[xsl] changing only one attribute of element copy

[Garvin Riensche <g.riensche@xxxxxxx>]

Hello,

It would be nice if someone could tell me the easiest way of changing 
just one (and keeping the other) attributes of an copied element.

I am using the following templates to copy elements from a source class 
to a target class:

<xsl:template match="node() | @*" mode="transform">
  <xsl:copy>
    <xsl:apply-templates select="node() | @*" mode="transform"/>
  </xsl:copy>
</xsl:template>
	
<xsl:template match="//class[@name=$TargetCName]" mode="transform">
  <xsl:copy>
    <xsl:copy-of select="@*"/>
    <xsl:copy-of select="//class[@name=$SrcCName]/field">
    <xsl:apply-templates mode="transform"/>
  </xsl:copy>
</xsl:template>

Inside "class" are elements like
<field id="31" class="3" type="type(basic,int,0)" name="publ_int_j" 
init="null"/>
which are going to be copied.

What, if I want to change the "id" of the copied element? (Suppose, I 
don't know that the element name is "field") Do I have to create a 
completely new element using <xsl:element> and to write down every 
single attribute explicitely using <xsl:attribute>? Or is there a 
possibility to copy the old attributes und to create just one new attribute?
My problem is, that <xsl:copy-of> doesn't allow <xsl:attribute> inside 
and <xsl:copy> which allows it doesn't allow a "select" inside...

Regards,
Garvin