ignore node using XSLT - node.js

Question

I have below xml file which is coming form my vendor.
<?xml version="1.0" encoding="UTF-8"?>
<nm:MT_employee xmlns:nm="http://firstscenario.com"xmlns:tl="http://secondscenario.com">
<EMployeeDetails>
<Name>Janardhan</Name>
<id>1234</id>
<Address>India</Address>
</EMployeeDetails>
<tl:Extension>
<tl:Number>5678</tl:Number>
<tl:Salary>2345678</tl:Salary>
</tl:Extension>
</nm:MT_employee>
In the above xml I want to ignore the entire tl:Extension node. the final output should be like below
<?xml version="1.0" encoding="UTF-8"?>
<nm:MT_employee xmlns:nm="http://firstscenario.com"xmlns:tl="http://secondscenario.com">
<EMployeeDetails>
<Name>Janardhan</Name>
<id>1234</id>
<Address>India</Address>
</EMployeeDetails>
</nm:MT_employee>
I tried to with different XSLT codes but it's not working. Could you please suggest how can I achieve this?
Regards,
Janardhan

Answer 1

The general rule to "ignore" an element from the source XML
is to write an "empty" template for this element, in your case:
<xsl:template match="tl:Extension"/>
As this template refers to tl namespace, it must be specified
in the xsl:transform tag.
Of course, to copy the rest of the source content, your script
must include the identity template.
Below you have an example script:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:tl="http://secondscenario.com" version="1.0">
<xsl:output method="xml" indent="yes" />
<xsl:template match="tl:Extension"/>
<xsl:template match="#*|node()">
<xsl:copy><xsl:apply-templates select="#*|node()"/></xsl:copy>
</xsl:template>
</xsl:transform>