Subject: Re: [xsl] Getting a distinct list of node names From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> Date: Wed, 06 Apr 2005 14:51:35 -0400 |
Hi all,
Back in december 2003 G. Ken Holman posted an excellent solution on this topic. However I got a question to his solution.
His solution was the following: http://www.biglist.com/lists/xsl-list/archives/200312/msg00649.html
Now my question is, what is the generate-id for? When I write .=$childnodes[name(.)=name(current())][1] this works just as fine. Is it advisable not to use comparison of objects, but rather compare strings or so?
Cheers, Wendell
Yours sincerely, Paul
====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ======================================================================
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