Subject: [xsl] Getting a distinct list of node names From: "Paul Koppen" <wpkoppen@xxxxxxxxxxxxxx> Date: Wed, 6 Apr 2005 20:47:58 +0200 (CEST) |
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? Yours sincerely, Paul
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