xslt - How to get the nearest ancestor or child of an ancestor with xpath -


look @ <bookmark/>. desired output want id value 5

<xsl:value-of select="//bookmark/ancestor::*[@id][1]/@id"/> gives me 4

<?xml version="1.0" encoding="utf-8"?> <content>    <section id="1">         <section id="2"/>         <section id="3"/>         <section id="9"/>     </section>     <section id="4">         <section>             <section id="10"/>             <section id="5"/>             <section>                 <bookmark/>                 <section id="6">                     <section id="7">                         <section id="8"/>                     </section>                 </section>             </section>         </section>     </section> </content>

the way understand question , provided xml document nearest id attribute can happen on ancestor (section) element.

in such case expressions using ony preceding:: axis (as specified in other answers question) don't select node.

one correct xpath expression, selects wanted id attribute is:

    (//bookmark/ancestor::*[@id][1]/@id  |      //bookmark/preceding::*[@id][1]/@id      )      [last()]

if id attribute allowed on bookmark element itself, above xpath expression needs modified accomodate this:

    (//bookmark/ancestor-or-self::*[@id][1]/@id  |      //bookmark/preceding::*[@id][1]/@id      )      [last()]

do note: ancestor:: , preceding:: axes not intersect (do not overlap).


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