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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![http://domain.com/servlet/pictures/image.jpg"/](http://domain.com/servlet/pictures/image.jpg"/){kind=link}