xsl check if a root node contains any child nodes - xslt

Question

Is there a way to determine if a a root level node contains any child nodes?
I have this code file that builds up a navigation menu for a drop-down menu, but for the root nodes that have no nodes below them I want to apply a different template to them:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html"/>
<xsl:template match="/Home">
<xsl:apply-templates select="root" />
</xsl:template>
<xsl:template match="root">
<ul class="nav navbar-nav">
<xsl:apply-templates select="node">
<xsl:with-param name="level" select="0"/>
</xsl:apply-templates>
</ul>
</xsl:template>
<xsl:template match="node">
<xsl:param name="level" />
<xsl:choose>
<xsl:when test="$level=0">
<li>
<xsl:attribute name="class">
<xsl:if test="#breadcrumb = 1">active</xsl:if>
<xsl:if test="node">
<xsl:text> dropdown</xsl:text>
</xsl:if>
</xsl:attribute>
<xsl:choose>
<xsl:when test="#enabled = 1">
<a href="#" class="dropdown-toggle" data-toggle="dropdown">
<xsl:attribute name="class">
<xsl:if test="node">
<xsl:text>dropdown-toggle</xsl:text>
</xsl:if>
</xsl:attribute>
<xsl:if test="node">
<xsl:attribute name="data-toggle">dropdown</xsl:attribute>
</xsl:if>
<xsl:value-of select="#text" />
<xsl:if test="node">
<b class="caret"></b>
</xsl:if>
</a>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="#text" />
</xsl:otherwise>
</xsl:choose>
<xsl:if test="node">
<ul class="dropdown-menu">
<xsl:apply-templates select="node">
<xsl:with-param name="level" select="$level + 1" />
</xsl:apply-templates>
</ul>
</xsl:if>
</li>
</xsl:when>
<xsl:otherwise>
<li>
<xsl:attribute name="class">
<xsl:if test="#breadcrumb = 1">active</xsl:if>
<xsl:if test="node">
<xsl:text> dropdown</xsl:text>
</xsl:if>
</xsl:attribute>
<xsl:choose>
<xsl:when test="#enabled = 1">
<a href="{#url}">
<xsl:value-of select="#text" />
</a>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="#text" />
</xsl:otherwise>
</xsl:choose>
</li>
<xsl:if test="node">
<!-- no extra level in default bootstrap -->
</xsl:if>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>

Answer 1

Your question is not clear. If the root node does not have any child nodes, then the XML document is empty. Perhaps you meant the root element; there will be exactly one element like that and it's easy to see if it has any child nodes by using:
test="/*/node()"
in an xsl:if or xsl:when instruction.
Alternatively, you could use two templates - one matching a root element with child nodes:
<xsl:template match="/*[node()]">
and one for the other case:
<xsl:template match="/*[not(node())]">

Answer 2

You can use this XSLT-file:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" />
<xsl:template match="/*[count(descendant::*) = 0]">
No siblings
</xsl:template>
<xsl:template match="/*[count(descendant::*) > 0]">
Has siblings
</xsl:template>
</xsl:stylesheet>
With an input XML file with one or more siblings of the root element like this
<?xml version="1.0"?>
<root>
<a />
</root>
it will output "Has siblings".
And with an input file with an empty root tag like this
<?xml version="1.0"?>
<root>
</root>
it will output "No siblings".

Answer 3

If I understand the question correctly, try adding the template rule
<xsl:template match="root[not(*)]"/>