I'm using version two and I got the correct output (each network address
once). Maybe something wrong with XMLSpy's parser? I'm using Saxon 8.1.
--Jim Neff
-----Original Message-----
From: Craig W [mailto:codecraig(_at_)gmail(_dot_)com]
Sent: Wednesday, March 02, 2005 9:12 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Sorting network addresses
well, i changed the xslt version from 2.0 to 1.0 and i get
the correct output....i wonder why? FYI, XMLSpy be default
uses its own XSLT engine (for either 1.0 or 2.0), or you can
specify that it uses Microsoft's XML parser, or you can have
it use some other XSL transformation program that you specify.
Anyway, thanks for the help....it would still be nice to know
why version 2.0 is acting differently.
-Craig
On Wed, 2 Mar 2005 14:05:20 GMT, David Carlisle
<davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:
....now do u get what i get?
no I get:
$ saxon ip2.xml ip2.xsl
<?xml version="1.0" encoding="UTF-8"?> <networks> <network>
<address>1.1.1.1</address> </network> <network>
<address>2.3.1.2</address> </network> <network>
<address>170.5.2.4</address> </network> </networks>
But I just noticed that you are specifying XSLT2 (although
you don't
have any xslt2 specific features) so I'll try with saxon8:
I get same:
$ java -jar ../saxon8/saxon8.jar ip2.xml ip2.xsl <?xml
version="1.0"
encoding="UTF-8"?> <networks> <network> <address>1.1.1.1</address>
</network> <network> <address>2.3.1.2</address> </network>
<network>
<address>170.5.2.4</address> </network> </networks>
To repeat how I did it,
open the XML document, go to the "XSL/XQuery"
sorry that would mean downloading xmlspy first which is too much:-)
What processor does it use for XSLT2 ?
David
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