701
votes

How can I upload files to server using JSP/Servlet? I tried this:

<form action="upload" method="post">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

However, I only get the file name, not the file content. When I add enctype="multipart/form-data" to the <form>, then request.getParameter() returns null.

During research I stumbled upon Apache Common FileUpload. I tried this:

FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = upload.parseRequest(request); // This line is where it died.

Unfortunately, the servlet threw an exception without a clear message and cause. Here is the stacktrace:

SEVERE: Servlet.service() for servlet UploadServlet threw exception
javax.servlet.ServletException: Servlet execution threw an exception
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:313)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
    at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
    at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
    at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
    at java.lang.Thread.run(Thread.java:637)
14
Perhaps this article will be helpful: baeldung.com/upload-file-servletAdam Gerard
@Adam: They copied from my answer and added a sleuth of advertising on top of it in an attempt to earn money with it. Yeah, great article ..BalusC
No, actually nothing was copied. I wrote the first draft of that article along with the supplemental code. The core reference documentation can be found here: commons.apache.org/proper/commons-fileupload/using.html (and is linked to and cited in the article). Examples are partly reprised from the core reference document (which is the point of reference documentation - i.e. to be a point of reference) but not in their entirety (note that the reference docs don't go into much detail). Thanks!Adam Gerard

14 Answers

1231
votes

Introduction

To browse and select a file for upload you need a HTML <input type="file"> field in the form. As stated in the HTML specification you have to use the POST method and the enctype attribute of the form has to be set to "multipart/form-data".

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

After submitting such a form, the binary multipart form data is available in the request body in a different format than when the enctype isn't set.

Before Servlet 3.0, the Servlet API didn't natively support multipart/form-data. It supports only the default form enctype of application/x-www-form-urlencoded. The request.getParameter() and consorts would all return null when using multipart form data. This is where the well known Apache Commons FileUpload came into the picture.

Don't manually parse it!

You can in theory parse the request body yourself based on ServletRequest#getInputStream(). However, this is a precise and tedious work which requires precise knowledge of RFC2388. You shouldn't try to do this on your own or copypaste some homegrown library-less code found elsewhere on the Internet. Many online sources have failed hard in this, such as roseindia.net. See also uploading of pdf file. You should rather use a real library which is used (and implicitly tested!) by millions of users for years. Such a library has proven its robustness.

When you're already on Servlet 3.0 or newer, use native API

If you're using at least Servlet 3.0 (Tomcat 7, Jetty 9, JBoss AS 6, GlassFish 3, etc), then you can just use standard API provided HttpServletRequest#getPart() to collect the individual multipart form data items (most Servlet 3.0 implementations actually use Apache Commons FileUpload under the covers for this!). Also, normal form fields are available by getParameter() the usual way.

First annotate your servlet with @MultipartConfig in order to let it recognize and support multipart/form-data requests and thus get getPart() to work:

@WebServlet("/upload")
@MultipartConfig
public class UploadServlet extends HttpServlet {
    // ...
}

Then, implement its doPost() as follows:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String description = request.getParameter("description"); // Retrieves <input type="text" name="description">
    Part filePart = request.getPart("file"); // Retrieves <input type="file" name="file">
    String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
    InputStream fileContent = filePart.getInputStream();
    // ... (do your job here)
}

Note the Path#getFileName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

In case you have a <input type="file" name="file" multiple="true" /> for multi-file upload, collect them as below (unfortunately there is no such method as request.getParts("file")):

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    // ...
    List<Part> fileParts = request.getParts().stream().filter(part -> "file".equals(part.getName()) && part.getSize() > 0).collect(Collectors.toList()); // Retrieves <input type="file" name="file" multiple="true">

    for (Part filePart : fileParts) {
        String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
        InputStream fileContent = filePart.getInputStream();
        // ... (do your job here)
    }
}

When you're not on Servlet 3.1 yet, manually get submitted file name

Note that Part#getSubmittedFileName() was introduced in Servlet 3.1 (Tomcat 8, Jetty 9, WildFly 8, GlassFish 4, etc). If you're not on Servlet 3.1 yet, then you need an additional utility method to obtain the submitted file name.

