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| 07.28.2008 at 01:06PM PDT, ID: 23602039 |
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The Solution Rating System
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With so many solutions, how can you tell which solutions are most likely to help you and which ones are not? To provide you with a tool to use, we rate our solutions based on various elements that most accurately determine if a solution is a quality solution. To explain what factors affect the solution rating, here are the elements we take into consideration when formulating our solution rating.
Your Input Matters If you have any suggestions that you would like to make for our rating system, please ask a question in the Suggestions Zone of Community Support. Thank you! |
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set ANSI_NULLS ON set QUOTED_IDENTIFIER ON go ALTER PROC [dbo].[AddUser]( @FirstName varchar(5) ,@LastName varchar(50) ,@Location int ,@ContactInfo_Tele varchar(50) ,@Email varchar(50) ) AS BEGIN --Steps --1. Get PK+1 from tbl UserS as @NewUserId. --3. Get PK+1 from tbl ContactInfo @NewContactID. --4. Insert values @NewUserId,@FirstName,@LastName,@Location,@NewContactID -- into tbl UserS --6. Insert values @NewContactID,@ContactInfo_Tele,@Email -- into tbl ContactInfo DECLARE @error int BEGIN TRAN --Do steps 1,2,3. DECLARE @NewUserId INT SELECT @NewUserId = CASE WHEN MAX(PK_User) IS NULL THEN 1 ELSE MAX(PK_User) + 1 END FROM UserS DECLARE @NewContactID INT SELECT @NewContactID = CASE WHEN MAX(PK_ContactInfo) IS NULL THEN 1 ELSE MAX(PK_ContactInfo) + 1 END FROM ContactInfo --Do step 4. BEGIN INSERT INTO UserS VALUES( @NewUserId ,@FirstName ,@LastName ,@Location ,@NewContactId) SET @error = @@ERROR END IF @error = 0 --Do step 6. BEGIN INSERT INTO ContactInfo VALUES ( @NewContactID ,@ContactInfo_Tele ,@Email) SET @error = @@ERROR END IF @error = 0 COMMIT ELSE ROLLBACK END |