Removing duplicate attribute in XSLT - xslt-2.0

Question

I understand this is one such common question posted many times, but unfortunately I am unable to find an exact solution with any of the proposed approaches:
Here's how my XML looks like:
<root>
<parent>
<table>
<attr ......>
<type t="enumone">
<info>
<name .....>
</info>
</attr>
<attr>
<type t="int">
<range min="1" max="255"/>
</type>
<info>
<name .....>
</info>
</attr>
<attr>
<type t="string">
<info>
<name .....>
</info>
</attr>
<attr ......>
<type t="enumtwo">
<info>
<name .....>
</info>
</attr>
<attr>
<type t="float">
<range min="1.0" max="25.5"/>
</type>
<info>
<name .....>
</info>
</attr>
<attr>
<type t="int">
<info>
<name .....>
</info>
</attr>
<attr>
<type t="enumone">
<info>
<name .....>
</info>
</attr>
<attr>
<type t="enumthree">
<info>
<name .....>
</info>
</attr>
<attr>
<type t="enumone">
<info>
<name .....>
</info>
</attr>
</parent>
</root>
The intention is to retrieve one occurrence of attribute "#t" from "type" element by using XSLT:
Using for-each-group:
<xsl-template match="/root/parent">
<xsl:for-each select="table">
<xsl:for-each-group select="//type" group-by="#t">
<xsl:copy-of select="current-group( )[1]"/>
</xsl:for-each-group>
</xsl:for-each>
<xsl-template>
But I got no output! Something flawed I believe.
Using distinct-values():
<xsl:for-each select="distinct-values(type/#t)">
<xsl:sort/>
<xsl:value-of select="."/> <xsl:call-template name="newline"/>
</xsl:for-each>
Still no desired output.
The expected output is:
enumone
int
string
enumtwo
float
enumthree
Appreciate any help in this regard.

Answer 1

Well the input sample is not well-formed XML but if you really only want the distinct t attribute values of all type elements then doing
<xsl:template match="/">
<xsl:value-of select="distinct-values(//type/#t)" separator="
"/>
</xsl:template>
suffices with XSLT 2.0.
If you additionally want to sort the distinct values then doing
<xsl:template match="/">
<xsl:value-of separator="
">
<xsl:perform-sort select="distinct-values(//type/#t)">
<xsl:sort select="."/>
</xsl:perform-sort>
</xsl:value-of>
</xsl:template>
To give a complete example, with the input being
<root>
<foo>
<type t="int"/>
</foo>
<bar>
<type t="enum"/>
</bar>
<foobar>
<foo>
<type t="enum">
<x/>
</type>
</foo>
</foobar>
<foo>
<type t="string"/>
</foo>
</root>
the stylesheet
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:value-of separator="
">
<xsl:perform-sort select="distinct-values(//type/#t)">
<xsl:sort select="."/>
</xsl:perform-sort>
</xsl:value-of>
</xsl:template>
</xsl:stylesheet>
outputs
enum
int
string

Answer 2

Just modify your code to this:
<xsl:template match="/">
<xsl:for-each-group select="//type" group-by="#t">
<xsl:copy-of select="current-group( )[1]"/>
</xsl:for-each-group>
</xsl:template>
The complete transformation:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<xsl:for-each-group select="//type" group-by="#t">
<xsl:copy-of select="current-group( )[1]"/>
</xsl:for-each-group>
</xsl:template>
</xsl:stylesheet>
When applied on a well-formed XML document:
<root>
<foo>
<type t="int"/>
</foo>
<bar>
<type t="enum"/>
</bar>
<foobar>
<foo>
<type t="enum">
<x/>
</type>
</foo>
</foobar>
<foo>
<type t="string"/>
</foo>
</root>
the correct result is produced:
<type t="int"/>
<type t="enum"/>
<type t="string"/>