debugger wrote:


> I have an xml source of the following type:
> <tag attrib="val1">
>       <tag2 attribute="something" />
> </tag>
> <tag attrib="val2">
>       <tag2 />
> </tag>
> <tag attrib="val3">
>       <tag2 attribute="zzz" />
> </tag>
> <tag1 attrib="val4">
>       <tag2 attribute="abc" />
> </tag>

That is not well-formed, but assuming you have the well-formed source 
document

<?xml version="1.0" encoding="UTF-8"?>
<root>
<tag attrib="val1">
       <tag2 attribute="something" />
</tag>
<tag attrib="val2">
       <tag2 />
</tag>
<tag attrib="val3">
       <tag2 attribute="zzz" />
</tag>
<tag1 attrib="val4">
       <tag2 attribute="abc" />
</tag1>
</root>

> I want to use xsl transform of this xml source into:
> 
> <val1>something</val1>
> <val2></val2>
> <val3>zzz</val3>
> <val4>abc</val4>

then you get your result (plus a root element) with

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   version="1.0">

<xsl:output method="xml" encoding="UTF-8" indent="yes"/>

<xsl:template match="tag | tag1">
   <xsl:element name="{@attrib}"><xsl:value-of select="tag2/@attribute" 
/></xsl:element>
</xsl:template>

<xsl:template match="root">
   <xsl:copy>
     <xsl:copy-of select="@*" />
     <xsl:apply-templates select="*" />
   </xsl:copy>
</xsl:template>

</xsl:stylesheet>

-- 

	Martin Honnen --- MVP XML
	http://JavaScript.FAQTs.com/