Question

I want to transform the XML using XSLT
Here is my sample XML

<EmployeList>
  <EMpDetails>
    <Name>Kiran</Name>
    <ID>ID001</ID>
    <city>Hyderabad</city>
    <Country>India</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Sunny</Name>
    <ID>ID002</ID>
    <city>Banglore</city>
    <Country>INDIA</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>John</Name>
    <ID>ID001</ID>
    <city>TEXAS</city>
    <Country>US</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Raj</Name>
    <ID>ID006</ID>
    <city>Dallas</city>
    <Country>US</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Nag</Name>
    <ID>ID007</ID>
    <city>ITALY</city>
    <Country>Rome</Country>
  </EMpDetails>
</EmployeList>

Required output using XSLT

<EmployeList>
  <EMpDetails>
    <Name>Kiran</Name>
    <ID>ID001</ID>
    <city>Hyderabad</city>
    <Country>India</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Sunny</Name>
    <ID>ID002</ID>
    <city>Banglore</city>
    <Country>INDIA</Country>
  </EMpDetails>   
</EmployeList>

Answer 1

That should work as you need (assuming that you want all EMpDetails elements where Country is equal to India with case-insensitivity):

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

    <xsl:template match="EmployeList">
        <EmployeList>
            <xsl:copy-of select="EMpDetails[lower-case(Country) = 'india']"/>
        </EmployeList>
    </xsl:template>
</xsl:stylesheet>