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Re: super basic xsl question
Subject: Re: super basic xsl question
From: Jeb Boniakowski <jeb@xxxxxxxxxxx>
Date: Thu, 13 Jan 2005 12:06:53 -0500
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Thanks a lot, that's what I was looking for.
jeb..
On Jan 13, 2005, at 12:09 PM, JBryant@xxxxxxxxx wrote:
Don't have time for a long reply, but try xsl:copy-of
Jay Bryant
Bryant Communication Services
(on contract at Synergistic Solution Technologies)
Jeb Boniakowski <jeb@xxxxxxxxxxx>
01/13/2005 10:58 AM
Please respond to
xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To
xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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Subject
super basic xsl question
Thanks to everyone on the list that helped me with my last problem. I
still have some fundamental misunderstanding of how xsl is supposed to
work, because I can't figure out how to do this:
<xmlroot>
<child>some text</child>
<child><a href="">a link</child>
</xmlroot>
and convert it to say...
<ul>
<li>some text</li>
<li><a href="">a link</child>
</ul>
I've tried various combos like:
<xsl:template match="child">
<xsl:value-of select="."/>
</xsl:template>
but that means I lose the <a> tags.
I thought something with <xsl:copy> would work, but that gives me the
context node, and furthermore, it doesn't give me the children, so it
does the opposite. Then I thought fiddling with the 'select' expr in
the value-of tag would do it, but I can't figure out if there's a
magical combination of XPath slang that means, "whatever the hell is
below here, be it tags or text, i want them".
jeb.
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