Re: How to output the unused namespaces in the XML do

<?xml version="1.0"?>
<N1:NumberList>
        <Number>23</Number>
        <Number>41</Number>
        <Number xmlns:N3="http://www.example3.org">70</Number>
        <N3:Number xmlns:N3="http://www.example3.org">103</Number>
        <Number>99</Number>
        <Number>6</Number>
</N1:NumberList>

Michael Kay
Saxonica

Hi Folks,

Consider this XML document:

<?xml version="1.0"?>
<N1:NumberList xmlns:N1="http://www.example1.org"
                xmlns:N2="http://www.example2.org">
         <Number>23</Number>
         <Number>41</Number>
         <Number xmlns:N3="http://www.example3.org">70</Number>
         <Number>103</Number>
         <Number>99</Number>
         <Number>6</Number>
</N1:NumberList>

Notice that there are three (3) namespaces, but two of them are unused:

http://www.example2.org

http://www.example3.org

I want an XSLT transform that will output all the unused namespaces in the input XML document.

Here's the solution I came up with:

     <xsl:template match="*">
         <xsl:variable name="elem" select="." />
         <xsl:for-each select="namespace::*[. != 'http://www.w3.org/XML/1998/namespace']">
             <xsl:variable name="ns" select="." />
             <xsl:if test="not($elem/ancestor::*[namespace::* = $ns])">
                 <xsl:if test="not($elem/descendant-or-self::*[namespace-uri() = $ns])">
                     This namespace is unused:<xsl:value-of select="$ns" />
                 </xsl:if>
             </xsl:if>
         </xsl:for-each>
         <xsl:apply-templates select="*" />
     </xsl:template>

Is there a better solution? (I am using XSLT 1.0)

/Roger