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resource file location?

 
grant meng
Greenhorn
Posts: 4
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hi :
I have got a question when I try to practice how to use getOutputStream in the following example. The question is where shall I put the file bart.gif. I tried several location such as /WEB-INF/classes , docment-root. It always complains to me "file not found". Any thoughts on that. Thank you.

public void doGet(HttpServletRequest req,
HttpServletResponse res)
throws ServletException, IOException
{
res.setContentType("images/gif");
File f = new File("bart.gif");
byte[] bytearray = new byte[(int) f.length()];
FileInputStream is = new FileInputStream(f);
is.read(bytearray);
OutputStream os = res.getOutputStream();
os.write(bytearray);
os.flush();
}
 
Madan, Gopal
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Posts: 86
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Grant:
I tried this with a jpg (image/jpg - mime type).
The file was directly under the Web application DocumentRoot. I also got the same error. Only thing i changed is used a forward slash just before the image file name. So I guess in your case you would use
File f = new File ("/bart.gif")

HTH
 
Madan, Gopal
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Grant:
I need to recant my suggestions in the previous post.
It then worked because my file was in the root drive (d:\). I realized and tested out that when you create a File instance the argument must be a path to the file you want to create File instance upon. So in your case it should be as follows:
File f = new File ("d:\\whateverdir\\whateversubfolder\\bart.gif");
I am using W2K hence the above file path.
I think your best bet for portablity and all is to use the getRequestDispatcher() and/or getNamedDispatcher() methods of ServletContext object.
HTH
 
Bhushan Jawle
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You need to use ServletContext's getResource() or getResourceAsStream() to do your job.
You can get ServletContext as getServletContext() of GenericServlet.
Thanks,
Bhushan
 
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