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Emiliano pSubject: please help! simple XML transformation with xquery!
Author: Emiliano p
Date: 13 Apr 2009 06:26 PM
Hi All,

I'm a Xquery newbie. I don't know how to do this simple XML transformation:

FROM:

<GetCatalog>
<ListOfCatalog>
<Catalog>
<Id><param name="id">default</param></Id>
<Catalog>
<ListOfCatalog>
</GetCatalog>

TO:

<GetCatalog>
<ListOfCatalog>
<Catalog>
<Id>NewValue</Id>
<Catalog>
<ListOfCatalog>
</GetCatalog>

In other words, I have to search for param element with name attribute = "id" (something like that: //*[local-name()='param'][@name='id']) remove it and then:

1) or leave the same value ("default")
2) or update it with a new one ("NewValue")

Please, consider that all elements name could change. Only "param" will be a constant.

Thanks a lot in advance.
emiliano

Posttop
Minollo I.Subject: please help! simple XML transformation with xquery!
Author: Minollo I.
Date: 17 Apr 2009 05:08 PM
Something like this:

copy $input :=
<GetCatalog>
<ListOfCatalog>
<Catalog>
<Id><param name="id">default</param></Id>
</Catalog>
</ListOfCatalog>
</GetCatalog>

modify (
for $id in $input//Id
let $param := $id/param[@name="id"]
where $param
return
if (true()) then
replace value of node $id with "NewValue"
else
replace value of node $id with $param/string()
) return $input
   
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