Subject: RE: [xsl] Returning the file name of the input file From: Cams Ismael <Ismael.Cams@xxxxxxxxxxxxxxx> Date: Wed, 28 Aug 2002 13:01:01 +0200 |
Thanks all for trying to help, but I think I havn't explained myself well enough. What I try to do is the following: I have an xml file which refers to other xml files that can refer to other xml files, etc. E.g. <package type="expr" href="../../package1.xml"/> <package href="../../package2.xml"/> <package type="expr" href="../../package3.xml"/> Out of this xml file I generate a list of all references of type 'expr'. So I search through all xml files and generate a list of all references of type 'expr'. The result file is used by a Java program to execute another process. The problem is because relative paths are used the Java program can't find the files (the Java program is started from another place). Therefore I would like to translate the relative paths into absolute paths. But I don't find any function in xsl to do this, nor do I see a solution to solve this with xsl. Hopefully somebody can help me. Kind regards, Ismaël -----Original Message----- From: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx] Sent: Wednesday, August 28, 2002 9:57 AM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: RE: [xsl] Returning the file name of the input file Hi, > Subject: [xsl] Returning the file name of the input file > > > Hello, > > does anybody if the following is possible: > > I want to use the name of the inputfile in the outputfile. So > if I want to > transform the file c:\temp\test.xml, I want to get in my output > c:\temp\test.xml. Pass the input filename as a parameter. Jarno XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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