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Re: [xsl] getting xml file info
Subject: Re: [xsl] getting xml file info From: Stephane Bortzmeyer <bortzmeyer@xxxxxx> Date: Wed, 30 Apr 2003 12:48:32 +0200 |
On Tue, Apr 29, 2003 at 11:16:09AM -0700, Nathan Shaw <n8_shaw@xxxxxxxxx> wrote a message of 34 lines which said: > A while back, I wrote an XSLT that uses the saxon > extension function system-id() to get the file path > info of the xml file that I am parsing. What does system-id() return when the XML input is not a file? For instance, the standard input on an Unix machine? Or a DOM tree parsed a long time ago? I believe that the difficulty to reply to this question is one of the reasons why system-id() is not standard. > how else could I get this information? If they do not > exist natively in the parsers, am I looking at having > to either write my own extension function to do this > or pass the info in from the XML file (which kind of > defeats the purpose)? The way I do it, which is standard (it does not rely on extensions): In the XSL stylesheet: <xsl:param name="current_dir">(UNSPECIFIED)</xsl:param> <xsl:param name="current_host">(UNSPECIFIED)</xsl:param> ... <xsl:text>Generated on </xsl:text><xsl:value-of select="$current_host"/> In the Makefile which calls the XSL processor: xsltproc -o $@ \ --stringparam current_dir "`pwd`" \ --stringparam current_host "`hostname -f`" \ ${STYLESHEET} $< If you use something else than xsltproc, you just have to change the Makefile, not the stylesheet. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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