[xsl] File URI / EXPath / Saxon

[Martynas Jusevicius <martynas.jusevicius@xxxxxxxxx>]

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