Subject: Re: [xsl] Xsl:variable content should be empty From: Ronan Klyne <ronan.klyne@xxxxxxxxxxx> Date: Wed, 04 Apr 2007 10:36:29 +0100 |
This line: <xsl:variable name="link" select="file_link"> </xsl:variable> specifies the value of the variable twice - as the result of evaluating 'file_link', and as " " (the contents of the variable element). You probably want to use: <xsl:variable name="link" select="file_link" /> and to read up on xsl:variable in the XSL specification. # r Alan Hale wrote: > I'm a relative newcomer to xslt and I wonder if someone could kindly > explain to me what is the issue with the following use of xsl:variable > and how I can correct it: > > <xsl:for-each select="/fieldguide/account"> > > <xsl:variable name="link" select="file_link"> </xsl:variable> > . > . > <td><a target = "_parent" href="{$link}"><xsl:value-of > select="species_name"/></a></td> > . > . > </xsl:for-each> > > Here is a fragment of the XML: > > <fieldguide> > <account> > <species_name>Octodiceras fontanum</species_name> > <file_link>../accounts/mosses/Octodiceras > fontanum_DMT.pdf</file_link> > . > . > </account> > </fieldguide> > > > 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. > > Glad of any help. > > Alan Hale > Aberystwyth > Wales > > > > > -- Ronan Klyne Business Collaborator Developer Tel: +44 (0)870 163 2555 ronan.klyne@xxxxxxxxxxx www.groupbc.com
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