Skip to content
Related Articles

Related Articles

Asynchronous file transfer in AJAX

View Discussion
Improve Article
Save Article
  • Last Updated : 27 Apr, 2020
View Discussion
Improve Article
Save Article

To transfer Files in jQuery using Ajax we follow the following steps and an example demonstrating the same:

  1. Using HTML5, we can upload files using a form. The content-type is specified as multipart/form-data.
  2. An input field is created for files to be uploaded with input type as file.
  3. We can use the multiple attribute to allow more than one file to be uploaded and can also filter out file types by using the accept attribute.
  4. On upload a listener appends the uploaded files to a file list iteratively.
  5. With the submit button an ajax request is created and the form data is sent across to the specified url.

Example: Here consider the case of sending videos asynchronously.

  • We have created an HTML5 form as below:

        <form method="POST" enctype="multipart/form-data"
                action="" id="capt">
            <div class="form-group">
                <label>Select Files</label>
                <input type="file" id="file" name="file[]"
                    accept="video/*" class="form-control"
                    multiple="multiple" required>
            <input type="submit" class="btn btn-info"

  • Using jQuery, we create appropriate methods for uploading and sending file asynchronously.

    var fileList = [];
    $('#file').on('change', function (event) {
        fileList = [];
        // Append files to fileList
        for (var i = 0; i < this.files.length; i++) {
    sendFile = function (file) {
        // Create Ajax Request
            url: 'notify.php',
            type: 'POST',
            data: new FormData($('form')[0]),
            cache: false,
            contentType: false,
            processData: false
    $('#capt').on('submit', function (event) {
        //Asynchronous Transfer

  • For the sake of demonstrating the output, we create a minimal PHP file as below:

    $video = $_FILES['file']['name'];
    foreach($video as $vd){
        echo $vd."<br>";

  • Output(notify.php):

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!