Subject: [dssslist] Any way to tell a node's file name? From: tmcd@xxxxxxxxx Date: Thu, 3 Mar 2005 22:11:51 -0600 (CST) |
In my project, there's a main file named like 05-03.xml, which has the overall structure, like <?xml version="1.0" encoding="ISO-8859-1"?> <!DOCTYPE loar PUBLIC "-//SCA COA//DTD LoAR 1.1//EN" "loar.dtd"[ ... <!ENTITY ANSTEORRA SYSTEM "ANSTEORRA.xml"> <!ENTITY ARTEMISIA SYSTEM "ARTEMISIA.xml"> <!ENTITY ATLANTIA SYSTEM "ATLANTIA.xml"> ... ]> <loar file="05-03" ...> ... &ANSTEORRA; &ARTEMISIA; &ATLANTIA; ... </loar> <loar> is the root node of the whole thing. The file= attribute tells my code the file name of the input, because the HTML output code generates a file name (in this example, like "05-03.html"). I'd like to get rid of the file= attribute. Is there any way for my code to find out the name of the file in which the single <loar> node occurred? If it has something to do with pages 146-148 of the standard, then I'm lost in a twisty maze of little functions like entity-public-id, entity-system-id, entity-generated-system-id, et al. -- Tim McDaniel; Reply-To: tmcd@xxxxxxxxx
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