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XML root copy from InputRoot |
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crojas |
Posted: Tue Jun 18, 2002 7:07 am Post subject: XML root copy from InputRoot |
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Apprentice
Joined: 10 Mar 2002 Posts: 28
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I want to copy an child from InputRoot.XML to the root of the OutputRoot, but when i do the next:
SET OutputRoot.XML = InputRoot.XML.trx.trx1;
It uses the 'XML' tag as root, how do i do it? |
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kirani |
Posted: Tue Jun 18, 2002 7:21 am Post subject: |
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Jedi Knight
Joined: 05 Sep 2001 Posts: 3779 Location: Torrance, CA, USA
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Have you tried this,
SET OutputRoot.XML.trx1 = InputRoot.XML.trx.trx1; _________________ Kiran
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crojas |
Posted: Tue Jun 18, 2002 7:56 am Post subject: |
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Apprentice
Joined: 10 Mar 2002 Posts: 28
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Yes, but i have messages that can change, i mean trx1 could be trx2 and so on. |
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CodeCraft |
Posted: Tue Jun 18, 2002 1:55 pm Post subject: |
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Disciple
Joined: 05 Sep 2001 Posts: 195
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Not sure exactly what you want to do here: You need to give example input and output XML messages.
Look up the Item Keyword for removing the name from name/value items. Another way is to set first to a local variable, which can only hold a value, not name+value, and then map this to your destination.
If you need anonymous references, use:
Set OutputRoot.XML.*[1] = InputRoot.XML.trx.*[1]; |
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crojas |
Posted: Wed Jun 19, 2002 6:15 am Post subject: |
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Apprentice
Joined: 10 Mar 2002 Posts: 28
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That was exactly what i want to do, i didn't know that you can use '*' in the OutputRoot.
Thanks |
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