XSLT with XML source that has a default namespace set to xmlns

Question

I have an XML document with a default namespace indicated at the root. Something like this:

<MyRoot xmlns="http://www.mysite.com"&gt;
   <MyChild1>
       <MyData>1234</MyData> 
   </MyChild1> 
</MyRoot>

The XSLT to parse the XML does not work as expected because of the default namespace, i.e. when I remove the namespace, everything works as expected.

Here is my XSLT:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   xmlns:xs="http://www.w3.org/2001/XMLSchema"&gt;
 <xsl:template match="/" >
  <soap:Envelope xsl:version="1.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"  xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"&gt;
   <soap:Body>
     <NewRoot xmlns="http://wherever.com"&gt;
       <NewChild>
         <ChildID>ABCD</ChildID>
         <ChildData>
            <xsl:value-of select="/MyRoot/MyChild1/MyData"/>
         </ChildData>
       </NewChild>
     </NewRoot>
   </soap:Body>
  </soap:Envelope>
 </xsl:template>
</xsl:stylesheet>

What needs to be done with XSLT document so that translation works properly? What exactly needs to be done in XSLT document?

Answer 1

You need to declare the namespace in your XSLT, and use it in XPath expressions. E.g.:

<xsl:stylesheet ... xmlns:my="http://www.mysite.com"&gt;

   <xsl:template match="/my:MyRoot"> ... </xsl:template>

</xsl:stylesheet>

Note that you must provide some prefix if you want to refer to elements from that namespace in XPath. While you can just do xmlns="..." without the prefix, and it will work for literal result elements, it won't work for XPath - in XPath, an unprefixed name is always considered to be in namespace with blank URI, regardless of any xmlns="..." in scope.