You're setting the value after you call the insert. Do this instead... set @Requested = 1 <--- I put it here, but it doesn't work Insert into CallingCardInfo(CustomerUID, Requested) Values(@CustomerUID, @Requested)
On 11/8/05, sas0riza <[EMAIL PROTECTED]> wrote: > HI, > > I have a SP where I do 2 insertions. The first one inserts the first > record and the second one, inserts the second record using the PK > from the first record (which goes into a second table). I have all > of this, but my question is: > > How/where do I set the value for a column in the second table? > > Here's my code: > > CREATE procedure sp_InsertNewCard > @Title varchar(5), > @LastName varchar(50), > @FirstName varchar(50), > @MiddleInitial char(1), > @Email varchar(200), > @Phone varchar(12), > @PhoneExt varchar(10), > @Error_Code int OUTPUT > > as > declare > @SQLErrorCode int, > @CustomerUID int, > @Requested bit > > begin tran InsertNewCard > > -- Step 1: Insert the first record > Insert into Cardholders(Title, LastName, FirstName, MiddleInitial, > Email, Phone, PhoneExt) > Values(@Title, @LastName, @FirstName, @MiddleInitial, @Email, > @Phone, @PhoneExt) > > -- Now check for any errors > Select @SQLErrorCode = @@Error > > -- Get the PK for this new record > Select @CustomerUID = @@identity > > -- If we have an error, set the variable and return it, and rollback > transaction > if @SQLErrorCode <> 0 > begin > rollback tran InsertNewCard > set @Error_Code = 1 > return > end > > -- Step 2: Insert the second record using the PK from the first > record inserted above > Insert into CallingCardInfo(CustomerUID, Requested) > Values(@CustomerUID, @Requested) > > set @Requested = 1 <--- I put it here, but it doesn't work > > -- Now check for any errors > Select @SQLErrorCode = @@Error > > -- If we have an error, set the variable and return it, and rollback > transaction > if @SQLErrorCode <> 0 > begin > rollback tran InsertNewCard > set @Error_Code = 2 > return > end > > -- If we have it this far, then the transaction was successful > commit tran InsertNewCard > set @Error_Code = 0 > return > GO > > > > > Any help is greatly appreciated. > > Thanks! > > > > > > > > > Yahoo! Groups Links > > > > > > > > -- Dean Fiala Very Practical Software, Inc http://www.vpsw.com ------------------------ Yahoo! Groups Sponsor --------------------~--> Most low income households are not online. Help bridge the digital divide today! http://us.click.yahoo.com/cd_AJB/QnQLAA/TtwFAA/saFolB/TM --------------------------------------------------------------------~-> Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/AspNetAnyQuestionIsOk/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
