Get current node's local-name in XSL
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by green67
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Published on 2010-05-13T21:43:54Z
Indexed on
2010/05/13
22:04 UTC
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xsl
Here's the structure of my XML
<FileRoot>
<UserSet1>
<User>
<FirstName></FirstName>
<LastName></LastName>
</User>
<User>
<FirstName></FirstName>
<LastName></LastName>
</User>
...
</UserSet1>
<InactiveUsers>
<User>
<FirstName></FirstName>
<LastName></LastName>
</User>
<User>
<FirstName></FirstName>
<LastName></LastName>
</User>
...
</InactiveUsers>
</FileRoot>
In my XSL template
<xsl:template match="/*/*">
<File>
<xsl attribute name="Name">
<xsl:value-of select="local-name(/*/*)"/>
</xsl:attribute>
</File>
</xsl>
After transforming, for both UserSet1 and InactiveUsers, gave me "UserSet1". The expected results should be "UserSet1" for UserSet1, and "InactiveUsers" for InactiveUsers. How do I correctly retrieve the value?
Thanks
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