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}
String fileName = getSubmittedFileName(filePart);

Note the MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

When you're not on Servlet 3.0 yet, use Apache Commons FileUpload

If you're not on Servlet 3.0 yet (isn't it about time to upgrade?), the common practice is to make use of Apache Commons FileUpload to parse the multpart form data requests. It has an excellent User Guide and FAQ (carefully go through both). There's also the O'Reilly ("cos") MultipartRequest, but it has some (minor) bugs and isn't actively maintained anymore for years. I wouldn't recommend using it. Apache Commons FileUpload is still actively maintained and currently very mature.

In order to use Apache Commons FileUpload, you need to have at least the following files in your webapp's /WEB-INF/lib:

Your initial attempt failed most likely because you forgot the commons IO.

Here's a kickoff example how the doPost() of your UploadServlet may look like when using Apache Commons FileUpload:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
        for (FileItem item : items) {
            if (item.isFormField()) {
                // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
                String fieldName = item.getFieldName();
                String fieldValue = item.getString();
                // ... (do your job here)
            } else {
                // Process form file field (input type="file").
                String fieldName = item.getFieldName();
                String fileName = FilenameUtils.getName(item.getName());
                InputStream fileContent = item.getInputStream();
                // ... (do your job here)
            }
        }
    } catch (FileUploadException e) {
        throw new ServletException("Cannot parse multipart request.", e);
    }

    // ...
}

It's very important that you don't call getParameter(), getParameterMap(), getParameterValues(), getInputStream(), getReader(), etc on the same request beforehand. Otherwise the servlet container will read and parse the request body and thus Apache Commons FileUpload will get an empty request body. See also a.o. ServletFileUpload#parseRequest(request) returns an empty list.

Note the FilenameUtils#getName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

Alternatively you can also wrap this all in a Filter which parses it all automagically and put the stuff back in the parametermap of the request so that you can continue using request.getParameter() the usual way and retrieve the uploaded file by request.getAttribute(). You can find an example in this blog article.

Workaround for GlassFish3 bug of getParameter() still returning null

Note that Glassfish versions older than 3.1.2 had a bug wherein the getParameter() still returns null. If you are targeting such a container and can't upgrade it, then you need to extract the value from getPart() with help of this utility method:

private static String getValue(Part part) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(part.getInputStream(), "UTF-8"));
    StringBuilder value = new StringBuilder();
    char[] buffer = new char[1024];
    for (int length = 0; (length = reader.read(buffer)) > 0;) {
        value.append(buffer, 0, length);
    }
    return value.toString();
}
String description = getValue(request.getPart("description")); // Retrieves <input type="text" name="description">

Saving uploaded file (don't use getRealPath() nor part.write()!)

Head to the following answers for detail on properly saving the obtained InputStream (the fileContent variable as shown in the above code snippets) to disk or database:

Serving uploaded file

Head to the following answers for detail on properly serving the saved file from disk or database back to the client:

Ajaxifying the form

Head to the following answers how to upload using Ajax (and jQuery). Do note that the servlet code to collect the form data does not need to be changed for this! Only the way how you respond may be changed, but this is rather trivial (i.e. instead of forwarding to JSP, just print some JSON or XML or even plain text depending on whatever the script responsible for the Ajax call is expecting).


Hope this all helps :)

26
votes

If you happen to use Spring MVC, this is how to: (I'm leaving this here in case someone find it useful).

Use a form with enctype attribute set to "multipart/form-data" (Same as BalusC's Answer)

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file" />
    <input type="submit" value="Upload"/>
</form>

In your controller, map the request parameter file to MultipartFile type as follows:

@RequestMapping(value = "/upload", method = RequestMethod.POST)
public void handleUpload(@RequestParam("file") MultipartFile file) throws IOException {
    if (!file.isEmpty()) {
            byte[] bytes = file.getBytes(); // alternatively, file.getInputStream();
            // application logic
    }
}

You can get the filename and size using MultipartFile's getOriginalFilename() and getSize().

I've tested this with Spring version 4.1.1.RELEASE.

13
votes

Without component or external Library in Tomcat 6 o 7

Enabling Upload in the web.xml file:

http://joseluisbz.wordpress.com/2014/01/17/manually-installing-php-tomcat-and-httpd-lounge/#Enabling%20File%20Uploads.

