Retrieve XML Variable in XSL file and say YES if found - html

Question

I have problems with the Xpath expression test="$roles/roles/role='HOBSCS1GB'" . Can anyone help in solving. Thanks
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:template match="/">
<xsl:variable name="roles">
<roles>
<role>HOBSCS1ROI</role>
<role>HOBSCS1GB</role>
<role>HOBSCS1FT</role>
</roles>
</xsl:variable>
<xsl:if test="$roles/roles/role='HOBSCS1GB'">
<xsl:value-of select="'YES'"/>
</xsl:if>
</xsl:template>
</xsl:stylesheet>

Answer 1

Assuming you want to find if a the roles element has one or more role elements with text = 'HOBSCS1GB': (Works in Saxon)
<xsl:if test="$roles/roles[role='HOBSCS1GB']">
<xsl:value-of select="'YES'"/>
</xsl:if>
Note that certain parsers like Microsoft may require you to tell the parser that $roles is a result tree fragment, by using node-set(), like so: (Works in msxsl)
<xsl:stylesheet ... xmlns:msxsl="urn:schemas-microsoft-com:xslt" ... />
<xsl:if test="msxsl:node-set($roles)/roles[role='HOBSCS1GB']">
<xsl:value-of select="'YES'"/>
</xsl:if>
Or in xsltproc:
<xsl:stylesheet ... xmlns:exsl="http://exslt.org/common" ... />
<xsl:if test="exsl:node-set($roles)/roles[role='HOBSCS1GB']">
<xsl:value-of select="'YES'"/>
</xsl:if>