Sending multipart/formdata with jQuery.ajax

Starting with Safari 5/Firefox 4, it’s easiest to use the FormData class:

var data = new FormData();
jQuery.each(jQuery('#file')[0].files, function(i, file) {
    data.append('file-'+i, file);
});

So now you have a FormData object, ready to be sent along with the XMLHttpRequest.

jQuery.ajax({
    url: 'php/upload.php',
    data: data,
    cache: false,
    contentType: false,
    processData: false,
    method: 'POST',
    type: 'POST', // For jQuery < 1.9
    success: function(data){
        alert(data);
    }
});

It’s imperative that you set the contentType option to false, forcing jQuery not to add a Content-Type header for you, otherwise, the boundary string will be missing from it. Also, you must leave the processData flag set to false, otherwise, jQuery will try to convert your FormData into a string, which will fail.

You may now retrieve the file in PHP using:

$_FILES['file-0']

(There is only one file, file-0, unless you specified the multiple attribute on your file input, in which case, the numbers will increment with each file.)

Using the FormData emulation for older browsers

var opts = {
    url: 'php/upload.php',
    data: data,
    cache: false,
    contentType: false,
    processData: false,
    method: 'POST',
    type: 'POST', // For jQuery < 1.9
    success: function(data){
        alert(data);
    }
};
if(data.fake) {
    // Make sure no text encoding stuff is done by xhr
    opts.xhr = function() { var xhr = jQuery.ajaxSettings.xhr(); xhr.send = xhr.sendAsBinary; return xhr; }
    opts.contentType = "multipart/form-data; boundary="+data.boundary;
    opts.data = data.toString();
}
jQuery.ajax(opts);

Create FormData from an existing form

Instead of manually iterating the files, the FormData object can also be created with the contents of an existing form object:

var data = new FormData(jQuery('form')[0]);

Use a PHP native array instead of a counter

Just name your file elements the same and end the name in brackets:

jQuery.each(jQuery('#file')[0].files, function(i, file) {
    data.append('file[]', file);
});

$_FILES['file'] will then be an array containing the file upload fields for every file uploaded. I actually recommend this over my initial solution as it’s simpler to iterate over.

Look at my code, it does the job for me

$( '#formId' )
  .submit( function( e ) {
    $.ajax( {
      url: 'FormSubmitUrl',
      type: 'POST',
      data: new FormData( this ),
      processData: false,
      contentType: false
    } );
    e.preventDefault();
  } );