<servlet>
    <servlet-name>jsp</servlet-name>
    <servlet-class>org.apache.jasper.servlet.JspServlet</servlet-class>
    <multipart-config>
      <max-file-size>3145728</max-file-size>
      <max-request-size>5242880</max-request-size>
    </multipart-config>
    <init-param>
        <param-name>fork</param-name>
        <param-value>false</param-value>
    </init-param>
    <init-param>
        <param-name>xpoweredBy</param-name>
        <param-value>false</param-value>
    </init-param>
    <load-on-startup>3</load-on-startup>
</servlet>

AS YOU CAN SEE:

    <multipart-config>
      <max-file-size>3145728</max-file-size>
      <max-request-size>5242880</max-request-size>
    </multipart-config>

Uploading Files using JSP. Files:

In the html file

<form method="post" enctype="multipart/form-data" name="Form" >

  <input type="file" name="fFoto" id="fFoto" value="" /></td>
  <input type="file" name="fResumen" id="fResumen" value=""/>

In the JSP File or Servlet

    InputStream isFoto = request.getPart("fFoto").getInputStream();
    InputStream isResu = request.getPart("fResumen").getInputStream();
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    byte buf[] = new byte[8192];
    int qt = 0;
    while ((qt = isResu.read(buf)) != -1) {
      baos.write(buf, 0, qt);
    }
    String sResumen = baos.toString();

Edit your code to servlet requirements, like max-file-size, max-request-size and other options that you can to set...

11
votes

You need the common-io.1.4.jar file to be included in your lib directory, or if you're working in any editor, like NetBeans, then you need to go to project properties and just add the JAR file and you will be done.

To get the common.io.jar file just google it or just go to the Apache Tomcat website where you get the option for a free download of this file. But remember one thing: download the binary ZIP file if you're a Windows user.

9
votes

I am Using common Servlet for every Html Form whether it has attachments or not. This Servlet returns a TreeMap where the keys are jsp name Parameters and values are User Inputs and saves all attachments in fixed directory and later you rename the directory of your choice.Here Connections is our custom interface having connection object. I think this will help you

public class ServletCommonfunctions extends HttpServlet implements
        Connections {

    private static final long serialVersionUID = 1L;

    public ServletCommonfunctions() {}

    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException,
            IOException {}

    public SortedMap<String, String> savefilesindirectory(
            HttpServletRequest request, HttpServletResponse response)
            throws IOException {
        // Map<String, String> key_values = Collections.synchronizedMap( new
        // TreeMap<String, String>());
        SortedMap<String, String> key_values = new TreeMap<String, String>();
        String dist = null, fact = null;
        PrintWriter out = response.getWriter();
        File file;
        String filePath = "E:\\FSPATH1\\2KL06CS048\\";
        System.out.println("Directory Created   ????????????"
            + new File(filePath).mkdir());
        int maxFileSize = 5000 * 1024;
        int maxMemSize = 5000 * 1024;
        // Verify the content type
        String contentType = request.getContentType();
        if ((contentType.indexOf("multipart/form-data") >= 0)) {
            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(filePath));
            // 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.
                @SuppressWarnings("unchecked")
                List<FileItem> fileItems = upload.parseRequest(request);
                // Process the uploaded file items
                Iterator<FileItem> i = fileItems.iterator();
                while (i.hasNext()) {
                    FileItem fi = (FileItem) i.next();
                    if (!fi.isFormField()) {
                        // Get the uploaded file parameters
                        String fileName = fi.getName();
                        // 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);
                    } else {
                        key_values.put(fi.getFieldName(), fi.getString());
                    }
                }
            } catch (Exception ex) {
                System.out.println(ex);
            }
        }
        return key_values;
    }
}
8
votes

For Spring MVC I have been trying for hours to do this and managed to have a simpler version that worked for taking form input both data and image.

