I'm trying to do a simple copy and rename:

Assume the following XML:
<food name="joe">
    <pizza name="Large">
    <topping name="mushroom"/>
    </pizza>
</food>

I want to :
1) make a copy of the pizza node
2) change the name of copied version to "Big"
3) change the name (in the copied version) of "mushroom" to "onion"

so the output would be
<food name="joe">
    <pizza name="Large">
    <topping name="mushroom"/>
    </pizza>
    <pizza name="Big">
    <topping name="onion"/>
    </pizza>
</food>

I tried using a template match with a copy, but I got my #1 and #2 but I couldn't reference the newly created pizza node using XPath.

Any help would really be appreciated.
Thanks!

It would be useful to show your code so we can see where you went wrong.

Referencing a newly created node using XPath can't be done in XSLT 1.0 except with the xx:node-set() extension function = but it's not something you need to do to solve this problem

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Here's a portion of my XSL: Thanks for looking!


<xsl:template match="node() | @*">
    <xsl:copy>
        <xsl:apply-templates select="node() | @*"/>
    </xsl:copy>
  </xsl:template>


<xsl:template match="//food/pizza[@name='Large']">
    <xsl:copy>
        <xsl:apply-templates select="node() | @*"/>
    </xsl:copy>
    <xsl:copy>
        <xsl:attribute name="name">Big</xsl:attribute>
        <xsl:copy-of select="node()"/>
    </xsl:copy>
  </xsl:template>


<xsl:template match="//food/pizza[@name='Big']">
    <xsl:element name="topping">
         <xsl:attribute name="name">onion</xsl:attribute>
    </xsl:element>
  </xsl:template>

I looked at node-set() but I'm sure there's an easy way to do this. Suggestions?