Retrieve XML using XSL transformation

Wysłane przez dnia 30-11-2011

Problem: you use WebService which returns result of XSL transformation of some XML data. You don’t know what is the structure of the XML so you want to retrieve it to be able to see it and create XSL for it.

Solution: to do that you need to create special XSL transformation which will return the original XML file.

This transformation looks as follows:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:msxsl="urn:schemas-microsoft-com:xslt"
  xmlns:insxsl="urn:insxsl">;
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:msxsl="urn:schemas-microsoft-com:xslt"
  xmlns:insxsl="urn:insxsl">;
  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>
SAP, , , ,

Komentowanie jest zakończone.