<form action="/handleform" method="post" enctype="multipart/form-data">
  <input type="text" name="name" />
  <input type="text" name="age" />
  <input type="file" name="file" />
  <input type="submit" />
</form>

Controller to handle

@Controller
public class FormController {
    @RequestMapping(value="/handleform",method= RequestMethod.POST)
    ModelAndView register(@RequestParam String name, @RequestParam int age, @RequestParam MultipartFile file)
            throws ServletException, IOException {

        System.out.println(name);
        System.out.println(age);
        if(!file.isEmpty()){
            byte[] bytes = file.getBytes();
            String filename = file.getOriginalFilename();
            BufferedOutputStream stream =new BufferedOutputStream(new FileOutputStream(new File("D:/" + filename)));
            stream.write(bytes);
            stream.flush();
            stream.close();
        }
        return new ModelAndView("index");
    }
}

Hope it helps :)

6
votes

Another source of this problem occurs if you are using Geronimo with its embedded Tomcat. In this case, after many iterations of testing commons-io and commons-fileupload, the problem arises from a parent classloader handling the commons-xxx jars. This has to be prevented. The crash always occurred at:

fileItems = uploader.parseRequest(request);

Note that the List type of fileItems has changed with the current version of commons-fileupload to be specifically List<FileItem> as opposed to prior versions where it was generic List.

I added the source code for commons-fileupload and commons-io into my Eclipse project to trace the actual error and finally got some insight. First, the exception thrown is of type Throwable not the stated FileIOException nor even Exception (these will not be trapped). Second, the error message is obfuscatory in that it stated class not found because axis2 could not find commons-io. Axis2 is not used in my project at all but exists as a folder in the Geronimo repository subdirectory as part of standard installation.

Finally, I found 1 place that posed a working solution which successfully solved my problem. You must hide the jars from parent loader in the deployment plan. This was put into geronimo-web.xml with my full file shown below.

Pasted from <http://osdir.com/ml/user-geronimo-apache/2011-03/msg00026.html> 



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<web:web-app xmlns:app="http://geronimo.apache.org/xml/ns/j2ee/application-2.0" xmlns:client="http://geronimo.apache.org/xml/ns/j2ee/application-client-2.0" xmlns:conn="http://geronimo.apache.org/xml/ns/j2ee/connector-1.2" xmlns:dep="http://geronimo.apache.org/xml/ns/deployment-1.2" xmlns:ejb="http://openejb.apache.org/xml/ns/openejb-jar-2.2" xmlns:log="http://geronimo.apache.org/xml/ns/loginconfig-2.0" xmlns:name="http://geronimo.apache.org/xml/ns/naming-1.2" xmlns:pers="http://java.sun.com/xml/ns/persistence" xmlns:pkgen="http://openejb.apache.org/xml/ns/pkgen-2.1" xmlns:sec="http://geronimo.apache.org/xml/ns/security-2.0" xmlns:web="http://geronimo.apache.org/xml/ns/j2ee/web-2.0.1">
    <dep:environment>
        <dep:moduleId>
            <dep:groupId>DataStar</dep:groupId>
            <dep:artifactId>DataStar</dep:artifactId>
            <dep:version>1.0</dep:version>
            <dep:type>car</dep:type>
        </dep:moduleId>

<!--Don't load commons-io or fileupload from parent classloaders-->
        <dep:hidden-classes>
            <dep:filter>org.apache.commons.io</dep:filter>
            <dep:filter>org.apache.commons.fileupload</dep:filter>
        </dep:hidden-classes>
        <dep:inverse-classloading/>        


    </dep:environment>
    <web:context-root>/DataStar</web:context-root>
</web:web-app>
0
votes

Here's an example using apache commons-fileupload:

// apache commons-fileupload to handle file upload
DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setRepository(new File(DataSources.TORRENTS_DIR()));
ServletFileUpload fileUpload = new ServletFileUpload(factory);

List<FileItem> items = fileUpload.parseRequest(req.raw());
FileItem item = items.stream()
  .filter(e ->
  "the_upload_name".equals(e.getFieldName()))
  .findFirst().get();
String fileName = item.getName();

item.write(new File(dir, fileName));
log.info(fileName);
0
votes

You first have to set the enctype attribute of the form to "multipart/form-data"

This is shown below.

<form action="Controller" method="post" enctype="multipart/form-data">
     <label class="file-upload"> Click here to upload an Image </label>
     <input type="file" name="file" id="file" required> 
</form>

And then, in the Servlet "Controller" add the Annotation for a Multi-part to indicate multipart data is processed in the servlet.

After doing this, retrieve the part sent through the form and then retrieve the file name (with path)of the submitted file. Use this to create a new file in the desired path and write the parts of the file to the newly created file to recreate the file.

