Servlet can not show the proper photo from database

Hi,
I have successfully uploaded the images to my database.
But , when I am trying to show the images , it is always showing the first image ( image from the first row ).

This is my values from the database table :

But, when I am trying to retrieve the image of UPLOADID=1002, I am still getting the image with UPLOADID=1000.

Here's my code in servlet :

package test_Shounak;

import java.io.IOException;
import java.io.InputStream;

import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.sql.*;

/**
 * Servlet implementation class PhotoShow
 */
public class PhotoShow extends HttpServlet {
	private static final long serialVersionUID = 1L;
       
    /**
     * @see HttpServlet#HttpServlet()
     */
    public PhotoShow() {
        super();
        // TODO Auto-generated constructor stub
    }

/**
	 * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
	 */
	protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		// TODO Auto-generated method stub
		
		
		Blob image = null;
	    Connection con = null;
	    Statement stmt = null;
	    ResultSet rs = null;
	    ServletOutputStream out = response.getOutputStream();
	    try {
	      Class.forName("com.ibm.db2.jcc.DB2Driver");
	      con = DriverManager.getConnection("jdbc:db2://localhost:50000/DB2_TEST","db2inst1","9231827765");
	      stmt = con.createStatement();
	      rs = stmt.executeQuery("select BINARYFILE from UPLOADS where  UPLOADID = '1002' ");
	      //out.flush();
	      String length="";
	      			if (rs.next()) 
	      			{
	      				length=rs.getString(1);
	      				System.out.println(length.length());
	      			} 
	      		else {
	      			response.setContentType("text/html");
	      			out.println("<html><head><title>Display Blob Example</title></head>");
	      			out.println("<body><h4><font color='red'>image not found for given id</font></h4></body></html>");
	      			return;
	      		}
	      			
	      rs = stmt.executeQuery("select BINARYFILE from UPLOADS where  UPLOADID = '1002' ");
	      		
	      		if(rs.next())
	      		{	  
	      
	      
	      					int len = (int) length.length();
	      					byte[] buffer = new byte[len];
	      					InputStream in = rs.getBinaryStream(1);
	      					int index = in.read(buffer, 0, len);
	      					stmt.close();
	      					response.reset();
	      					response.setContentType("image/jpg");
	      					response.getOutputStream().write(buffer, 0, len);
	      					response.getOutputStream().flush();
	      					out.println("\n Image length is :  "+len);
	      					
	      		}
	      		
	      		
	    } catch (Exception e) {
	      response.setContentType("text/html");
	      out.println("<html><head><title>Unable To Display image</title></head>");
	      out.println("<body><h4><font color='red'>Image Display Error=" + e.getMessage() +
	       "</font></h4></body></html>");
	      return;
	    } finally {
	      try {
	        //rs.close();
	        //stmt.close();
	        con.close();
	      } catch (SQLException e) {
	        e.printStackTrace();
	      }
	    }
	}

/**
	 * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
	 */
	protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		// TODO Auto-generated method stub
	}

}

Please help....

Most likely this is because the web browser you use caches the image locally and if the servlet you use uses the exact url you used to create the first image then the cached image is shown instead.

You can verify if this is the case for example by adding some random parameter at the end of the servlet url. So for example if this is you normal url
http://localhost:8080/webapp/ServletName
try this
http://localhost:8080/webapp/ServletName?random=8585

If this solves it then you can start searching a way to either prevent the browser caching the image or by adding some randomly changing parameter in the end of the url programmatically.

Hi, you are right .

I tried this URL : http://localhost:8080/new_test/PhotoShow?rand=100

Still it's showing the same photo.

I am using Google Chrome.

How can I overcome this problem, advise me please.....

How am I right?

If adding some random parameter at the end of the url did not change the image then I probably was wrong and the problem is somewhere else.

Sorry I could not help you. Maybe someone else has a better idea.

EDIT: Maybe describing your code/environment more could help solving your problem. For some reason you do the query to database twice in you code. Why? In your code you have put a static uploadid in your query but how are you going to change it in the real version (to fetch different images)? etc etc

Sorry, I misinterpreted your statement previously. Actually , I am now testing the database table, that's why I gave the static uploadid. In actual code, I would replace it with dynamic uploadid.

So, is there any way out ? Please help me.....

Not sure what the problem could be but just to eliminate browser caching could you switch the image in row 1 with the one from row 2. That way you will know if the issue on the browser end or the DB end.

~Nauman

Shounak Kundu wrote:Actually , I am now testing the database table, that's why I gave the static uploadid. In actual code, I would replace it with dynamic uploadid.

Don't spend a lot of time trying to test code which isn't going to be the actual code. And spend more time testing code which is going to be the actual code.