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Servlet – Uploading File

Servlets are the Java programs that run on the Java-enabled web server or application server. They are used to handle the request obtained from the webserver, process the request, produce the response, then send a response back to the webserver.

Now let us learn how to upload a file to a server in this section. In an HTML file, the method must be posted and the enctype must be multipart/form-data when uploading a file to the server.

Creating a File Upload Form

  • The following HTML code below creates an uploader form. The following are some key factors to remember:
  • The form method should be set to POST, and the GET method should not be utilized.
  • The multipart/form-data enctype property should be used.
  • Set the form action property to a servlet file that will handle file uploading on the backend server. To upload a file, the following example uses the UploadServlet servlet.
  • To upload a single file, use a single <input…/> element with the type=”file” attribute. Include several input tags with distinct names for the name attribute to allow various files to be uploaded. Each of them has a Browse button associated with it in the browser.

Servlet - Creating a File Upload Form

Implementation:

Step 1: We will create a dynamic web project in Eclipse and the project structure will look like the below image.

Remember: There are several options for uploading a file to the server. However, we are going to utilize O’Reilly’s MultipartRequest class. We will need the cos.jar file to use this class.

A. File: index.html

HTML




<html
<body
<form action="GoGfg" method="post" enctype="multipart/form-data"
Select File:<input type="file" name="fname"/><br/> 
<input type="submit" value="upload"/> 
</form
</body
</html


B. File: GfgFileUpload.java

Make sure you have created directories C:\\temp (it is with reference to Windows operating systems)

Example:

Java




// Java Program to Illustrate File Uploading 
// Via Servlets
 
// Importing required classes
import java.io.*;
import javax.servlet.ServletException;
import javax.servlet.http.*;
import com.oreilly.servlet.MultipartRequest;
 
// Class
// Extending HttpServlet class
public class GfgFileUpload extends HttpServlet {
 
    // Method
    // To handle request response mechanism
    public void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {
 
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
 
        MultipartRequest m = new MultipartRequest(request, "C:\\temp");
 
        out.print("File uploaded successfully");
    }
}


 

 

C. File: web.xml

 

XML




<?xml version="1.0" encoding="UTF-8"?>
<web-app>
<servlet>
   <servlet-name>GfgFileUpload</servlet-name>
   <servlet-class>GfgFileUpload</servlet-class>
</servlet>
 
<servlet-mapping>
   <servlet-name>GfgFileUpload</servlet-name>
   <url-pattern>/GoGfg</url-pattern>
</servlet-mapping>
</web-app>


 

 

Output:

 

 

Step 2: After Clicking on the upload button file will be uploaded to the C:\\temp location

 

 

Dominic
Dominichttp://wardslaus.com
infosec,malicious & dos attacks generator, boot rom exploit philanthropist , wild hacker , game developer,
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