Hi
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Jesse M. Heines
Sent: Wednesday, July 30, 2003 12:38 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] RE: Using a reference in a sort
Try <xsl:sort
select="document($filePeople)/people/person[(_at_)id=current()/pers
on/@id]/@l
ast"
order="descending"/> instead
Thanks for your reply, but I'm sorry to report that that
doesn't work either. The context of current() is still the
It should. I've tried on this input:
<tasks>
<task>
<person id="p001">1 ... </person>
</task>
<task>
<person id="p002">2 ... </person>
</task>
<task>
<person id="p003">3 ... </person>
</task>
</tasks>
With this 'people.xml':
<people>
<person first="Jesse" id="p001" last="Heines"> ... </person>
<person first="Americo" id="p002" last="Albuquerque"> ... </person>
<person first="Me" id="p003" last="Myself"> ... </person>
</people>
And got this result:
Myself, Me: 3 ...
Heines, Jesse: 1 ...
Albuquerque, Americo: 2 ...
node in the other XML file. The documentation I have says
that current() is the same as ".", which I tried and does not
help, either.
No, the current() is the same as "." outside of the select, in this case
will be the same used in the <xsl:apply-templates> So current() will be the
"task" node
Show us the template where you call this
Regards,
Américo Albuquerque
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