Subject: Re: [xsl] Copying Namespace Nodes From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Sun, 27 Nov 2011 16:43:58 -0500 |
I have an element I want to copy and in the source it looks like this:
<a:elem xmlns:a="example.com/ns/a" xsi:type="nsc:foo" xmlns:nsc="example.com/ns/sensitive" xmlns:b="example.com/ns/b" xmlns:c="example.com/ns/c" ... xmlns:z="example.com/ns/z">info</a:elem>
What I want to achieve at the same time:
1) Keep the namespace declaration for the namespace sensitive content of the xsi:type attribute (with prefix nsc).
2) Get rid of all namespace declarations that are not needed here (prefix b, c, ..., z) and that clutter up my result document.
If I use copy-of with copy-namespaces set to 'no' the declaration for namespace sensitive content will be lost. If I say 'yes' everything stays the same. So is there a short way I have not seen yet or is the solution some verbose template? :)
<xsl:template match="*"> <xsl:copy copy-namespaces="no"> <xsl:for-each select="@xsi:type"> <xsl:copy-of select="../namespace::*[name(.)=substring-before(current(),':')]"/> <xsl:copy/> </xsl:for-each> </xsl:copy> </xsl:template>
</xsl:stylesheet> T:\ftemp>
Thanks for help, Heiko
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