Xquery - using variable to find node with namespace prefix

Question

I am trying to find a node (prefixed with a namespace) using variables in my xquery like so:

declare variable $libx_ns as xs:string external;
declare namespace libx=$libx_ns;
declare variable $type as xs:string external;

return doc('xyz')/entry[libx:{$type}][1]

Parameter values: $libx_ns = 'http://libx.org/xml/libx2', $type = libapp. When I run this xquery, I see the following error:

[XPST0003] Incomplete expression.

The above code works if I modify the last line like so:

return doc('xyz')/entry[libx:libapp][1]

The purpose is to use a string variable(e.g. $type) for node-name rather than hard-wiring it (e.g. libapp). How would I do this?

Sample xml document:

<entry xmlns:libx="http://libx.org/xml/libx2">
  <libx:libapp> First libapp </libx:libapp>
  <libx:libapp> Second libapp </libx:libapp>
</entry>

Thanks, Sony

Answer 1

This XQuery:

declare variable $namespace-uri as xs:string external;
declare variable $local-name as xs:string external;

/entry[*[node-name(.) eq QName($namespace-uri,$local-name)]]

With this input:

<entry xmlns:libx="http://libx.org/xml/libx2">
    <libx:libapp> First libapp </libx:libapp>
    <libx:libapp> Second libapp </libx:libapp>
</entry>

Setting $namespace-uri as 'http://libx.org/xml/libx2' and $local-name as libapp, output:

<entry xmlns:libx="http://libx.org/xml/libx2">
    <libx:libapp> First libapp </libx:libapp>
    <libx:libapp> Second libapp </libx:libapp>
</entry>

Answer 2

If the namespace is not a critical difference, you could try:

return doc('xyz')/entry[*[local-name() = $type]][1]

If it is:

return doc('xyz')/entry[*[local-name() = $type and namespace-uri() = $libx_ns]][1]