Question

I am fairly new to XSL and need help with a transformation issue. I have an XML file that is described by an XSD. I use an XSL file to transform the XML into HTML. I want to reference the XSD in the XML file, but when I do the XML doesn’t get transformed.

Example XML:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="example.xsl"?>

<root>
<!--
<root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://localhost" xsi:schemaLocation="http://localhost example.xsd">
-->
  <element>Element 1</element>
  <element>Element 2</element>
  <element>Element 3</element>
</root>

Example XSL:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:template match="/">
    <ul>
      <xsl:for-each select="root/element">
        <li><xsl:value-of select="."/></li>
      </xsl:for-each>
    </ul>
  </xsl:template>
</xsl:stylesheet>

Example XSD:

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema 
  xmlns:xs="http://www.w3.org/2001/XMLSchema"
  targetNamespace="http://localhost"
  xmlns="http://localhost"
  elementFormDefault="qualified">
  <xs:element name="root">
    <xs:complexType>
      <xs:sequence>
        <xs:element name="element" type="xs:string" minOccurs="0" maxOccurs="unbounded"/>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
</xs:schema>

In the XML, if I use the commented out root tag, Firefox and Chrome do not transform the xml. If I just use the plain <root> tag, however, the transformation happens fine.

Can anyone explain why the XSL transformation doesn’t happen if I reference the XSD in my XML? Any help is appreciated!

Answer 1

<!-- <root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://localhost" xsi:schemaLocation="http://localhost example.xsd"> -->

This has nothing to do with using an XML Schema. The problem is that you specify a default namespace.

Using XPath expressions for node names in a default namespace is the biggest XPath FAQ.

Please, search the xpath and xslt tags for “default namespace” and you’ll find many good answers.

The solution for XSLT is to declare a namespace with some prefix (say “x”) and namespace-uri that is the same as the namespace-uri of the default namespace in the XML document. Then in any XPath expression use x:name instead of name.

Thus your XSLT code becomes:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:x="http://localhost" exclude-result-prefixes="x" >

    <xsl:template match="/">
        <ul>
            <xsl:for-each select="x:root/x:element">
                <li>
                    <xsl:value-of select="."/>
                </li>
            </xsl:for-each>
        </ul>
    </xsl:template>
</xsl:stylesheet>

and when applied on the provided XML document with uncommented <root> element:

<root xmlns="http://localhost"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://localhost example.xsd"> 
    <element>Element 1</element>
    <element>Element 2</element>
    <element>Element 3</element>
</root>

the wanted, correct result is produced:

<ul>
    <li>Element 1</li>
    <li>Element 2</li>
    <li>Element 3</li>
</ul>