• Hello again...

    As you can see, posting XML strings is stripping everything out.

    I've recoded this to build the string in a more complex method. Hopefully, it will post okay.

    This code, when I run it, returns 2 records. What do you get?

    declare @start char(1), @end char(1), @Root char(1), @Node char(8), @Element char(15)

    select @Start = char(60), @end = char(62), @Root = 'C', @Node = 'CUSTOMER', @Element = 'CUSTOMER_NUMBER'

    declare @C XML

    set @C = @Start + @Root + @End +

    @Start + @Node + @End + @Start + @Element + @End + '1' + @Start + '/' + @Element + @End + @Start + '/' + @Node + @End +

    @Start + @Node + @End + @Start + @Element + @End + '2' + @Start + '/' + @Element + @End + @Start + '/' + @Node + @End +

    @Start + '/' + @Root + @End

    select m.item.value('CUSTOMER_NUMBER[1]','integer') [Customer_Number]

    from @C.nodes('/C/CUSTOMER') as m(item)

    Wayne
    Microsoft Certified Master: SQL Server 2008
    Author - SQL Server T-SQL Recipes


    If you can't explain to another person how the code that you're copying from the internet works, then DON'T USE IT on a production system! After all, you will be the one supporting it!
    Links:
    For better assistance in answering your questions
    Performance Problems
    Common date/time routines
    Understanding and Using APPLY Part 1 & Part 2