How to copy everything as is and only remove a specific element - xslt

Question

<?xml version="1.0" encoding="UTF-8"?>
<Emp:Employee xmlns:Emp="http://Emp.com">
<Emp:EmpName>XYZ</Emp:EmpName>
<Emp:EmpAddres>AAAA</Emp:EmpAddres>
<Det:EmpDetails xmlns:Det="http://Det.com">
<Det:EmpDesignation>SE</Det:EmpDesignation>
<Det:EmpExperience>4</Det:EmpExperience>
</Det:EmpDetails>
</Emp:Employee>
I am just trying to copy all the elements including the namespace but without <Det:EmpExperience>4</Det:EmpExperience>
so the final output should be :
<?xml version="1.0" encoding="UTF-8"?>
<Emp:Employee xmlns:Emp="http://Emp.com">
<Emp:EmpName>XYZ</Emp:EmpName>
<Emp:EmpAddres>AAAA</Emp:EmpAddres>
<Det:EmpDetails xmlns:Det="http://Det.com">
<Det:EmpDesignation>SE</Det:EmpDesignation>
</Det:EmpDetails>
</Emp:Employee>
I used
<xsl:template match='/'>
<xsl:copy-of select='#*[not(Det:EmpExperience)]'/>
</xsl:template>
its not working :-( ... any solution for this plz.
how to remove only <Det:EmpExperience> element and copy rest of the elements including namespace ?

Answer 1

Try this (adapted from here):
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:Det="http://Det.com">
<xsl:output omit-xml-declaration="yes"/>
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Det:EmpExperience"/>
</xsl:stylesheet>
The second template overrides the identity transformation and the empty template uses your matching logic (selecting Det:EmpExperience nodes).