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- Dave
On 5/31/07, David Carlisle <davidc@xxxxxxxxx> wrote:
Re: [xsl] xsl:element will not create an output element, in any context
Subject: Re: [xsl] xsl:element will not create an output element, in any context From: "N David Brown" <orieldave@xxxxxxxxxxxxxx> Date: Thu, 31 May 2007 14:21:07 +0100 |
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)?
- Dave
On 5/31/07, David Carlisle <davidc@xxxxxxxxx> wrote:
> 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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