Java File Upload REST service

In this tutorial I will explain how to build Java REST web-service to upload files from any client over HTTP.

Uploading files to web-apps is a common task nowadays. A lot of services support uploading pictures or documents on their sites. With Java web services this is easy accomplished. What we need aside form the java web container (provided by your application server like Tomcat, GlassFish or JBoss) is the jersey framework to make it run. First I will show you how to implement the web-service and than give you two examples of clients to use the service.

Java file upload form

Java file upload form

Building the File Upload REST Service

The file is pushed over HTTP POST with encoding type “multipart/form-data” from the client to our web-service. This way you can add multiple parameters to the POST request in addition to the file. Lets start with the requirements. You will need an web/application server like Tomcat, GlassFish or JBoss to deploy the service. In addition we will use jersey framework to build our service endpoint. Please note, GlassFish 4.x version requires jersey version 2 libraries, so if you are using GlassFish 4 use jersey 2.x dependencies in your POM file instead.

For quick reference you can find the entire project in our GitHub repository under

I will post the entire POM file here, but what you need to take into account are the jersey dependencies

<project xmlns="" xmlns:xsi=""


	<name>File Uploader Rest Service</name>



							<!-- this is relative to the pom.xml directory -->


Having all necessary libraries now, lets go ahead and implement the REST service. There are several places in the code below I want to point your attention to. First note the usage of @Consumes(MediaType.MULTIPART_FORM_DATA) as requested encoding type. Second you may want to add additional parameters to the method if you like. For example you may want to pass some description or another text data with your upload. Finally Java will throw an Exception if you try to upload a file into a directory which not exists.  To avoid this issue I created the method createFolderIfNotExists(String dirName).




import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;

 * This example shows how to build Java REST web-service to upload files
 * accepting POST requests with encoding type "multipart/form-data". For more
 * details please read the full tutorial on
 * @author
public class FileUploadService {

	/** The path to the folder where we want to store the uploaded files */
	private static final String UPLOAD_FOLDER = "c:/uploadedFiles/";

	public FileUploadService() {

	private UriInfo context;

	 * Returns text response to caller containing uploaded file location
	 * @return error response in case of missing parameters an internal
	 *         exception or success response if file has been stored
	 *         successfully
	public Response uploadFile(
			@FormDataParam("file") InputStream uploadedInputStream,
			@FormDataParam("file") FormDataContentDisposition fileDetail) {

		// check if all form parameters are provided
		if (uploadedInputStream == null || fileDetail == null)
			return Response.status(400).entity("Invalid form data").build();

		// create our destination folder, if it not exists
		try {
		} catch (SecurityException se) {
			return Response.status(500)
					.entity("Can not create destination folder on server")

		String uploadedFileLocation = UPLOAD_FOLDER + fileDetail.getFileName();
		try {
			saveToFile(uploadedInputStream, uploadedFileLocation);
		} catch (IOException e) {
			return Response.status(500).entity("Can not save file").build();

		return Response.status(200)
				.entity("File saved to " + uploadedFileLocation).build();

	 * Utility method to save InputStream data to target location/file
	 * @param inStream
	 *            - InputStream to be saved
	 * @param target
	 *            - full path to destination file
	private void saveToFile(InputStream inStream, String target)
			throws IOException {
		OutputStream out = null;
		int read = 0;
		byte[] bytes = new byte[1024];

		out = new FileOutputStream(new File(target));
		while ((read = != -1) {
			out.write(bytes, 0, read);

	 * Creates a folder to desired location if it not already exists
	 * @param dirName
	 *            - full path to the folder
	 * @throws SecurityException
	 *             - in case you don't have permission to create the folder
	private void createFolderIfNotExists(String dirName)
			throws SecurityException {
		File theDir = new File(dirName);
		if (!theDir.exists()) {


Finally we need to configure our web.xml to register our class as web-service and make it run on startup.

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi=""
	xmlns="" xmlns:web=""
	id="WebApp_ID" version="2.5">
		<servlet-name>Jersey REST Service</servlet-name>
		<servlet-name>Jersey REST Service</servlet-name>

That’s it! Now you can build and deploy the WAR file. If you use the exact same names provided in the code above, your service URL (given you run on localhost) will be: http://localhost:8080/FileUploaderRESTService-1/rest/upload

File Upload HTML Form

You can use a very simple HTML post form as client to send files to the server.