As shown below:

@MultipartConfig

public class Controller extends HttpServlet {

    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        insertImage(request, response);
    }

    private void addProduct(HttpServletRequest request, HttpServletResponse response) {
        Part filePart = request.getPart("file");
        String imageName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString();

        String imageSavePath = "specify image path to save image"; //path to save image
        FileOutputStream outputStream = null;
        InputStream fileContent = null;

        try {
            outputStream = new FileOutputStream(new File(imageSavePath + File.separator + imageName));
            //creating a new file with file path and the file name
            fileContent = filePart.getInputStream();
            //getting the input stream
            int readBytes = 0;
            byte[] readArray = new byte[1024];
            //initializing a byte array with size 1024

            while ((readBytes = fileContent.read(readArray)) != -1) {
                outputStream.write(readArray, 0, readBytes);
            }//this loop will write the contents of the byte array unitl the end to the output stream
        } catch (Exception ex) {
            System.out.println("Error Writing File: " + ex);
        } finally {
            if (outputStream != null) {
                outputStream.close();
                //closing the output stream
            }
            if (fileContent != null) {
                fileContent.close();
                //clocsing the input stream
            }
        }
    }
}
-1
votes

you can upload file using jsp /servlet.

<form action="UploadFileServlet" method="post">
  <input type="text" name="description" />
  <input type="file" name="file" />
  <input type="submit" />
</form>

on the other hand server side. use following code.

     package com.abc..servlet;

import java.io.File;
---------
--------


/**
 * Servlet implementation class UploadFileServlet
 */
public class UploadFileServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    public UploadFileServlet() {
        super();
        // TODO Auto-generated constructor stub
    }
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        response.sendRedirect("../jsp/ErrorPage.jsp");
    }

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub

            PrintWriter out = response.getWriter();
            HttpSession httpSession = request.getSession();
            String filePathUpload = (String) httpSession.getAttribute("path")!=null ? httpSession.getAttribute("path").toString() : "" ;

            String path1 =  filePathUpload;
            String filename = null;
            File path = null;
            FileItem item=null;


            boolean isMultipart = ServletFileUpload.isMultipartContent(request);

            if (isMultipart) {
                FileItemFactory factory = new DiskFileItemFactory();
                ServletFileUpload upload = new ServletFileUpload(factory);
                String FieldName = "";
                try {
                    List items = upload.parseRequest(request);
                    Iterator iterator = items.iterator();
                    while (iterator.hasNext()) {
                         item = (FileItem) iterator.next();

                            if (fieldname.equals("description")) {
                                description = item.getString();
                            }
                        }
                        if (!item.isFormField()) {
                            filename = item.getName();
                            path = new File(path1 + File.separator);
                            if (!path.exists()) {
                                boolean status = path.mkdirs();
                            }
                            /* START OF CODE FRO PRIVILEDGE*/

                            File uploadedFile = new File(path + Filename);  // for copy file
                            item.write(uploadedFile);
                            }
                        } else {
                            f1 = item.getName();
                        }

                    } // END OF WHILE 
                    response.sendRedirect("welcome.jsp");
                } catch (FileUploadException e) {
                    e.printStackTrace();
                } catch (Exception e) {
                    e.printStackTrace();
                } 
            }   
    }

}
-1
votes
DiskFileUpload upload=new DiskFileUpload();

From this object you have to get file items and fields then yo can store into server like followed:

String loc="./webapps/prjct name/server folder/"+contentid+extension;
File uploadFile=new File(loc);
item.write(uploadFile);
-1
votes

Simplest way could come up with for files and input controls, w/out a billion libraries:

  <%
  if (request.getContentType()==null) return;
  // for input type=text controls
  String v_Text = 
  (new BufferedReader(new InputStreamReader(request.getPart("Text1").getInputStream()))).readLine();    

  // for input type=file controls
  InputStream inStr = request.getPart("File1").getInputStream();
  char charArray[] = new char[inStr.available()];
  new InputStreamReader(inStr).read(charArray);
  String contents = new String(charArray);
  %>
-2
votes

HTML PAGE

<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> 

SERVLET FILE

// 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.");
   } 
}

web.xml

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>
-3
votes

Sending multiple file for file we have to use enctype="multipart/form-data"
and to send multiple file use multiple="multiple" in input tag

<form action="upload" method="post" enctype="multipart/form-data">
 <input type="file" name="fileattachments"  multiple="multiple"/>
 <input type="submit" />
</form>