XPath 1.0 closest preceding and/or ancestor node w

Question

Here are three XML-trees

(1)

<?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 id="5">
            <section>
                <bookmark/> <!-- here's the bookmark-->
                <section id="6">
                    <section id="7">
                        <section id="8"/>
                    </section>
               </section>
            </section>
        </section>
    </section>
</content>

(2)

<?xml version="1.0" encoding="UTF-8"?>
<content>
    <section id="1"/>
    <section id="2">
        <section id="9">
            <section id="10">
                <section id="11"/>
            </section>            
        </section>
        <section>
            <section id="4">
                <section id="5"/>
            </section>            
        </section>
        <section/>
        <bookmark/> <!-- here's the bookmark-->
        <section id="6">
            <section id="7">
                <section id="8"/>
            </section>
        </section>
    </section>
</content>

The desired result is in both cases the id 5.

With XSLT 1.0 and XPath 1.0 I can either get the ancestor from (1)

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

or the preceding node from (2)

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

How do I get the nearest ancestor or preceding node with an id from my bookmark?
I need a single xsl:value-of which matches both cases. Thanks.

EDIT:

The solution should also cover this structure. Desired id is still 5.

(3)

<?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/> <!-- here's the bookmark-->
                <section id="6">
                    <section id="7">
                        <section id="8"/>
                    </section>
                </section>
            </section>
        </section>
    </section>
</content>

Answer 1

Try with :

<xsl:value-of 
    select="//bookmark/ancestor::*[1]/descendant-or-self::*[last()-1]/@id"/>

It returns 5 for both XML documents.

EDIT:

In such conditions you could use simple xsl:choose:

<xsl:variable name="lastSibling"
    select="//bookmark/preceding-sibling::*[1]"/>
<xsl:choose>
    <xsl:when test="$lastSibling">
        <xsl:value-of
            select="$lastSibling/descendant-or-self::*[last()]/@id"/>
    </xsl:when>
    <xsl:otherwise>
        <xsl:value-of select="//bookmark/ancestor::*[@id][1]/@id"/>
    </xsl:otherwise>
</xsl:choose>

Another general solution:

<xsl:for-each
    select="//section[following::bookmark or descendant::bookmark][@id]">
    <xsl:if test="position() = last()">
        <xsl:value-of select="./@id"/>
    </xsl:if>
</xsl:for-each>

Answer 2

Use:

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

Verification: Using XSLT as host of XPath, the following transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/">
  <xsl:value-of select=
  "(//bookmark/ancestor::*[@id][1]/@id
  |
   //bookmark/preceding::*[@id][1]/@id
   )
    [last()]"/>
 </xsl:template>
</xsl:stylesheet>

when applied on any of the provided three XML documents, produces the wanted, correct result:

5

I strongly recomment using the XPath Visualizer for playing with / learning XPath.