How to use FormData for AJAX file upload?

Question

This is my HTML which I'm generating dynamically using drag and drop functionality.

<form method="POST" id="contact" name="13" class="form-horizontal wpc_contact" novalidate="novalidate" enctype="multipart/form-data">
<fieldset>
    <div id="legend" class="">
        <legend class="">file demoe 1</legend>
        <div id="alert-message" class="alert hidden"></div>
    </div>

    <div class="control-group">
        <!-- Text input-->
        <label class="control-label" for="input01">Text input</label>
        <div class="controls">
            <input type="text" placeholder="placeholder" class="input-xlarge" name="name">
            <p class="help-block" style="display:none;">text_input</p>
        </div>
        <div class="control-group">  </div>
        <label class="control-label">File Button</label>

        <!-- File Upload --> 
        <div class="controls">
            <input class="input-file" id="fileInput" type="file" name="file">
        </div>
    </div>
    <div class="control-group">    

        <!-- Button --> 
        <div class="controls">
            <button class="btn btn-success">Button</button>
        </div>
    </div>
</fieldset>
</form>

This is my JavaScript code:

<script>
    $('.wpc_contact').submit(function(event){
        var formname = $('.wpc_contact').attr('name');
        var form = $('.wpc_contact').serialize();               
        var FormData = new FormData($(form)[1]);

        $.ajax({
            url : '<?php echo plugins_url(); ?>'+'/wpc-contact-form/resources/js/tinymce.php',
            data : {form:form,formname:formname,ipadd:ipadd,FormData:FormData},
            type : 'POST',
            processData: false,
            contentType: false,
            success : function(data){
            alert(data); 
            }
        });
   }

You should read this (developer.mozilla.org/en-US/docs/Web/API/FormData/append) the formData(); append method has an optional third parameter for a file. — www139
– www139, Commented Dec 26, 2015 at 14:36

Community · Accepted Answer · 2018-01-22 16:26:42Z

581

For correct form data usage you need to do 2 steps.

Preparations

You can give your whole form to FormData() for processing

var form = $('form')[0]; // You need to use standard javascript object here
var formData = new FormData(form);

or specify exact data for FormData()

var formData = new FormData();
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Attach file
formData.append('image', $('input[type=file]')[0].files[0]);

Sending form

Ajax request with jquery will looks like this:

$.ajax({
    url: 'Your url here',
    data: formData,
    type: 'POST',
    contentType: false, // NEEDED, DON'T OMIT THIS (requires jQuery 1.6+)
    processData: false, // NEEDED, DON'T OMIT THIS
    // ... Other options like success and etc
});

After this it will send ajax request like you submit regular form with enctype="multipart/form-data"

Update: This request cannot work without type:"POST" in options since all files must be sent via POST request.

Note: contentType: false only available from jQuery 1.6 onwards

edited Jan 22, 2018 at 16:26

CommunityBot

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answered Jan 10, 2014 at 13:00

Spell

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23 Comments

Wouter Over a year ago

Can I set the "enctype" in the Ajax call? I think I may have an issue with it. Or, can I set it on the FormData object?

Spell Over a year ago

You can. For this see lines after THIS MUST BE DONE FOR FILE UPLOADING in my code.

Юрий Светлов Over a year ago

@Spell How get data in controller? Do need send getCsrfToken?

Spell Over a year ago

@ЮрийСветлов This is depends of what type of controller you use. Is it server side or front side controller? You trying to solve CSRF protection here?

Spell Over a year ago

@ManthanJamdagni When you get $('form'), it will return jQuery object. But we need regular js object here without jQuery functionality. That's why we get regular object with [0] notation. Instead of this construction you can call document.getElementById() or simular call.

|

Quentin · Accepted Answer · 2016-04-12 10:56:52Z

28

<form id="upload_form" enctype="multipart/form-data">

jQuery with CodeIgniter file upload:

var formData = new FormData($('#upload_form')[0]);

formData.append('tax_file', $('input[type=file]')[0].files[0]);

$.ajax({
    type: "POST",
    url: base_url + "member/upload/",
    data: formData,
    //use contentType, processData for sure.
    contentType: false,
    processData: false,
    beforeSend: function() {
        $('.modal .ajax_data').prepend('<img src="' +
            base_url +
            '"asset/images/ajax-loader.gif" />');
        //$(".modal .ajax_data").html("<pre>Hold on...</pre>");
        $(".modal").modal("show");
    },
    success: function(msg) {
        $(".modal .ajax_data").html("<pre>" + msg +
            "</pre>");
        $('#close').hide();
    },
    error: function() {
        $(".modal .ajax_data").html(
            "<pre>Sorry! Couldn't process your request.</pre>"
        ); // 
        $('#done').hide();
    }
});

you can use.

var form = $('form')[0]; 
var formData = new FormData(form);     
formData.append('tax_file', $('input[type=file]')[0].files[0]);

or

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

Both will work.

edited Apr 12, 2016 at 10:56

Quentin

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answered Aug 27, 2015 at 6:47

chandoo

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2 Comments

26vivek Over a year ago

If we are getting dom with id, why we need to type 0 zero like $(#formId)[0] ?

andreszs Over a year ago

@26vivek only the jQuery developers would know the answer to that question.

