Sending Form via AJAX in plain javascript (no jQuery)

I have the following Form which I would like to send via an AJAX request. I am unsure how to proceed in the line 'xmlhttp.open'. I am trying to upload a video file to a third party video hosting site (using their API)and they have provided me with a URL ('upload_link_secure')to upload the file to. Please can someone advise?

my HTML:

<form id="upload" action="'+upload_link_secure+'" method="PUT" enctype="multipart/form-data">
  <input type="file" id="vidInput">
  <button type="submit" id="submitFile" onclick="uploadMyVid(\''+upload_link_secure+'\')">Upload Video</button>
</form>

my javascript:

 var uploadMyVid = function(upload_link_secure){

        var form = document.getElementById('upload')

        // FETCH FILEIST OBJECTS
        var vidFile = document.getElementById('vidInput').files;

        var xmlhttp = new XMLHttpRequest;

        xmlhttp.open('PUT', );  // NOT SURE HOW TO COMPLETE THIS LINE???
        xmlhttp.send(vidFile);

    }

my HTML:

<form id="upload" action="'+upload_link_secure+'" method="PUT" enctype="multipart/form-data">
  <input type="file" id="vidInput">
  <button type="submit" id="submitFile" onclick="uploadMyVid(\''+upload_link_secure+'\')">Upload Video</button>
</form>

my javascript:

 var uploadMyVid = function(upload_link_secure){

        var form = document.getElementById('upload')

        // FETCH FILEIST OBJECTS
        var vidFile = document.getElementById('vidInput').files;

        var xmlhttp = new XMLHttpRequest;

        xmlhttp.open('PUT', );  // NOT SURE HOW TO COMPLETE THIS LINE???
        xmlhttp.send(vidFile);

    }

Share Improve this question edited Oct 8, 2016 at 22:32 asked Oct 8, 2016 at 22:05 dave 4756 silver badges20 bronze badges

It's a bit more plicated than just sending a plete fileList, but basically the arguments are .open(method, url) so you have to add the URL to upload to. – adeneo Commented Oct 8, 2016 at 22:10
i think you can plete it based on documentation developer.mozilla/en-US/docs/Web/API/XMLHttpRequest/open – Smile0ff Commented Oct 8, 2016 at 22:12
i'm a little confused with the 'action' attribute in the Form element. My thoughts are this element would send the file to the url. Would i still need to duplicate this request at the xmlhttp.open line... – dave Commented Oct 8, 2016 at 22:12
if you make sync request you don't need to specify action url, but in your case, you did async request, so XHR don't know anything about your endpoint, yes you need to set it explicitly – Smile0ff Commented Oct 8, 2016 at 22:15
Are you trying to send the files using <form> or XMLHttpRequest()? – guest271314 Commented Oct 8, 2016 at 22:25

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2 Answers 2

Sorted by: Reset to default 6

First of all your action attribute not correct, change to some endpoint like /upload for example.

Here is a simple example without server side.

var form = document.getElementById("upload-form"),
        actionPath = "";
        formData = null;

    var xhr = new XMLHttpRequest();

    form.addEventListener("submit", (e) => {
        e.preventDefault();
        
        formData = new FormData(form);
        actionPath = form.getAttribute("action");

        xhr.open("POST", actionPath);
        xhr.send(formData);

    }, false);

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

Substitute <span> element for <button> element, use click event handler attached to #submitFile element; FormData() at XMLHttpRequest() to send File object within <input type="file"> .files object; remove action attribute at <form> element, set URL of XMLHttpRequest.prototype.open() to URL provided for PUT request

<body>
  <form id="upload">
    <input type="file" id="vidInput">
    <span id="submitFile" 
      style="-webkit-appearance:button;-moz-appearance:button;padding:4px;font-family:arial;font-size:12px">Upload Video</span>
  </form>
  <script>
    function uploadMyVid(event) {

      // FETCH FILEIST OBJECTS
      var vidFile = document.getElementById('vidInput').files;

      var xmlhttp = new XMLHttpRequest;

      xmlhttp.open('PUT', "upload_link_secure");

      var data = new FormData();
      data.append("file", vidFile[0], vidFile[0].name);

      xmlhttp.send(data);

    }

    var button = document.getElementById("submitFile");
    button.addEventListener("click", uploadMyVid);
  </script>
</body>

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Sending Form via AJAX in plain javascript (no jQuery) - Stack Overflow

2 Answers 2

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