Thanks David.
The code you wrote for me, did exactly what I asked for. Unfortunately, I
didn't ask a specific enough question.
Not all documents in the source xml file contain links and I want the
output xml to only contain documents that have links. How can the template
below be changed to only output documents that contain links.
<xsl:template match="document">
<document>
<xsl:copy-of select="form|.//link"/>
</document>
</xsl:template>
David
=======================================
This is probably easy for someone who knows what they are doing, but I
haven't been able to figure this out.
I'm trying to transform XML to XML. The end result I want is something
like:
<document>
<form>forminfo</form>
<link>linkinfo</link>
</document>
The problem that I'm having is this: in the source xml file, the <form>
information is always at the same level in the source document, but the
<link> information can be any number of levels deep.
For example it could look like this:
<document>
<form>forminfo</form>
<link>linkinfo</link>
</document>
or it could look like this:
<document>
<form>forminfo</form>
<table>
<link>linkinfo</link>
</table>
</document>
or like this:
<document>
<form>forminfo</form>
<table>
<table>
<link>linkinfo</link>
</table>
</table>
</document>
I need to find a way to ignore all of the levels of table information and
just get the form and the link information in my output file. Any help is
appreciated.
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