Subject: Passing an URL as a parameter with Xalan From: Marco.Mistroni@xxxxxxxxx Date: Mon, 31 Jul 2000 11:22:15 +0300 |
hi all, i have a problem: i have defined in my stylesheet an xsl:param now i want at run-time to pass an URL to that parameter i tried with file:///c:\mydir\layout.xml as value of hte parameter however, the XSLProcessor continues to give me the error XSL Error: pattern = 'c:\mydir\layout.xml' Extra illegal tokens: 'mydir', '\', 'layout.xml', style tree node: org.ap ache.xalan.xslt.StylesheetRoot@2a5ce2b7 XSL Error: SAX Exception There was a SAX Exception! org.apache.xalan.xslt.XSLProcessorException: pattern = 'c:\mydir\layout.x ml' Extra illegal tokens: 'marcodev', '\', 'layouthtml.xml' at org.apache.xalan.xslt.XSLTEngineImpl.error(XSLTEngineImpl.java:1630) at org.apache.xalan.xslt.XSLTEngineImpl.error(XSLTEngineImpl.java:1594) at org.apache.xalan.xslt.XSLTEngineImpl.process(XSLTEngineImpl.java, Com i am using the latest version of Xalan can anybody help me?? thanx in advance & regards marco XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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