Ankush Kumar · Accepted Answer · 2019-09-17 10:55:32Z

6

$(document).ready(function () {
    $(".submit_btn").click(function (event) {
        event.preventDefault();
        var form = $('#fileUploadForm')[0];
        var data = new FormData(form);
        data.append("CustomField", "This is some extra data, testing");
        $("#btnSubmit").prop("disabled", true);
        $.ajax({
            type: "POST",
            enctype: 'multipart/form-data',
            url: "upload.php",
            data: data,
            processData: false,
            contentType: false,
            cache: false,
            timeout: 600000,
            success: function (data) {
                console.log();
            },
        });
    });
});

answered Sep 17, 2019 at 10:55

Ankush Kumar

611 silver badge1 bronze badge

Comments

Ranch Camal · Accepted Answer · 2020-01-10 03:04:14Z

3

Better to use the native javascript to find the element by id like: document.getElementById("yourFormElementID").

$.ajax( {
      url: "http://yourlocationtopost/",
      type: 'POST',
      data: new FormData(document.getElementById("yourFormElementID")),
      processData: false,
      contentType: false
    } ).done(function(d) {
           console.log('done');
    });

answered Jan 10, 2020 at 3:04

Ranch Camal

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Comments

Shaiful Ezani · Accepted Answer · 2017-05-23 04:23:39Z

$('#form-withdraw').submit(function(event) {

    //prevent the form from submitting by default
    event.preventDefault();



    var formData = new FormData($(this)[0]);

    $.ajax({
        url: 'function/ajax/topup.php',
        type: 'POST',
        data: formData,
        async: false,
        cache: false,
        contentType: false,
        processData: false,
        success: function (returndata) {
          if(returndata == 'success')
          {
            swal({
              title: "Great",
              text: "Your Form has Been Transfer, We will comfirm the amount you reload in 3 hours",
              type: "success",
              showCancelButton: false,
              confirmButtonColor: "#DD6B55",
              confirmButtonText: "OK",
              closeOnConfirm: false
            },
            function(){
              window.location.href = '/transaction.php';
            });
          }

          else if(returndata == 'Offline')
          {
              sweetAlert("Offline", "Please use other payment method", "error");
          }
        }
    });



});

Vkl125 · Accepted Answer · 2016-10-27 19:23:37Z

View:
<label class="btn btn-info btn-file">
Import <input type="file" style="display: none;">
</label>
<Script>
$(document).ready(function () {
                $(document).on('change', ':file', function () {
                    var fileUpload = $(this).get(0);
                    var files = fileUpload.files;
                    var bid = 0;
                    if (files.length != 0) {
                        var data = new FormData();
                        for (var i = 0; i < files.length ; i++) {
                            data.append(files[i].name, files[i]);
                        }
                        $.ajax({
                            xhr: function () {
                                var xhr = $.ajaxSettings.xhr();
                                xhr.upload.onprogress = function (e) {
                                    console.log(Math.floor(e.loaded / e.total * 100) + '%');
                                };
                                return xhr;
                            },
                            contentType: false,
                            processData: false,
                            type: 'POST',
                            data: data,
                            url: '/ControllerX/' + bid,
                            success: function (response) {
                                location.href = 'xxx/Index/';
                            }
                        });
                    }
                });
            });
</Script>
Controller:
[HttpPost]
        public ActionResult ControllerX(string id)
        {
            var files = Request.Form.Files;
...

It is normally considered good form to provide an explanation along with an answer.

Onic Team · Accepted Answer · 2017-07-29 20:23:15Z

0

Actually The documentation shows that you can use XMLHttpRequest().send() to simply send multiform data in case jquery sucks

edited Jul 29, 2017 at 20:23

Onic Team

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answered Jul 29, 2017 at 19:19

Richie

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Comments

E. Coelho · Accepted Answer · 2019-04-24 15:14:39Z

-7

Good morning.

I was have the same problem with upload of multiple images. Solution was more simple than I had imagined: include [] in the name field.

<input type="file" name="files[]" multiple>

I did not make any modification on FormData.

answered Apr 24, 2019 at 15:14

E. Coelho

112 bronze badges

1 Comment

Quentin Over a year ago

This has nothing to do with the problem the question is asking about and is just a peculiarity of how PHP handles form data with multiple values that have the same name.

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