Subject: RE: [xsl] [XSL] How can I seek info in XML files and tranform it using XSLT to answer a query From: "Paul Brown" <prb@xxxxxxxxxxxxx> Date: Wed, 11 Jul 2001 08:36:56 -0500 |
edouard panie: > Thanks for answering. I should give some more details: > -money is not the issue, I dont have any (it's a > school project!) Since you're a student, you're a very rich man indeed -- you have lots of time. > -the pages will only be accessible to a few people, > therefore there won't be more than a few hits a day. If you don't want the implementation necessarily to be fast, you can spoof an XSLT processor into thinking that your collection of files is one large document, and that will allow you to use XPath as your query language. Disclaimer 1: an XML database is going to do a MUCH, MUCH better job than the hack below. Here's how you do it. 1) Take a look at the old SQL extensions for XalanJ2 or the XSLT-based flatfile mapper that I wrote as an example for a Java/XSLT class. (http://www.fivesight.com/downloads/xsltflatfile.asp) 2) Implement a spoof DOM that mirrors the concatenation of your directory/file structure and your individual XML files. For example, this means that the implementation of Node that represents a directory has to return a list of the files it contains (as wrapping Nodes) when getChildNodes() is called. I would suggest implementing an equivalent<directory name="foo"><file name="bar.xml" >[... ...]</file><file name="baz.xml" >[... ...]</file></directory> Where the children of file are the children of the document element of the XML document contained in the file. 3) Do something really, really naughty and make use of how Xalan actually traverses XML documents on an xsl:apply-templates call. (You will NOT be able to use xsl:for-each!) This is present in the examples that I mention in (1) above. Cheers, - Paul XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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