Subject: [xsl] File URI / EXPath / Saxon From: Martynas Jusevicius <martynas.jusevicius@xxxxxxxxx> Date: Sat, 21 Nov 2009 01:23:14 +0100 |
Hey list, in my XSLT, I'm passing an absolute URI of the file which I want EXPath to create for me as a parameter. Using getRealPath() in Java gives backslashes on Windows, so I replace them with forward ones and this is the value I pass: file:///C:/Users/Martynas/WebRoot/odf2epub/build/web/epub/test.epub But then running the stylesheet gives errors: net.sf.saxon.trans.XPathException: Invalid URI: file:///C:\Users\Martynas\WebRoot\odf2epub\build\web\epub\test.epub at org.expath.saxon.Zip.fileFromURI(Zip.java:261) at org.expath.saxon.Zip.analyseHref(Zip.java:239) at org.expath.saxon.Zip.zipFile(Zip.java:305) And the file path is shown with backslashes again. Backslashes are not allowed in URIs as far as I know, but the parameter value is without them. Any ideas on what's going wrong? I'm running Saxon on Windows. Martynas
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