RE: multiple input files to one output file

[James Garriss <jgarriss at mitre dot org>]

LotusXSL 0.19.2 has a very elegant solution for passing parameters into an 
XSL stylesheet.  After instantiating the processor, you call 
processor.setStylesheetParam to replace a placeholder with some 
value.  Here at MITRE, we replaced a placeholder with a URI of an XML file, 
thus we were able to access the contents of it directly, just as if it was 
a variable.  This is done pretty much in the same manner as passing a 
parameter to a template.

Note that we are calling LotusXSL from a Java servlet.  Here's the function 
we used (note that much of this is code right out of the LotusXSL 
documentation):

***************

public static void transform(String xmlSourceURL, String xslURL, String 
outputURL, String userDirectoryURL, String userName)
  throws java.io.IOException, java.net.MalformedURLException, 
org.xml.sax.SAXException
{
   // Instantiate an XSLTProcessor.
   org.apache.xalan.xslt.XSLTProcessor processor =
   org.apache.xalan.xslt.XSLTProcessorFactory.getProcessor();

   // Set a param named "userdirectory", that the stylesheet can obtain.
   processor.setStylesheetParam("userdirectory", "'" + userDirectoryURL + "'");

   // Create the 3 objects the XSLTProcessor needs to perform the 
transformation.
   org.apache.xalan.xslt.XSLTInputSource xmlSource =
     new org.apache.xalan.xslt.XSLTInputSource (xmlSourceURL);
   org.apache.xalan.xslt.XSLTInputSource xslSheet =
     new org.apache.xalan.xslt.XSLTInputSource (xslURL);
   org.apache.xalan.xslt.XSLTResultTarget xmlResult =
     new org.apache.xalan.xslt.XSLTResultTarget (outputURL);

   // Perform the transformation.
   processor.process(xmlSource, xslSheet, xmlResult);
}

***************

Our stylesheet has this line in it....

   <xsl:param name="userdirectory" select="'default'"/>

and the placeholder "default" is what gets replaced.  Pretty nifty, eh?

--J

***************

At 02:54 AM 2/2/2000 , Linda van den Brink wrote:
 >I've used the document() function before, but this time I don't know the
 >paths and filenames of the documents beforehand. These are thousands of
 >files in hundreds of different directories.
 >
 >I do have XML files that specify the filenames and paths of all the
 >documents in a specific subdirectory. Could I somehow read such a
 >filenames-file and call the document function for each of the files listed,
 >and then extract the information I want from each document?
 >
 >Maybe my best option is to use a Perl script or something similar...
 >
 >-----Original Message-----
 >From: Steve Tinney [mailto:stinney@sas.upenn.edu]
 >Sent: Tuesday, February 01, 2000 3:58 PM
 >To: xsl-list@mulberrytech.com
 >Subject: Re: multiple input files to one output file
 >
 >
 >Linda van den Brink wrote:
 >> Is it possible with XSLT to have lots of input files, and create one
 >output
 >> file listing specific data from each of the files?
 >>
 >> What are my options to achieve this?
 >
 >XSLT spec under document().  You will need to manage the actual
 >integration of your list of documents into the XSL script, either
 >manually, by use of some kind of XML control file, or with a Java
 >extension function.
 >
 > Steve
 >
 >--
 >----------------------------------------------------------------------
 >Steve Tinney                                        Babylonian Section
 >                                 *   University of Pennsylvania Museum
 >stinney@sas.upenn.edu                          Phila, PA. 215-898-4047
 >
 >
 > XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
 >
 >
 > XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


   James Garriss | The MITRE Corporation | jgarriss @ mitre.org



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list