CODEWITHSUNDEEP
htmljqueryajaxfile-uploadform-data
saadovsky
saadovsky

Reputation: 11

Uploading file using ajax and handling it with a java servlet server-side

I have been stuck on this problem for the last couple of days and no other posts have been helpful. I'm trying to upload a file on a HTML form which looks like this (it contains other fields):

       <input type="text" id="prenom" name="fname" placeholder="Prenom"><br><br>
       <input type="text" id="nom" name="fname" placeholder="Nom"><br><br>
       <input type="email" id="mail" name="fname" placeholder="Adresse mail"><br><br>
       <input type="file" id="file" name="file" />
   </form>
   <br><button  id="bouton-inscription" >s'inscrire</button>```

and then upload it to using Ajax through a FormData:

    var formData = new FormData(); 
    formData.append('file', $('#file')[0].files[0]);

    $.ajax({
        url: './upload',
        cache: false,
        processData: false,
        contentType: false,
        method: 'POST',
        data: formData,
        dataType: 'json'
           })

Then, server-side, I want to retrieve the file and save it using a httpServlet, I'm using this code I "adapted" from Oracle docs:

@WebServlet(name = "FileUploadServlet", urlPatterns = {"/upload"})
@MultipartConfig
public class FileUploadServlet extends HttpServlet {
    
    protected void processRequest(HttpServletRequest request,HttpServletResponse response)
        throws ServletException, IOException {
        
    response.setContentType("text/html;charset=UTF-8");

    // Create path components to save the file
    final String path = "C:\\Users\\me\\Desktop";
    final Part filePart = request.getPart("file");
    final String fileName = "myFile";

    OutputStream out = null;
    InputStream filecontent = null;
    final PrintWriter writer = response.getWriter();

    try {
        out = new FileOutputStream(new File(path + File.separator
                + fileName));
        filecontent = filePart.getInputStream();

        int read = 0;
        final byte[] bytes = new byte[1024];

        while ((read = filecontent.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        writer.println("New file " + fileName + " created at " + path);
    } catch (FileNotFoundException fne) {
        writer.println("You either did not specify a file to upload or are "
                + "trying to upload a file to a protected or nonexistent "
                + "location.");
        writer.println("<br/> ERROR: " + fne.getMessage());
    } finally {
        if (out != null) {
            out.close();
        }
        if (filecontent != null) {
            filecontent.close();
        }
        if (writer != null) {
            writer.close();
        }
    }
}

But I'm getting a an error in the Ajax call (I'm handling the other fields with another ajax call to another servlet which works just fine since they're text fields).

I feel like the problem must be server-side because the code I'm using client-side I've seen everywhere.

Any hints would be much appreciated!

Thanks!

Edit: Actually with this code the file is saved correctly but there is still an Ajax error

Upvotes: 1

Views: 934

Answers (1)

saadovsky
saadovsky

Reputation: 11

Problem was the double ajax call with a .done for each, fixed it by not expecting an answer from the file upload ajax call, and expecting one (that would redirect the user to another page) from the rest of the submitted info.

Upvotes: 0

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