Subject: [xsl] Re: How can I get the XPATH of the current node ? From: Dimitre Novatchev <dnovatchev@xxxxxxxxx> Date: Thu, 10 Oct 2002 13:50:58 -0700 (PDT) |
--- Louis Meigret wrote: > First of all, thanks a lot to Jörg for pointing me to the Muenchian > method.Relentlessly learning. > > I have another question : I would like to get the Xpath value of the > current node. I am done some homework and tried this method suggested > by Jeni Tennison : > > « To get an XPath that's guaranteed to be unique, you need to use the > index of the element within its parent rather than its > attributevalues. > An easy way is to use recursion with the ancestor-or-self axis: > > <xsl:for-each select="ancestor-or-self::*"> > <xsl:text />/<xsl:value-of select="name()" /> > <xsl:text />[<xsl:number />]<xsl:text /> > </xsl:for-each> > > Cheers, > > Jeni » > > Unfortunately, this does not print any number between the square > brackets (with Xalan). I thought of adding value="position()", but > this > would be the position within this context (the for-each), not very > useful. > > Any suggestion to solve what would be useful debugging information > (in > my case) ? > > Thanks > > Louis Hi Louis, Do have a look at: http://www.topxml.com/code/default.asp?p=3&id=v20010323001030 or at: http://www.dpawson.co.uk/xsl/sect2/N6077.html#d177e20 This is a more complete solution, which will generate the XPath expression not only for an element but for any type of node -- text, PI, comment, root, attribute and namespace. Also, positional predicates are not used when they are not necessary, which results in a more compact and readable XPath expression ===== Cheers, Dimitre Novatchev. http://fxsl.sourceforge.net/ -- the home of FXSL __________________________________________________ Do you Yahoo!? Faith Hill - Exclusive Performances, Videos & More http://faith.yahoo.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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