Subject: Re: [xsl] getting associated file name for element From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Wed, 6 Mar 2002 22:44:11 +0000 |
Hi Matthew, > If I am doing something like: > > <xsl:for-each select="document($all-questions//quiz/@href)//question"> > <xsl:sort order="ascending" select="faa-num"/> > ...do stuff with each question... > <xsl:value-of select="faa-num"/> > Here's where I want to grab the name of the file, i.e. @href, the > question came from > </xsl:for-each> > > where quiz/@href contains the filenames containing questions to be > grabbed and then sorted by a question child element called faa-num. > How can I grab the value of @href as each question is evaluated? Since you need to sort the questions, you can't. I think that the best way to do what you're trying to do here is to create an intermediate node tree and use a node-set() extension function to get at the nodes. So something like: <xsl:variable name="questions-rtf"> <xsl:for-each select="$all-questions//quiz/@href"> <quiz href="{.}"> <xsl:copy-of select="document(.)//question" /> </quiz> </xsl:for-each> </xsl:variable> <xsl:for-each select="exsl:node-set($questions-rtf)/quiz/question"> <xsl:sort order="ascending" select="faa-num" /> ... do stuff with each question ... <xsl:value-of select="faa-num" /> ... <xsl:value-of select="../@href" /> </xsl:for-each> Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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