Subject: Re: [xsl] Replacing default namespace From: Martin Honnen <Martin.Honnen@xxxxxx> Date: Mon, 02 Jul 2007 18:42:07 +0200 |
In any case, a reduced version of the input WSDL is below. I want the output to be exactly the same except for the changes to the root tag:
<definitions targetNamespace="http://my.new.namespace/" xmlns:tns="http://my.new.namespace/"
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" xmlns:tns="http://my.new.namespace/" xmlns:mime="http://schemas.xmlsoap.org/wsdl/mime/" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:ro="urn:Databox" xmlns:n1="http://schemas.xmlsoap.org/wsdl/" xmlns="http://schemas.xmlsoap.org/wsdl/">
<xsl:template match="*"> <xsl:element name="{name()}" namespace="{namespace-uri()}"> <xsl:apply-templates select="@* | node()"/> </xsl:element> </xsl:template>
<xsl:template match="/wsdl:definitions"> <xsl:element name="{name()}" namespace="{namespace-uri()}"> <xsl:copy-of select="document('')/*/namespace::*"/> <xsl:apply-templates select="@* | node()"/> </xsl:element> </xsl:template>
<xsl:template match="@* | comment() | processing-instruction()"> <xsl:copy-of select="."/> </xsl:template>
Martin Honnen http://JavaScript.FAQTs.com/
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