Please note the usage of “multipart/form-data” as encoding type. You also need to add an <input> of type file with the name “file”

Choose file to upload<br>
<form action="http://localhost:8080/FileUploaderRESTService-1/rest/upload" method="post" enctype="multipart/form-data">
	<input name="file" id="filename" type="file" /><br><br>
	<button name="submit" type="submit">Upload</button>

As I already mentioned you can add additional data to your request. In this case just don’t forget to handle it in the web-service 🙂

Java File Upload Client

You can create a file upload client for you Android or stand-alone programs in java. I the example below I will use Apache http libraries, you will need this five:

  • commons-logging
  • httpclient
  • httpclient-cache
  • httpcore
  • httpmime
package net.javatutorial.tutorials.clienst;


import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.mime.MultipartEntity;
import org.apache.http.entity.mime.content.InputStreamBody;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.params.BasicHttpParams;
import org.apache.http.util.EntityUtils;

 * This example shows how to upload files using POST requests 
 * with encoding type "multipart/form-data".
 * For more details please read the full tutorial
 * on
 * @author
public class FileUploaderClient {
	public static void main(String[] args) {
		// the file we want to upload
		File inFile = new File("C:\\Users\\admin\\Desktop\\Yana-make-up.jpg");
		FileInputStream fis = null;
		try {
			fis = new FileInputStream(inFile);
			DefaultHttpClient httpclient = new DefaultHttpClient(new BasicHttpParams());
			// server back-end URL
			HttpPost httppost = new HttpPost("http://localhost:8080/FileUploaderRESTService-1/rest/upload");
			MultipartEntity entity = new MultipartEntity();

			// set the file input stream and file name as arguments
			entity.addPart("file", new InputStreamBody(fis, inFile.getName()));

			// execute the request
			HttpResponse response = httpclient.execute(httppost);
			int statusCode = response.getStatusLine().getStatusCode();
			HttpEntity responseEntity = response.getEntity();
			String responseString = EntityUtils.toString(responseEntity, "UTF-8");
			System.out.println("[" + statusCode + "] " + responseString);
		} catch (ClientProtocolException e) {
			System.err.println("Unable to make connection");
		} catch (IOException e) {
			System.err.println("Unable to read file");
		} finally {
			try {
				if (fis != null) fis.close();
			} catch (IOException e) {}

You will find the project files in our GitHub repository

Thanks for reading. As always comments are welcome 🙂

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21 Comments on "Java File Upload REST service"

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Two things:

First, if I use I the URL
that is in the form and the text of the example I get a 404 error. That was from the github version of this.

If I take out the “-1” at the end of the service name and use
then I got an upload. Is the “-1” just a typo or was it supposed to mean something?

Secondly, I started looking at this tutorial because of the use of multipart, as I want to add parameters as part of the upload. Is there somewhere I could find an example of that usage?


Sarith NOB
Sarith NOB

Thank you for you useful code!

And I have another problem, it’s about pom file because I am in use of Glassfish which have no pom file as a rest web service.

How can I add dependency in Glassfish web service?

I hope you have solution for me!

Many thank!


Hi Flipp,
The article is amazing. I have few questions. I have already created a Rest web service(URL). Now I only want to call that Api using java. Can I use just the chunck of code you have mentioned above to call that Web service and upload file to a web service?
My requirement is to call a Post method web service and upload entire file to a web service. What code of chunck will be used for that?


Thankyou Flipp.
Does that work for JSON data as well.
I am creating a JSON file from Informatica (ETL) and the file store in ETL server. Now I need to call the web service to upload file from ETL server to an API (web service) using Java.
The issue is I don’t have same format of request when creating JSON file from ETL so do I need a string manipulation and read file from that server when writing code in Java?

Hi Filip, So this is the modification of code I am using (JavaClient). I am getting this error: [401] This request is not authorized. Even thou I put username & pass of URL: http://localhost:8001/customer (because it had basic authentication). This is my code: package net.javatutorial.tutorials.clienst; import; import; import; import org.apache.http.HttpEntity; import org.apache.http.HttpResponse; import org.apache.http.client.ClientProtocolException; import org.apache.http.client.methods.HttpPost; import org.apache.http.entity.mime.MultipartEntity; import org.apache.http.entity.mime.content.ContentBody; import org.apache.http.entity.mime.content.InputStreamBody; import org.apache.http.entity.mime.content.StringBody; import org.apache.http.impl.client.DefaultHttpClient; import org.apache.http.message.BasicNameValuePair; import org.apache.http.params.BasicHttpParams; import org.apache.http.util.EntityUtils; /** * This example shows how to upload files using POST requests * with encryption type "multipart/form-data". * For more details please read the full… Read more »

when running the client or server of the downloaded project I get error 404 and that the resource is not available

Petr Šponer
Petr Šponer

There is an error in your method saveToFile in git (it is ok here in the article). The while loop belong to the try block.
However, thank you for the example.


i’m trying this code on my code but i have this exception:
נוב 26, 2017 4:33:46 PM com.sun.jersey.spi.container.ContainerRequest getEntity
SEVERE: A message body reader for Java class com.sun.jersey.core.header.FormDataContentDisposition, and Java type class com.sun.jersey.core.header.FormDataContentDisposition, and MIME media type multipart/form-data; boundary=—-WebKitFormBoundarykzdzsKUpAFNpL25b was not found.
The registered message body readers compatible with the MIME media type are:
*/* ->

please help