File upload using servlet returning NULL value as exception

Question

hi i am uploading images in to data base using servlets. when i submit my form i am getting a exception "NULL" here is my code any help are accepted.And can any one suggest me which way to be used to upload files in to database such that there wont be much of load on database and lessen httprequest. thank you in advance.

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1" import="requirementlogic.*,java.util.*"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<body>
<form  method="post" action="../InsertImage">
<table>
<tr><TD ><B>Upload Image</B></TD>
<td><input type="file" name="Image" size="20" value=""></TD>
</tr>
<tr>
<td><input type="submit" height="30" width="62"> </td>
</tr>
<tr>

</table>
</form>
</body>
</html>

here is my JDBC:

package requirementlogic;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.SQLException;

public class DataBase {
    Connection con = null;
    String url = "jdbc:mysql://localhost:3306/";
    String dbname= "deshopa";
    String driver ="com.mysql.jdbc.Driver";
    String Username="root";
    String Password ="";
    public Connection getConnection() throws InstantiationException, IllegalAccessException, ClassNotFoundException, SQLException{
        Class.forName(driver).newInstance();
        con = DriverManager.getConnection(url+dbname,Username,Password);
        return con;
    }
}

here is my servlet:

package requirementlogic;

import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.sql.PreparedStatement;

import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

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

    /**
     * @see Servlet#init(ServletConfig)
     */
    public void init(ServletConfig config) throws ServletException {
        // TODO Auto-generated method stub
        super.init(config);
        db = new DataBase();
    }

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

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html;charset=UTF-8");
          PrintWriter pw = response.getWriter();

        try
        {

        String strpath=request.getParameter("image");

        String filepath=strpath.substring(strpath.lastIndexOf(".")+1);
        pw.println(filepath);
        File imgfile = new File(strpath);
        FileInputStream fin = new FileInputStream(imgfile);
        String sql = "insert into upload_image(image,image_name,image_length)";
        PreparedStatement pre = db.getConnection().prepareStatement(sql);

        pre.setBinaryStream(1,fin,(int)imgfile.length());
        pre.setString(2,filepath);
        pre.setLong(3,imgfile.length());
        pre.executeUpdate();
        pre.close();
        pw.println("done");
        }
        catch(Exception e){
            pw.println("direct exception");
        pw.println("Exception is "+ e.getMessage());
        }
    }

}

Sql query used to create table:

CREATE TABLE `upload_image` (
`Image_id` int(11) NOT NULL AUTO_INCREMENT ,
`IMAGE` longblob NULL DEFAULT NULL ,
`Image_name` varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,
`image_length` varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL ,
PRIMARY KEY (`Image_id`)
)

By the way your form type is not multipart. I believe you will encounter problems with that, and then you shall handle a multipart request (which differs from a simple request). — Amir Pashazadeh
– Amir Pashazadeh, Commented Dec 25, 2013 at 6:28

Deepak · Accepted Answer · 2013-12-25 05:13:31Z

1

change to this

String strpath=request.getParameter("Image");

as in your html form the name of the file field is Image not image

<td><input type="file" name="Image" size="20" value=""></TD>

also you are dealing with multipart request so you have to chenge your code

if(ServletFileUpload.isMultipartContent(request)){
      private final String UPLOAD_DIRECTORY = "C:/uploads";
        try {
            List<FileItem> multiparts = new ServletFileUpload(
                                     new DiskFileItemFactory()).parseRequest(request);
            for(FileItem item : multiparts){
                if(!item.isFormField()){
                    String name = new File(item.getName()).getName();
                    item.write( new File(UPLOAD_DIRECTORY + File.separator + name));
               }

           }  
      } catch (Exception ex) {
         request.setAttribute("message", "File Upload Failed due to " + ex);
     }         
 }

See the below link for further details

http://javarevisited.blogspot.in/2013/07/ile-upload-example-in-servlet-and-jsp-java-web-tutorial-example.html

edited Dec 25, 2013 at 5:13

answered Dec 25, 2013 at 4:53

Deepak

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2 Comments

3bu1 Over a year ago

i was trying to upload "01_arrival.jpg" it says: Exception is 01_arrival.jpg (The system cannot find the file specified)

Deepak Over a year ago

@3bu1 if you will not use enctype="multipart/form-data" in your form then you will get only the file Name no the whole image so use this and parse as i suggested

Prateek · Accepted Answer · 2013-12-25 05:20:40Z

0

You are trying this:

String strpath=request.getParameter("image");

Try this:

String strpath=request.getParameter("Image");

Error is name which you are trying. Name in capital letter.

<input type="file" name="Image" size="20" value="">

In case

File file = new File("yourPath");
file.mkdirs();

You can try this Link1 and Link2

edited Dec 25, 2013 at 5:20

answered Dec 25, 2013 at 4:56

Prateek

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1 Comment

3bu1 Over a year ago

i was trying to upload "01_arrival.jpg" it says: Exception is 01_arrival.jpg (The system cannot find the file specified)

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