My code given below works fine. U can try if u want public void sendImageFile(HttpServletResponse response) { FileInputStream in=null; File f = null; try { f = new File("webapps/Plasma/graphs", "image.gif"); in = new FileInputStream(f); response.setContentType("image/gif"); int size = (int)f.length(); response.setContentLength(size); byte buffer[] = new byte[size]; in.read(buffer); response.getOutputStream().write(buffer); } catch(NullPointerException e) { System.out.println(" NP xpn in sendImageFile "+ e.getMessage()); } catch(FileNotFoundException e) { System.out.println(" FNF Xpn in SendImageFile "+ e.getMessage()); } catch(IOException ee2){ log("Trouble: " +ee2);} finally { try { response.getOutputStream().close(); } catch(IOException ee2){ log("Trouble: " +ee2);} } } On Sat, Sep 27, 2008 at 7:03 AM, Dave <[EMAIL PROTECTED]> wrote: > For <img src="http://domain.com/servlet/pictures/image.jpg"/> > > in servlet get method, > > InputStream is = new FileInputStream("/apphome/pictures/image.jpg"); > OutputStream os = response.getOutputStream(); > > byte[] buffer = new byte[256*1024]; //256k > while (true) { > int n = is.read(buffer); > if (n < 0) > return; > os.write(buffer, 0, n); > } > > is.close(); > os.close(); > > > Is this the right way? Sometimes only the half image is shown on web page. Is > there a more efficient and robust way? How about for audio/video files? > > I want to take at how Tomcat does it. Could anyone tell me which class? > thanks > Dave > > >
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