Question

In XSLT 2.0 I would like to get the unique nodes with 2 parameters..

For example I have:

    <items>
        <item x="1" y="1" z="a"/>
        <item x="1" y="2" z="b"/>
        <item x="2" y="1" z="c"/>
        <item x="1" y="2" z="d"/>
        <item x="2" y="2" z="e"/>
    </items>

I would like the result to be:

    <items>
        <item x="1" y="1"/>
        <item x="1" y="2"/>
        <item x="2" y="1"/>
        <item x="2" y="2"/>
    </items>

Is there an easier way to get this?
My current code is really redundant.

Answer 1

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="xml" indent="yes"/>

    <xsl:template match="/items">
        <xsl:copy>
            <xsl:for-each-group select="item" group-by="concat(@x,'#',@y)">
                <item x="{@x}" y="{@y}"/>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

Edit: The context is the first item of the group, so no need for current-group()[1]. Thanks Michael Kay.