Thanks Chris, it works if I gives the absolute path of the file to
FileReader constructor, such as
C:\\jakarta-tomcat\\webapps\\ROOT\\WEB-INF\\classes\\normal.xml
Peter
>From: "Christopher K. St. John" <[EMAIL PROTECTED]>
>Reply-To: "A mailing list for discussion about Sun Microsystem's Java
> Servlet API Technology." <[EMAIL PROTECTED]>
>To: [EMAIL PROTECTED]
>Subject: Re: file path question in servlet
>Date: Mon, 8 Jan 2001 12:39:12 -0600
>
>Peter Smith wrote:
> >
> > The two input source files, normal.xml and normal.xsl are
> > located at the same directory as XSLServlet.class is.
> >
>
> <URL:http://java.sun.com/products/servlet/index.html>
>
> 9.4 Directory Structure
>
> A special directory exists within the
> application hierarchy named WEB-INF . This
> directory contains all things related to the
> application that aren't in the document root of
> the application. It is important to note that
> the WEB-INF node is not part of the public
> document tree of the application. No file
> contained in the WEB-INF directory may be served
> directly to a client.
>
>
> > FileReader xslReader = new FileReader("normal.xsl");
> > FileReader xmlReader = new FileReader("normal.xml");
>
> Try:
>
> File f = new File("normal.xsl");
> out.println(f.getAbsolutePath());
> out.println(this.getServletContext().getRealPath("normal.xml"));
>
> and you'll see what's happening. If you want
>normal.{xsl,xml} to be within the webapp, then you can
>try something like ServletContext::getResourceAsStream()
>(refer to the docs, path is relative to the context root)
>or, depending on the container settings, you can give the
>FileReader an absolute path (if you know the absolute
>path, that is). Easiest is probably to put it within the
>webapp, and use the ServletContext path routines to get a
>location you can give to FileReader, but that's not always
>possible...
>
>
>-cks
>
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