572

The XMLHttpRequest Level 2 standard (still a working draft) defines the FormData interface. This interface enables appending File objects to XHR-requests (Ajax-requests).

Btw, this is a new feature - in the past, the "hidden-iframe-trick" was used (read about that in my other question).

This is how it works (example):

var xhr = new XMLHttpRequest(), fd = new FormData(); fd.append( 'file', input.files[0] ); xhr.open( 'POST', 'http://example.com/script.php', true ); xhr.onreadystatechange = handler; xhr.send( fd ); 

where input is a <input type="file"> field, and handler is the success-handler for the Ajax-request.

This works beautifully in all browsers (again, except IE).

Now, I would like to make this functionality work with jQuery. I tried this:

var fd = new FormData(); fd.append( 'file', input.files[0] ); $.post( 'http://example.com/script.php', fd, handler ); 

Unfortunately, that won't work (an "Illegal invocation" error is thrown - screenshot is here). I assume jQuery expects a simple key-value object representing form-field-names / values, and the FormData instance that I'm passing in is apparently incompatible.

Now, since it is possible to pass a FormData instance into xhr.send(), I hope that it is also possible to make it work with jQuery.


Update:

I've created a "feature ticket" over at jQuery's Bug Tracker. It's here: http://bugs.jquery.com/ticket/9995

I was suggested to use an "Ajax prefilter"...


Update:

First, let me give a demo demonstrating what behavior I would like to achieve.

HTML:

<form> <input type="file" id="file" name="file"> <input type="submit"> </form> 

JavaScript:

$( 'form' ).submit(function ( e ) { var data, xhr; data = new FormData(); data.append( 'file', $( '#file' )[0].files[0] ); xhr = new XMLHttpRequest(); xhr.open( 'POST', 'http://hacheck.tel.fer.hr/xml.pl', true ); xhr.onreadystatechange = function ( response ) {}; xhr.send( data ); e.preventDefault(); }); 

The above code results in this HTTP-request:

multipartformdata

This is what I need - I want that "multipart/form-data" content-type!


The proposed solution would be like so:

$( 'form' ).submit(function ( e ) { var data; data = new FormData(); data.append( 'file', $( '#file' )[0].files[0] ); $.ajax({ url: 'http://hacheck.tel.fer.hr/xml.pl', data: data, processData: false, type: 'POST', success: function ( data ) { alert( data ); } }); e.preventDefault(); }); 

However, this results in:

wrongcontenttype

As you can see, the content type is wrong...

2
  • @zaf Setting processData:false is a step in the right direction (see pradeek's proposed solution). Now if I could manually set the "Content-Type" HTTP-request header... See my updated question for details.CommentedNov 28, 2011 at 16:52
  • jQuery 4.0.0-beta now officially supports binary data. See this answer.
    – Mahozad
    CommentedFeb 8, 2024 at 8:39

8 Answers 8

1031
+100

I believe you could do it like this :

var fd = new FormData(); fd.append( 'file', input.files[0] ); $.ajax({ url: 'http://example.com/script.php', data: fd, processData: false, contentType: false, type: 'POST', success: function(data){ alert(data); } }); 

Notes:

  • Setting processData to false lets you prevent jQuery from automatically transforming the data into a query string. See the docs for more info.

  • Setting the contentType to false is imperative, since otherwise jQuery will set it incorrectly.

13
  • Unfortunately, this results in the wrong "Content-Type" HTTP-request header being set. See my question for detailed explanation what I want to achieve and why your current solution doesn't satisfy. Maybe I can set the Content-Type manually?CommentedNov 28, 2011 at 16:50
  • 12
    Yes,I believe you could manually set the contentType to 'multipart/form-data' by adding a key-value pair in $.ajax argument.
    – pradeek
    CommentedNov 29, 2011 at 15:05
  • Hi @pradeek, we tried this solution here (stackoverflow.com/questions/10215425/…), but it wasn't clear how to pass the image data in the body of the POST request. Any clues? Thanks!
    – Crashalot
    CommentedApr 19, 2012 at 1:18
  • 2
    Tested in Firefox, and Chrome. This is the solution.CommentedJan 11, 2013 at 1:44
  • 2
    Setting processData and contentType to false fixed the issue for me. Thank you, that was unpleasant.CommentedNov 6, 2013 at 16:58
35

You can send the FormData object in ajax request using the following code,

$("form#formElement").submit(function(){ var formData = new FormData($(this)[0]); }); 

This is very similar to the accepted answer but an actual answer to the question topic. This will submit the form elements automatically in the FormData and you don't need to manually append the data to FormData variable.

The ajax method looks like this,

$("form#formElement").submit(function(){ var formData = new FormData($(this)[0]); //append some non-form data also formData.append('other_data',$("#someInputData").val()); $.ajax({ type: "POST", url: postDataUrl, data: formData, processData: false, contentType: false, dataType: "json", success: function(data, textStatus, jqXHR) { //process data }, error: function(data, textStatus, jqXHR) { //process error msg }, }); 

You can also manually pass the form element inside the FormData object as a parameter like this

var formElem = $("#formId"); var formdata = new FormData(formElem[0]); 

Hope it helps. ;)

3
  • 13
    Isn't $(this)[0] the same as this ?
    – dfl
    CommentedAug 11, 2016 at 15:53
  • Why $(this) returns an array?
    – halbano
    CommentedJul 18, 2017 at 0:49
  • 4
    @halbano $(this) returns not array, but jQuery collection object which can be iterated over just like array. form[0] does almost exactly the same as form.first(), which just returns the first, real HTML element.
    – SteveB
    CommentedOct 26, 2017 at 12:48
31

There are a few yet to be mentioned techniques available for you. Start with setting the contentType property in your ajax params.

