Re: xsl:element will not create an output element, in

Ok, thanks David and Abel for your latest emails; I'll have to wait a
few hours until I'm back home from work to test things in the browser
now.

I have, up until now, been viewing the source within the browser; I
realised that the rendered content isn't very indicative of the
elements output. When my PHP is doing the processing, the produced
XHTML is output to a file, so I can see everything output, for
definite. I didn't know that 'view source' would not show everything
when the browser is performing the transformation though, so thanks
for that tip Abel =)

One last question for David, before I test things when back at home.
What you've written below produces 'ul' and 'li' elements in no
namespace, according to what you said earlier. I'm assuming this means
they will NOT be processed as XHTML elements, and hence will not
produce the result I want (which is XHTML ul and li elements)?

> Any other suggestions?

I'm sorry I can only comment on what I can see, and all the examples you
have posted would generate non-html which would render in a browser just
as text. So that seems to be the most likely cause. If your actual
examples are not like the posted exampels the problem may be something
else, but I can't guess.

I can comment on the requested transformation though:

root -- <ul> ... </ul>

<xsl:template match="root">
 <ul>
   <xsl:apply-templates/>
 </ul>
</xsl:template>

dir -- <li> dirname <ul> ... </ul> </li>

<xsl:template match="dir">
 <li><xsl:value-od=f select="text()[1]"/>
   <ul><xsl:apply-templates/></il>
 </li>
</xsl:template>

file -- <li> filename </li>

<xsl:template match="file">
 <li>
   <xsl:value-of select="."/>
 </li>
</xsl:template>

David

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