A Servlet can be used with an HTML form tag to allow users to upload files to the server. An uploaded file could be a text file or image file or any document.

Creating a File Upload Form

The following HTM code below creates an uploader form. Following are the important points to be noted down −

The form method attribute should be set to POST method and GET method can not be used
The form enctype attribute should be set to multipart/form-data.
The form action attribute should be set to a servlet file which would handle file uploading at backend server. Following example is using UploadServlet servlet to upload file.
To upload a single file you should use a single <input …/> tag with attribute type=”file”. To allow multiple files uploading, include more than one input tags with different values for the name attribute. The browser associates a Browse button with each of them.

<html>

   <head>

      <title>File Uploading Form</title>

   </head>

  

   <body>

      <h3>File Upload:</h3>

      Select a file to upload: <br />

      <form action = "UploadServlet" method = "post" enctype = "multipart/form-data">

         <input type = "file" name = "file" size = "50" />

         <br />

         <input type = "submit" value = "Upload File" />

      </form>

   </body>

</html>

This will display following result which would allow to select a file from local PC and when user would click at “Upload File”, form would be submitted along with the selected fil −

File Upload:

Select a file to upload:

NOTE: This is just dummy form and would not work.

Writing Backend Servlet

Following is the servlet UploadServlet which would take care of accepting uploaded file and to store it in directory <Tomcat-installation-directory>/webapps/data. This directory name could also be added using an external configuration such as a context-param element in web.xml as follows −


<web-app>

   ....

   <context-param>

      <description>Location to store uploaded file</description>

      <param-name>file-upload</param-name>

      <param-value>

         c:\apache-tomcat-5.5.29\webapps\data\

     </param-value>

   </context-param>

   ....

</web-app>

Following is the source code for UploadServlet which can handle multiple file uploading at a time. Before proceeding you have make sure the followings −

Following example depends on FileUpload, so make sure you have the latest version of commons-fileupload.x.x.jar file in your classpath. You can download it from https://commons.apache.org/fileupload/.
FileUpload depends on Commons IO, so make sure you have the latest version of commons-io-x.x.jar file in your classpath. You can download it from https://commons.apache.org/io/.
While testing following example, you should upload a file which has less size than maxFileSize otherwise file would not be uploaded.
Make sure you have created directories c:\temp and c:\apache-tomcat8.0.28\webapps\data well in advance.

// Import required java libraries

import java.io.*;

import java.util.*;




import javax.servlet.ServletConfig;

import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;




import org.apache.commons.fileupload.FileItem;

import org.apache.commons.fileupload.FileUploadException;

import org.apache.commons.fileupload.disk.DiskFileItemFactory;

import org.apache.commons.fileupload.servlet.ServletFileUpload;

import org.apache.commons.io.output.*;




public class UploadServlet extends HttpServlet {

  

   private boolean isMultipart;

   private String filePath;

   private int maxFileSize = 50 * 1024;

   private int maxMemSize = 4 * 1024;

   private File file ;




   public void init( ){

      // Get the file location where it would be stored.

      filePath = getServletContext().getInitParameter("file-upload");

   }

  

   public void doPost(HttpServletRequest request, HttpServletResponse response)

      throws ServletException, java.io.IOException {

  

      // Check that we have a file upload request

      isMultipart = ServletFileUpload.isMultipartContent(request);

      response.setContentType("text/html");

      java.io.PrintWriter out = response.getWriter( );

  

      if( !isMultipart ) {

         out.println("<html>");

         out.println("<head>");

         out.println("<title>Servlet upload</title>"); 

         out.println("</head>");

         out.println("<body>");

         out.println("<p>No file uploaded</p>");

         out.println("</body>");

         out.println("</html>");

         return;

      }




      DiskFileItemFactory factory = new DiskFileItemFactory();

  

      // maximum size that will be stored in memory

      factory.setSizeThreshold(maxMemSize);

  

      // Location to save data that is larger than maxMemSize.

      factory.setRepository(new File("c:\\temp"));




      // Create a new file upload handler

      ServletFileUpload upload = new ServletFileUpload(factory);

  

      // maximum file size to be uploaded.

      upload.setSizeMax( maxFileSize );




      try {

         // Parse the request to get file items.

         List fileItems = upload.parseRequest(request);

              

         // Process the uploaded file items

         Iterator i = fileItems.iterator();




         out.println("<html>");

         out.println("<head>");

         out.println("<title>Servlet upload</title>"); 

         out.println("</head>");

         out.println("<body>");

  

         while ( i.hasNext () ) {

            FileItem fi = (FileItem)i.next();

            if ( !fi.isFormField () ) {

               // Get the uploaded file parameters

               String fieldName = fi.getFieldName();

               String fileName = fi.getName();

               String contentType = fi.getContentType();

               boolean isInMemory = fi.isInMemory();

               long sizeInBytes = fi.getSize();

           

               // Write the file

               if( fileName.lastIndexOf("\\") >= 0 ) {

                  file = new File( filePath + fileName.substring( fileName.lastIndexOf("\\"))) ;

               } else {

                  file = new File( filePath + fileName.substring(fileName.lastIndexOf("\\")+1)) ;

               }

               fi.write( file ) ;

               out.println("Uploaded Filename: " + fileName + "<br>");

            }

         }

         out.println("</body>");

         out.println("</html>");

         } catch(Exception ex) {

            System.out.println(ex);

         }

      }

     

      public void doGet(HttpServletRequest request, HttpServletResponse response)

         throws ServletException, java.io.IOException {




         throw new ServletException("GET method used with " +

            getClass( ).getName( )+": POST method required.");

      }

   }

}

Compile and Running Servlet

Compile above servlet UploadServlet and create required entry in web.xml file as follows.

<servlet>

   <servlet-name>UploadServlet</servlet-name>

   <servlet-class>UploadServlet</servlet-class>

</servlet>




<servlet-mapping>

   <servlet-name>UploadServlet</servlet-name>

   <url-pattern>/UploadServlet</url-pattern>

</servlet-mapping>

Now try to upload files using the HTML form which you created above. When you would try http://localhost:8080/UploadFile.htm, it would display following result which would help you uploading any file from your local machine.

File Upload:

Select a file to upload:

If your servlet script works fine, your file should be uploaded in c:\apache-tomcat8.0.28\webapps\data\ directory.

So, this brings us to the end of blog. This Tecklearn ‘File Uploading in Servlets’ blog helps you with commonly asked questions if you are looking out for a job in Java Programming. If you wish to learn Servlets and build a career Java Programming domain, then check out our interactive, Java and JEE Training, that comes with 24*7 support to guide you throughout your learning period. Please find the link for course details:

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