Hey Roger,
Sorry, unfortunately it doesnt work that way.
When you get a resource for a page, it gives you the source of that page,
not the actual results of it.
>From servlet 2.3 specs:
"These methods are not used to obtain dynamic content. For example,
in a
container supporting the JavaServer PagesTM specification1, a method
call of the
form getResource("/index.jsp") would return the JSP source code and
not the
processed output. See Chapter SRV.8, "Dispatching Requests" for more
information about accessing dynamic content."
I have used in my code:
URL url = getServletContext().getResource("/flow.xml");
URLConnection conn = (URLConnection) url.openConnection();
InputStream in = conn.getInputStream();
This, however, reads the actual source of the resource.
To actually get the dynmaic content of it, you might consider not using
getResource and simply using the net package.
You might look at sample code for Applet/Servlet communication to find out
more.
http://java.sun.com/docs/books/tutorial/networking/urls/index.html
Also, depending on what you need to do, it might be sufficient to use the
include method of the requestDispatcher.
-Tim
-----Original Message-----
From: Varley, Roger [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 20, 2002 10:40 AM
To: [EMAIL PROTECTED]
Subject: Re: Your Message Sent on Tue, 19 Mar 2002 15:35:33 -0000
>
>
> You can only go back to the root of your project when using
> getResource
>
> For the servlet 2.3 specs:
>
> "The getResource and getResourceAsStream methods take
> a String with
> a
> leading "/" as argument which gives the path of the resource
> relative to the root of
> the context. This hierarchy of documents may exist in
> the server's
> file system, in a
> web application archive file, on a remote server, or
> at some other
> location."
>
> As you can see beginning with the / is a requirement.
>
> for mine, i use getServletContext().getResource("/flow.xml");
> This refers to the /flow.xml in my root directory.
> If your Register1.asp is not in the root directory or its
> subfolders then
> you need to move it.
Hi,
Thanks for your response. I've tried that and moved the register1.asp file.
The code now looks like;
ServletContext sc = config.getServletContext();
url = sc.getResource("/register1.asp");
System.out.println(url.toString());
I must be missing something obvious though because now the problem is that
getResource() is returning a file:// url not an Http url so when I try to
open the connection, the servlet crunches again. The system prints the
content of the URL as file:e:\tomcat\webapps\Application\register1.asp. I
need an HTTP connection.
Regards
Roger
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