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In this tutorial, we discuss how to upload files with Spring MVC.

If you’re more interested in doing it the old way using Servlets, then check “How to upload files to a Servlet” tutorial.

1- Upload.html

Before digging into the controller, let’s see how would the HTML form looks like:

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

The above is a very simple HTML form which only holds an input file along with a submit button.

As noticed, we set the encoding type to “multipart/form-data” and define “/upload” API to handle the request.

2- Controller

I n order to parse the uploaded files at the server-side, you have to create a controller method that maps the “/upload” requests, the method should:

  1. Consume “multipart/form-data” request type.
  2. Has an input argument of type MultipartFile[].

The below method parses the uploaded files and saves them to the file system:

@RequestMapping(value = "/upload", method = RequestMethod.POST, consumes = "multipart/form-data")
public void upload(@RequestParam("file") MultipartFile[] filesToBeUploaded, ModelMap modelMap, HttpServletResponse response) {
    for(MultipartFile fileToBeUploaded : filesToBeUploaded)
        File output = new File("C:\\Users\\user\\Desktop\\output\\" + fileToBeUploaded.getOriginalFilename());
        try(FileOutputStream outputStream = new FileOutputStream(output); InputStream is = fileToBeUploaded.getInputStream();)
            int read = 0;
            byte[] bytes = new byte[1024];
            while ((read = != -1) {
                outputStream.write(bytes, 0, read);
        catch(Exception ex)

As noticed, we parse the uploaded file using getInputStream() method and we get its name through getOriginalFilename() method.

P.S: If you feel confused on the above try/catch syntax, check “Try with resources” tutorial.


In this tutorial, we discuss how to upload files with Spring MVC.

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