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![]() | ![]() | ![]() | Altova Mailing List Archives>Archive Index >xsl-list Archive Home >Recent entries >Thread Prev - SV: [xsl] XML Question [Thread Next] RE: [xsl] XML QuestionTo: Date: 10/4/2006 11:23:00 AM The name() function takes a single node as its argument, not a list of nodes. Change name(ancestor::*) to ancestor::*/name() Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Per Osnes [mailto:per.osnes@xxxxxx] > Sent: 04 October 2006 12:18 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: SV: [xsl] XML Question > > Hi folks, > This template gives the required output (in Altova). > > <xsl:template match="//*[not(*)]"> > <xsl:variable name="path"> > <xsl:for-each select="ancestor::*"><xsl:value-of > select="name()"/>/</xsl:for-each> > <xsl:value-of select="name()"/> > </xsl:variable> > <xsl:text> > </xsl:text> > <xsl:value-of select="$path"/> > <xsl:value-of select="' ===> '"/> > <xsl:value-of select="text()"/> > </xsl:template> > > Actually I wanted to replace the "for each" by something like > <xsl:value-of select="string-join(((name(ancestor::*)), > name()), '/')"/> > > but Altova gives error message "too many items"... > I have used this method earlier for producing the full path > for elements in an XML schema: > select="string-join((ancestor::xs:element/@name, ./@name), '.')" > > Any comments from the experts? > > - Per Osnes. > > > ----- LINKE Markus <markus.linke@xxxxxxxx> wrote: > > Hi, > > > > I am not sure if this is the right list for such a > question, but lets > > try :) > > > > I would like to create a list of xpaths via xsl from a given > > xml-structure like this: > > > > <a> > > <b> > > <c> > > content > > </c> > > <d> > > morecontent > > </d> > > </b> > > </a> > > > > it should show the full xpath like > > > > a/b/c ====> content > > a/b/d ====> morecontent > > > > and not just > > > > c => content > > d => morecontent > > > > Ideas anyone? I am looking for either a script or some xsl ... > > > > Cheers, > > Markus | ![]() | ![]() | ![]() |
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