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

Question

Look at <bookmark/>. The desired output I want is the id with value 5

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

Answer 1

The way I understand the question and the provided XML document is that the nearest id attribute can also happen on an ancestor (section) element.

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

One correct XPath expression, which selects the wanted id attribute is:

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

If the id attribute is also allowed on the bookmark element itself, the above XPath expression needs to be modified slightly to accomodate this:

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

Do note: The ancestor:: and the preceding:: axes do not intersect (do not overlap).