Alan Hale wrote:
<snip />



     <xsl:variable name="link" select="file_link"> </xsl:variable>



<snip />



As you can see, I am trying to construct an html link from the filename
and path held in the <file_link> element.

This works fine when I call the stylesheet in-line or from Javascript in
Internet Explorer, but when I run it with the PHP XSL extension, it
generates a warning: Xsl:variable content should be empty since select
is present. It's only a warning and the output is still as expected, but
clearly I'm doing something wrong.

>From reading up it seems to me this IS an xslt issue rather than a PHP
one. I just don't understand it.




You read up well. The warning is wrong. However, there's a little truth 
in it ;)



You can use an xsl:variable with either a select or with more complex 
content:



<!-- with select -->
<xsl:variable name="x" select="file_link" />

<!-- with sequence constructor -->
<xsl:variable name="x"><xsl:value-of select="file_link" /></xsl:variable>

<!-- or more complex, like this: -->
<xsl:variable name="x">
   <xsl:for-each select="bla">
      <xsl:call-template name="xyz" />
      <xsl:apply-templates select="." />
   </xsl:for-each>
</xsl:variable>

<!-- or with text, like this (discouraged, use select to prevent 
document node to be created) -->

<xsl:variable name="x">Some text here</xsl:variable>



etc.



But you can't mix the two methods. You PHP XSL extension is a bit too 
strict. In XSLT, the following:



<xsl:variable name="x" select="file_link" >  </xsl:variable>



should be regarded equal to:



<xsl:variable name="x" select="file_link" />




That is a lot of text for helping you to understand the issue at hand. 
To help PHP XSL a bit, you can remove the whitespace, or just make it an 
empty tag altogether (like above). Apparently the whitespace is regarded 
as a value, but PHP XSL shouldn't, really.



HTH,
-- Abel Braaksma
  http://www.nuntia.nl