Hi,

I'm trying to transform some XML that begins with the following:

<?xml version="1.0" encoding="UTF-8"?>
<release xmlns="[url of xsd]" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="[url of xsd]">

... this doesn't work. When I try transforming, I am left with an empty XML document, with only an empty root element.

If I change the tag to this:

<?xml version="1.0" encoding="UTF-8"?>
<release xmlns="[url of xsd]" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="[url of xsd]">

... then everything works fine.

My XSL looks like this:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.juno.co.uk/digital BulkImportSchema.xsd" xmlns:jd="[my domain]" xmlns:sc="[url of their XSD]">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="/">
    <xsl:element name="transformedRelease">
      <xsl:apply-templates select="release"></xsl:apply-templates>
    </xsl:element>
  </xsl:template>
  <xsl:template match="release">

     ..... etc ....


My question is, is the XML I've been given possible to transform, and if it is, how do I do it? Or do I need to request a different format?

cheers,
Alex

Sorry I messed up the second bit, it should be :
If I change the tag to this:

<?xml version="1.0" encoding="UTF-8"?>
<release xmlns:xsd="[url of xsd]" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="[url of xsd]">

Please follow the link labelled "default namespace" in the sticky post at the start of this forum. When you refer to an element in a namespace from an XPath expression, the name must be prefixed, even if in the source document it is your default namespace.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference