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
>
>
>



-- 
With Regards
T.K.Thiyagarajan

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