Subject: Re: [xsl] Xsl:variable content should be empty From: Abel Braaksma <abel.online@xxxxxxxxx> Date: Wed, 04 Apr 2007 11:43:50 +0200 |
<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.
<!-- 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>
HTH, -- Abel Braaksma http://www.nuntia.nl
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