Subject: RE: [xsl] Getting the XPath of a node From: "Michael Kay" <michael.h.kay@xxxxxxxxxxxx> Date: Wed, 4 Sep 2002 12:27:58 +0100 |
<xsl:for-each select="ancestor-or-self::*"> <xsl:value-of select="concat( '/', name(.), '[', count(preceding-sibling::*[name(.)=name(current())])-1, ']')"/> </xsl:for-each> (Saxon has a saxon:path extension function) Michael Kay Software AG home: Michael.H.Kay@xxxxxxxxxxxx work: Michael.Kay@xxxxxxxxxxxxxx > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Dennis > Sent: 04 September 2002 10:03 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] Getting the XPath of a node > > > Hi All, > > Is there any way to get the XPath of a particular > element and attribute in match template??? > > Say if I have following XML: > <Person id="12345"> > <Name>Dennis</Name> > <Company>Netscape</Company> > <Address>Mountain View</Address> > <Email>dennis@xxxxxxxxxxxx</Email> > </Person> > > ----The XSL to print XPath--- > <xsl:template match="Company"> > //Print the XPath of Company as /Person/Company > </xsl:template> > More templates corresponding to each element. > > How do I do this...any thoughts??? > > Thanks > Dennis > > __________________________________________________ > Do You Yahoo!? > Yahoo! Finance - Get real-time stock quotes http://finance.yahoo.com > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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