hey list,
i'm trying to process xml data that has a default namespace:
<--- snip: test.xml --->
<?xml version="1.0" encoding="utf-8"?>
<foo xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns="http://www.patexpert.org/ont/metaont#">
<bar rdf:about="#id">argh</bar>
</foo>
<--- snip --->
processing this data with the stylesheet
<--- snip: test1.xsl --->
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:pat="http://www.patexpert.org/ont/metaont#"
exclude-result-prefixes="pat"
version="2.0">
<xsl:output method="xml" version="1.0" indent="yes" encoding="utf-8" />
<xsl:template match="pat:foo">
<xsl:element name="bar"
namespace="http://www.patexpert.org/ont/metaont#">
</xsl:element>
</xsl:template>
</xsl:stylesheet>
<--- test1.xsl --->
yields the (expected) result
<--- snip:output1 --->
<?xml version="1.0" encoding="utf-8"?>
<bar xmlns="http://www.patexpert.org/ont/metaont#"/>
<--- snip --->
i was trying to replace the xsl:element tags with a more direct
<--- snip:test2.xsl --->
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:pat="http://www.patexpert.org/ont/metaont#"
exclude-result-prefixes="pat"
version="2.0">
<xsl:output method="xml" version="1.0" indent="yes" encoding="utf-8" />
<xsl:template match="pat:foo">
<pat:bar>
</pat:bar>
</xsl:template>
</xsl:stylesheet>
<--- snip --->
but i am getting
<--- snip: output2 --->
<?xml version="1.0" encoding="utf-8"?>
<pat:bar xmlns:pat="http://www.patexpert.org/ont/metaont#"/>
<--- snip --->
i was expecting to get the same result as in output1. why is pat:bar
apparently not the same as the xsl:element instruction above?
thanks,
achim.
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