Subject: Re: [xsl] count nodes which are previous to a given node and have to meet xpath condition From: "Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Tue, 14 Jul 2015 09:19:19 -0000 |
> Having obtained the node <instrText> whose text contains value 6 > I want to count all previous instrText nodes whose text starts with > "HYPERLINK" > and are buried at any level within the same record/Kommentar/body/section > tree as the node with value 6 First letbs do XSLT 2.0. Your requirement translates directly into syntax: count( preceding::instrText [starts-with(., bHYPERLINKb)] [ancestor::sect is current()/ancestor::sect] ) The only problem is the bisb operator which requires 2.0. In XSLT 1.0 you can substitute generate-id(): count( preceding::instrText [starts-with(., bHYPERLINKb)] [generate-id(ancestor::sect) = generate-id(current()/ancestor::sect)] ) > > > So in my example the count result should be b4" I donbt see how you get 4. Which 4? From your description, I would expect to select those labelled 3, 4, and 5. Michael Kay Saxonica > I stumble when I try to count nodes which are previous to a given node but > have to meet the above xpath condition > Thanks in advance for any advice. > Regards Markus > > > > <?xml version="1.0" encoding="UTF-8"?> > <file> > <record> > <Kommentar> > <body> > <sect> > <p> > <r> > <instrText>HYPERLINK \l "anchor" \t "1"</instrText> > </r> > </p> > <p> > <r> > <instrText>HYPERLINK \l "anchor" \t "2"</instrText> > </r> > </p> > </sect> > </body> > </Kommentar> > </record> > > > <record> > <Kommentar> > <body> > <sect> > <p> > <r> > <instrText>HYPERLINK \l "anchor" \t "3"</instrText> > <instrText>HYPERLINK \l "anchor" \t "4</instrText> > </r> > </p> > <p> > <r> > <rPr> > <rFonts ascii="BLG New Garamond" h-ansi="BLG > New Garamond"/> > <sz val="32"/> > </rPr> > <instrText>HYPERLINK \l "anchor" \t "5"</instrText> > </r> > </p> > <p> > <r> > <instrText>HYPERLINK \l "anchor" \t "6"</instrText> > </r> > </p> > </sect> > </body> > </Kommentar> > </record> > </file>
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