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From Chen Yang <cy...@hrsg.ca>
Subject RE: The resource path [...] is not valid
Date Fri, 02 Oct 2015 15:12:17 GMT
Hi Christ 


javax.servlet.ServletException: javax.xml.transform.TransformerException: org.xml.sax.SAXException:
The resource path [file:/C:/Program%20Files/Apache%20Software%20Foundation/Tomcat%208.0/]
is not valid
java.lang.IllegalArgumentException: The resource path [file:/C:/Program%20Files/Apache%20Software%20Foundation/Tomcat%208.0/]
is not valid
	org.apache.fop.servlet.FopServlet.doGet(FopServlet.java:162)
	javax.servlet.http.HttpServlet.service(HttpServlet.java:618)
	javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
	org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)


rootUrl is the site domain or IP. for example http://192.xxx.128.xx  

here is the doGet(), just in case you want to see

public void doGet(HttpServletRequest request,HttpServletResponse response) throws ServletException,
IOException {
        try {
            //Get parameters
            String foParam = request.getParameter(FO_REQUEST_PARAM);
            String xmlParam = request.getParameter(XML_REQUEST_PARAM);
            String xsltParam = request.getParameter(XSLT_REQUEST_PARAM);
			String rootParam = request.getParameter(ROOT_REQUEST_PARAM);

            //Analyze parameters and decide with method to use
            if (foParam != null) {
                renderFO(foParam, response);
            } else if ((xmlParam != null) && (xsltParam != null)) {
					renderXML(xmlParam,xsltParam,rootParam,response);
            } else {
                response.setContentType("text/html");
                PrintWriter out = response.getWriter();
                out.println("<html><head><title>Error</title></head>\n"
                          + "<body><h1>FopServlet Error</h1><h3>No
'fo' "
                          + "request param given.</body></html>");
            }
        } catch (Exception ex) {
            throw new ServletException(ex);
        }
    }

-----Original Message-----
From: Christopher Schultz [mailto:chris@christopherschultz.net] 
Sent: Friday, October 02, 2015 11:05 AM
To: Tomcat Users List <users@tomcat.apache.org>
Subject: Re: The resource path [...] is not valid

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Hash: SHA256

Chen,

On 10/2/15 10:42 AM, Chen Yang wrote:
> I'm transforming xml into pdf files, but getting this strange error in 
> my log. (tomcat 8.0.26)
> 
> The resource path
> [file:/C:/Program%20Files/Apache%20Software%20Foundation/Tomcat%208.0/
]
> is not valid
> 
> Here is my code
> 
> //Setup sources StreamSource xmlSrc = new StreamSource(new URL(rootUrl 
> + "/output/xml/" + xml).openStream()); StreamSource xsltSrc = new 
> StreamSource(new URL(rootUrl + "/output/xslt/" + xslt).openStream());

What's the value of "rootUrl"?

> //Setup the XSL transformation Transformer transformer =
> this.transFactory.newTransformer(xsltSrc);      //error out 
> transformer.setURIResolver(this.uriResolver);
> 
> XML & XSLT file are not stored locally, so I don't understand why it's 
> looking at the tomcat folder?

What's the stack trace?

- -chris
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