How to upload files to a server using JSP/Servlet? [Answered]

Sample problem:

How can I upload files to the 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" />

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(
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(
    at org.apache.catalina.core.StandardWrapperValve.invoke(
    at org.apache.catalina.core.StandardContextValve.invoke(
    at org.apache.catalina.core.StandardHostValve.invoke(
    at org.apache.catalina.valves.ErrorReportValve.invoke(
    at org.apache.catalina.core.StandardEngineValve.invoke(
    at org.apache.catalina.connector.CoyoteAdapter.service(
    at org.apache.coyote.http11.Http11Processor.process(
    at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(

How to upload files to a server using JSP/Servlet? Answer #1:


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" />

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 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:

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 want to upload multiple files via either multiple="true",

<input type="file" name="files" multiple="true" />

or the old-fashioned way with multiple inputs,

<input type="file" name="files" />
<input type="file" name="files" />
<input type="file" name="files" />

then you can collect them as below (unfortunately there is no such method as request.getParts("files")):

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

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().

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 = > 0;) {
        value.append(buffer, 0, length);
    return value.toString();
String description = getValue(request.getPart("description")); // Retrieves <input type="text" name="description">

Answer #2:

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“:

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

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.

Answer #3:

Without components or external libraries in Tomcat 6 or Tomcat 7

Enabling upload in the web.xml file:


As you can see:


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 = != -1) {
  baos.write(buf, 0, qt);
String sResumen = baos.toString();

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

Answer #4:

I am using a 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 a fixed directory and later you rename the directory of your choice. Here Connections is our custom interface having a connection object.

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
            // 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.
            try {
                // Parse the request to get file items.
                List<FileItem> fileItems = upload.parseRequest(request);
                // Process the uploaded file items
                Iterator<FileItem> i = fileItems.iterator();
                while (i.hasNext()) {
                    FileItem fi = (FileItem);
                    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
                        } else {
                            file = new File(filePath
                                + fileName.substring(fileName
                                        .lastIndexOf("\\") + 1));
                    } else {
                        key_values.put(fi.getFieldName(), fi.getString());
            } catch (Exception ex) {
        return key_values;

Answer #5:

For Spring MVC

I managed to have a simpler version that worked for taking form input, both data and images.

<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" />

Controller to handle

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

            byte[] bytes = file.getBytes();
            String filename = file.getOriginalFilename();
            BufferedOutputStream stream =new BufferedOutputStream(new FileOutputStream(new File("D:/" + filename)));
        return new ModelAndView("index");

Hope you learned something from this post.

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