If there's a limit on the nesting depth you can declare a series of sort
keys:

<xsl:sort select="(ancestor::*/@name)[1]"/> 
<xsl:sort select="(ancestor::*/@name)[2]"/>
<xsl:sort select="(ancestor::*/@name)[3]"/>
<xsl:sort select="(ancestor::*/@name)[4]"/>
<xsl:sort select="(ancestor::*/@name)[5]"/>

Otherwise you need a multi-phase approach: one phase to compute the sort
keys, the second phase to do the sort.

It's easy in XSLT 2.0, just do

<xsl:sort select="string-join(ancestor::*/@name, '.')"/>

Michael Kay

> -----Original Message-----
> From: Ian Lang [mailto:ianplang@xxxxxxxxx] 
> Sent: 09 April 2004 05:58
> To: XSLT List
> Subject:  Is it possible to sort the way I want
> 
> I am building a flat table from the contents of a
> hierarchical XML tree.  I want to sort on the fully
> qualified name of the element but the data does not
> have a qname attribute.  Is it possible?
> 
> For example the data might look like this (a
> representation of Java code perhaps):
> <package name="serviceb">
>   <package name="subpackageofb">
>   </pacakge>
>   <package name="anothersubpackagetofb">
>   </pacakge>
> </pacakge>
> <package name="servicea">
>   <package name="subpackageofa">
>     <package name="subsubpackageofa">
>     </pacakge>
>   </pacakge>
>   <package name="anothersubpacakgeofa">
>   </pacakge>
> </pacakge>
> <package name="servicec">
> </pacakge>
> 
> and I want the flat list to look like this:
> serviceA
> serviceA.anothersubpackageofa
> serviceA.subpackageofa
> serviceA.subpackageofa.subsubpackageofa
> serviceB
> serivceB.anothersubpackageofa
> serviceB.subpackageofa
> serivceC
> 
> when I process this using:
> <xsl:for-each select="//package">
>   <xsl:sort select="@name"/>
>     stuff to build the table...
> 
> I get this as my order
> serviceA
> serviceB
> serivceC
> serivceB.anothersubpackageofa
> serviceA.anothersubpackageofa
> serviceA.subpackageofa.subsubpackageofa
> serviceB.subpackageofa
> serviceA.subpackageofa
> 
> which is as expected since it is sorting on just the
> name of the package.
> 
> I also tried this:
> <xsl:for-each select="//package">
>   <xsl:sort select="ancestor-or-self::*/@name"/>
> but that seemed to give me document order or at least
> not what I want.
> 
> Can this be done directly?  Can I derive the fully
> qualified name using XPath for use in the select of
> the sort element?  I have a template that outputs the
> fully qualified name but it will not 'fit' in a select
> statement.  Or am I forced to insert the qualified
> name into the XML?
> 
> Any advice would be appricated.
> 
> Thanks,
> 
> IL
> 
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