Question

I am wondering if is possible to create an XSLT stylesheet that would extract XPATHs for all leaf elements in a given XML file.
E.g. for

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <item1>value1</item1>
    <subitem>
        <item2>value2</item2>
    </subitem>
</root>

The output would be

/root/item1
/root/subitem/item2

Answer 1

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="text" indent="no" />

    <xsl:template match="*[not(*)]">
        <xsl:for-each select="ancestor-or-self::*">
            <xsl:value-of select="concat('/', name())"/>

            <xsl:if test="count(preceding-sibling::*[name() = name(current())]) != 0">
                <xsl:value-of select="concat('[', count(preceding-sibling::*[name() = name(current())]) + 1, ']')"/>
            </xsl:if>
        </xsl:for-each>
        <xsl:text>&#xA;</xsl:text>
        <xsl:apply-templates select="*"/>
    </xsl:template>

    <xsl:template match="*">
        <xsl:apply-templates select="*"/>
    </xsl:template>

</xsl:stylesheet>

outputs:

/root/item1
/root/subitem/item2