RE: [xsl] xsl sorting (Follow-up question)

[Greg Faron <gfaron at integretechpub dot com>]

At 12:24 AM 4/17/2002, you wrote:
> > If there is several sorting orders, they are all applied and not
> > only the last one. For example if there is xml file that includes
> > severals sets like:
> >
> > <Subject id="0000000001" id2="001" id3="001">Juu juu</Subject>
> > <Subject id="0000000001" id2="002" id3="002">RE:Juu juu</Subject>
> > <Subject id="0000000001" id2="002" id3="001">RE:Juu juu</Subject>
> >
> > and sets are sorted:
> > <xsl:sort select="Subject/@id" order="ascending" />
> > <xsl:sort select="Subject/@id2" order="ascending" />
> > <xsl:sort select="Subject/@id3" order="ascending" />
> >
> > So the result would be:
> > <Subject id="0000000001" id2="001" id3="001">Juu juu</Subject>
> > <Subject id="0000000001" id2="002" id3="001">RE:Juu juu</Subject>
> > <Subject id="0000000001" id2="002" id3="002">RE:Juu juu</Subject>
> >
> > Right?
>Right

   In the case where a sorting key (e.g. the attributes in the above 
example) is not present, is it possible to specify that absentees should be 
at the front or at the back of the list?  For example, I want to sort a 
group of items in ascending alphabetical order based on child element "name".
   Since the name is optional (there may be a series/name and series/number 
instead of ./name), can I request that any items without a ./name element 
be at the top of the list, followed by the alphabetical sorting of all 
items with a ./name element?  Conversely, how do I request that they be at 
the end of the list?  (I forget which is the default at the moment.)


Greg Faron
Integre Technical Publishing Co.



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