How to perform intersection in nodeset using xpath 2.0 - xslt

Question

I have two different node-sets that contain one or more elements with the same name.
I want to select these same-named elements by using the intersect operation in XPath 2.0 and XPath 1.0 as well.
This is the sample code i have tried.
Input:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<child1>
<a />
<b />
<d />
</child1>
<child2>
<c />
<a />
<d />
</child2>
</root>
The code I tried
Xpath 1.0:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<!-- TODO: Auto-generated template -->
<xsl:variable name="ns2" select="/root/child2/child::*"/>
<xsl:copy-of select="/root/child1/child[.=$ns2]"/>
</xsl:template>
</xsl:stylesheet>
Xpath 2.0:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<!-- TODO: Auto-generated template -->
<xsl:variable name="ns1" select="/root/child1/child::*"/>
<xsl:variable name="ns2" select="/root/child2/child::*"/>
<xsl:copy-of select="$ns1 intersect $ns2"/>
</xsl:template>
</xsl:stylesheet>
Issue: I am getting the empty result.
Expected result:
<?xml version="1.0" encoding="UTF-8"?>
<a/>
<d/>
Please suggest what I am missing.
I have tried the set operation intersection for two different noteset.I have attached my code sample below,

Answer 1

You are not really doing "intersection" here, because all of the child elements of child1 are different to the child elements of child2. Just because two elements share the same name, that does not make them the same element.
What it looks like you are after is elements under child1 that have the same name as an element under child2
If you want your XSLT 1.0 solution to get results, you need to change the xsl:copy-of to this
<xsl:copy-of select="/root/child1/child::*[.=$ns2]"/>
Or this...
<xsl:copy-of select="/root/child1/*[.=$ns2]"/>
However, this will return all child1 child elements, because you are checking the value of the elements are the same, not the name. One way to do it is this...
<xsl:for-each select="/root/child1/*">
<xsl:variable name="name" select="local-name()" />
<xsl:copy-of select=".[$ns2[local-name() = $name]]" />
</xsl:for-each>
Alternatively, define a key like so:
<xsl:key name="child2" match="child2/*" use="local-name()" />
Then you could do this...
<xsl:copy-of select="/root/child1/*[key('child2', local-name())]"/>
In XSLT 2.0, you could do something like this....
<xsl:copy-of select="$ns1[some $name in $ns2/local-name() satisfies $name=local-name()]"/>

Answer 2

Not repeating Tim's correct remark, but just to the problem:
XSLT 1.0 (Not possible with a single XPath expression, unless an XSLT standard function, such as current() is used):
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="child1/*">
<xsl:copy-of select="self::*[../../child2/*[name() = name(current())]]"/>
</xsl:template>
</xsl:stylesheet>
XPath 2.0:
Use:
/*/child1/*[name() = /*/child2/*/name()]
This can be verified with the below XSLT 2.0 transformation which evaluates the expression and outputs the selected nodes:
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/">
<xsl:sequence select=
"/*/child1/*[name() = /*/child2/*/name()]"/>
</xsl:template>
</xsl:stylesheet>
When applied on the provided XML document:
<root>
<child1>
<a />
<b />
<d />
</child1>
<child2>
<c />
<a />
<d />
</child2>
</root>
both transformations produce the wanted, correct result:
<a/>
<d/>