File Upload without Form

Without using any forms whatsoever, can I just send a file/files from <input type="file"> to 'upload.php' using POST method using jQuery. The input tag is not inside any form tag. It stands individually. So I don't want to use jQuery plugins like 'ajaxForm' or 'ajaxSubmit'.



All answers here are still using the FormData API. It is like a "multipart/form-data" upload without a form. You can also upload the file directly as content inside the body of the POST request using xmlHttpRequest like this:

var xmlHttpRequest = new XMLHttpRequest();

var file = ...file handle...
var fileName = ...file name...
var target =
var mimeType = ...mime type...'POST', target, true);
xmlHttpRequest.setRequestHeader('Content-Type', mimeType);
xmlHttpRequest.setRequestHeader('Content-Disposition', 'attachment; filename="' + fileName + '"');

Content-Type and Content-Disposition headers are used for explaining what we are sending (mime-type and file name).

I posted similar answer also here.


Basing on this tutorial, here a very basic way to do that:

$('your_trigger_element_selector').on('click', function(){    
    var data = new FormData();
    data.append('input_file_name', $('your_file_input_selector').prop('files')[0]);
    // append other variables to data if you want: data.append('field_name_x', field_value_x);

        type: 'POST',               
        processData: false, // important
        contentType: false, // important
        data: data,
        url: your_ajax_path,
        dataType : 'json',  
        // in PHP you can call and process file in the same way as if it was submitted from a form:
        // $_FILES['input_file_name']
        success: function(jsonData){

Don't forget to add proper error handling


Step 1: Create HTML Page where to place the HTML Code.

Step 2: In the HTML Code Page Bottom(footer)Create Javascript: and put Jquery Code in Script tag.

Step 3: Create PHP File and php code copy past. after Jquery Code in $.ajax Code url apply which one on your php file name.


//$(document).on("change", "#avatar", function() {   // If you want to upload without a submit button 
$(document).on("click", "#upload", function() {
  var file_data = $("#avatar").prop("files")[0]; // Getting the properties of file from file field
  var form_data = new FormData(); // Creating object of FormData class
  form_data.append("file", file_data) // Appending parameter named file with properties of file_field to form_data
  form_data.append("user_id", 123) // Adding extra parameters to form_data
    url: "/upload_avatar", // Upload Script
    dataType: 'script',
    cache: false,
    contentType: false,
    processData: false,
    data: form_data, // Setting the data attribute of ajax with file_data
    type: 'post',
    success: function(data) {
      // Do something after Ajax completes 


<input id="avatar" type="file" name="avatar" />
<button id="upload" value="Upload" />



Sorry for being that guy but AngularJS offers a simple and elegant solution.

Here is the code I use:

ngApp.controller('ngController', ['$upload',
function($upload) {

  $scope.Upload = function($files, index) {
    for (var i = 0; i < $files.length; i++) {
      var file = $files[i];
      $scope.upload = $upload.upload({
        file: file,
        url: '/File/Upload',
        data: {
          id: 1 //some data you want to send along with the file,
          name: 'ABC' //some data you want to send along with the file,

      }).progress(function(evt) {

      }).success(function(data, status, headers, config) {
          alert('Upload done');
    .error(function(message) {
      alert('Upload failed');
.Hidden {
  display: none
<script src=""></script>
<script src=""></script>

<div data-ng-controller="ngController">
  <input type="button" value="Browse" onclick="$(this).next().click();" />
  <input type="file" ng-file-select="Upload($files, 1)" class="Hidden" />

On the server side I have an MVC controller with an action the saves the files uploaded found in the Request.Files collection and returning a JsonResult.

If you use AngularJS try this out, if you don't... sorry mate :-)


Try this puglin simpleUpload, no need form


<input type="file" name="arquivo" id="simpleUpload" multiple >
<button type="button" id="enviar">Enviar</button>


  url: 'upload.php',
  trigger: '#enviar',
  success: function(data){
    alert('Envio com sucesso');


You can use FormData to submit your data by a POST request. Here is a simple example:

var myFormData = new FormData();
myFormData.append('pictureFile', pictureInput.files[0]);

  url: 'upload.php',
  type: 'POST',
  processData: false, // important
  contentType: false, // important
  dataType : 'json',
  data: myFormData

You don't have to use a form to make an ajax request, as long as you know your request setting (like url, method and parameters data).


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