUES (@Guid,@FirstName,@LastName)
			
			-- Assign newly generated PersonID to output parameter @intPersonID
			SET @intPersonID = @@IDENTITY
		END
END

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