How to use a element variable in xpath of an xquery?

You can pass the node name: declare variable $doc_name as xs:string external; declare variable $userid as xs:string external; declare variable $update_node external; declare variable $new_value as xs:string external; let $users_doc := doc($doc_name)/users let $old_node := $users_doc/useruserid=$userid/*name()=$update_node return replace value of node $old_node with $new_value.

Alejandro: Thanks..yet again :) – sony Nov 14 '10 at 22:13 @sony: You are wellcome, ask any time. – user357812 Nov 14 '10 at 22:26 +1 for a correct answer. – Dimitre Novatchev Nov 15 '10 at 1:16.

I think you need to change the line let $old_node := $users_doc/useruserid=$userid/{$update_node} for let $old_node := $users_docuserid=$userid/{$update_node}.

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