Hi Michael
Thanks very much for that! Worked a treat.
I'm relatively new to xsl and to be honest a lot of it seems like a
black art. Can you briefly describe what your solution does? I feel
daft for asking seeing as it's 13 lines long but I'd have never come
up with that.
Many thanks
Jim
On Wed, Jul 1, 2009 at 9:20 AM, Michael Ludwig<mlu(_at_)as-guides(_dot_)com>
wrote:
jim mcgovern schrieb:
What I need to do is return a set of nodes that contain a certain
child node. In the case below I need to return all the nodes that
contain a child node which have a name of "CONTENT".
And what I need to end up with is:-
<folder name="dir2" id="x2">
<folder name="dir5" id="x5">
<folder name="dir7" id="x7">
</folder>
</folder>
</folder>
Hi Jim,
the following seems to work. Should be rather self-explanatory; if not,
feel free to come back with any questions you might have.
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:template match="@*|node()"><!-- identity template -->
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="*" priority="-0.4" ><!-- skip unwanted -->
<xsl:apply-templates select="*"/>
</xsl:template>
<xsl:template match="*[*[(_at_)name='CONTENT']]" ><!-- keep wanted -->
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Michael Ludwig
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