Building on pradeek's example:

$('form').submit(function (e) { var data; data = new FormData(); data.append('file', $('#file')[0].files[0]); $.ajax({ url: 'http://hacheck.tel.fer.hr/xml.pl', data: data, processData: false, type: 'POST', // This will override the content type header, // regardless of whether content is actually sent. // Defaults to 'application/x-www-form-urlencoded' contentType: 'multipart/form-data', //Before 1.5.1 you had to do this: beforeSend: function (x) { if (x && x.overrideMimeType) { x.overrideMimeType("multipart/form-data"); } }, // Now you should be able to do this: mimeType: 'multipart/form-data', //Property added in 1.5.1 success: function (data) { alert(data); } }); e.preventDefault(); }); 

In some cases when forcing jQuery ajax to do non-expected things, the beforeSend event is a great place to do it. For a while people were using beforeSend to override the mimeType before that was added into jQuery in 1.5.1. You should be able to modify just about anything on the jqXHR object in the before send event.

4
  • 5
    There is a problem with this: the spec requires that a multipart content-type includes the boundary parameter (specifically, it’s stated as a required parameter).CommentedFeb 29, 2012 at 19:58
  • 1
    @romkyns I was coding this from the top of my head. If you have some experience/knowledge that I don't, please go ahead and edit the answer! Now that you mentioned it, a quick search turned up this: stackoverflow.com/questions/5933949/… which should get you going in the right direction. (You could take from that example and incorporate it into the above jQuery if you are into that.)
    – BenSwayne
    CommentedMar 1, 2012 at 23:43
  • Actually the currently accepted answer (that I edited to add a crucial fix) does the right thing, at last. This was tricky :)CommentedMar 2, 2012 at 14:44
  • I had high hopes for this solution, but my $_FILES is still NULL...
    – socca1157
    CommentedAug 27, 2014 at 18:58
7

I do it like this and it's work for me, I hope this will help :)

 <div id="data"> <form> <input type="file" name="userfile" id="userfile" size="20" /> <br /><br /> <input type="button" id="upload" value="upload" /> </form> </div> <script> $(document).ready(function(){ $('#upload').click(function(){ console.log('upload button clicked!') var fd = new FormData(); fd.append( 'userfile', $('#userfile')[0].files[0]); $.ajax({ url: 'upload/do_upload', data: fd, processData: false, contentType: false, type: 'POST', success: function(data){ console.log('upload success!') $('#data').empty(); $('#data').append(data); } }); }); }); </script> 
3
  • How is this different from the accepted answer?CommentedMar 28, 2015 at 16:04
  • fd.append( 'userfile', $('#userfile')[0].files[0]); THIS is the different you need to add the id's for input file to work.CommentedMar 29, 2015 at 14:36
  • This one helped me than others..CommentedJan 13, 2020 at 10:37
5

JavaScript:

function submitForm() { var data1 = new FormData($('input[name^="file"]')); $.each($('input[name^="file"]')[0].files, function(i, file) { data1.append(i, file); }); $.ajax({ url: "<?php echo base_url() ?>employee/dashboard2/test2", type: "POST", data: data1, enctype: 'multipart/form-data', processData: false, // tell jQuery not to process the data contentType: false // tell jQuery not to set contentType }).done(function(data) { console.log("PHP Output:"); console.log(data); }); return false; } 

PHP:

public function upload_file() { foreach($_FILES as $key) { $name = time().$key['name']; $path = 'upload/'.$name; @move_uploaded_file($key['tmp_name'], $path); } } 
    4

    You can use the $.ajax beforeSend event to manipulate the header.

    beforeSend: function(xhr) { xhr.setRequestHeader('Content-Type', 'multipart/form-data'); } 

    See this link for additional information: http://msdn.microsoft.com/en-us/library/ms536752(v=vs.85).aspx

      1

      If you want to submit files using ajax use "jquery.form.js" This submits all form elements easily.

      Samples http://jquery.malsup.com/form/#ajaxSubmit

      rough view :

      <form id='AddPhotoForm' method='post' action='../photo/admin_save_photo.php' enctype='multipart/form-data'> 


      <script type="text/javascript"> function showResponseAfterAddPhoto(responseText, statusText) { information= responseText; callAjaxtolist(); $("#AddPhotoForm").resetForm(); $("#photo_msg").html('<div class="album_msg">Photo uploaded Successfully...</div>'); }; $(document).ready(function(){ $('.add_new_photo_div').live('click',function(){ var options = {success:showResponseAfterAddPhoto}; $("#AddPhotoForm").ajaxSubmit(options); }); }); </script> 
        1

        Instead of - fd.append( 'userfile', $('#userfile')[0].files[0]);

        Use - fd.append( 'file', $('#userfile')[0].files[0]);

        1
        • 9
          If you can explain the difference, I'll give you a cookie.
          – ggdx
          CommentedJan 16, 2017 at 11:23

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