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Answer 1

+2 A:

If you need both nodes, which are in different namespaces, use:

/root/item/*[local-name() = 'a']

However, this should seldomly happen, normally, you want a node from only one namespace:

<xsl:stylesheet version="1.0" 
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:example="http://www.w3.org/1999/xhtml"
  >
....
<xsl:for-each select="/root/item/example:a">

Wrikken 2010-10-04 20:28:28

this works - thanks you. the problem here is that not all of the links are marked with the name space.

Nick 2010-10-04 20:37:07

Setting the default namespace to the correct one in the stylesheet tag (`xmlns="http://www.w3.org/1999/xhtml"`) should work to make `/root/item/a` select all items if the other 'anonymous' nodes should be in that namespace too.

Wrikken 2010-10-04 20:39:31

@Wrikken: That last namespace declaration is usefull for output, not for select. XSLT 2.0 has a `xpath-default-namespace` attribute for that matter.

Alejandro 2010-10-04 20:42:45

@Alejandro: Damn, you're right, I was not thinking straight. Good advice to the OP if all nodes should be in the same namespace is to follow your advice then :)

Wrikken 2010-10-04 20:44:47

Answer 2

+1 A:

The a elements are in 2 different namespaces, the default namespace and the xhtml namespace. If you move the XPath outside of the xhtml formatting, you can use both namespaces to search:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;

  <xsl:template match="/">
    <xsl:variable name="links" xmlns:xhtml="http://www.w3.org/1999/xhtml"
                  select="//root/item/(a | xhtml:a)"/>

    <html xmlns="http://www.w3.org/1999/xhtml"&gt;
      <dl>
        <xsl:for-each select="$links">
          <dd><xsl:value-of select="."/></dd>
          <dt><xsl:value-of select="@href"/></dt>
        </xsl:for-each>
      </dl>
    </html>
  </xsl:template>

</xsl:stylesheet>

Harold L 2010-10-04 20:48:56

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