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);
}
});
}
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
I can't add a comment above as I do not have enough reputation, but the above answer was nearly perfect for me, except I had to add
type: "POST"
to the .ajax call. I was scratching my head for a few minutes trying to figure out what I had done wrong, that's all it needed and works a treat. So this is the whole snippet:
Full credit to the answer above me, this is just a small tweak to that. This is just in case anyone else gets stuck and can't see the obvious.
$.ajax({
url: 'Your url here',
data: formData,
type: "POST", //ADDED THIS LINE
// THIS MUST BE DONE FOR FILE UPLOADING
contentType: false,
processData: false,
// ... Other options like success and etc
})
<form id="upload_form" enctype="multipart/form-data">
jQuery with CodeIgniter file upload:
var formData = new FormData($('#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.
$(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();
},
});
});
});
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');
});
$('#form-withdraw').submit(function(event) {
//prevent the form from submitting by default
event.preventDefault();
var formData = new FormData($(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");
}
}
});
});
Actually The documentation shows that you can use XMLHttpRequest().send()
to simply send multiform data
in case jquery sucks
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;
...
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.
I have following html
<form id="submit-form">
<input type="file" id="resume" name="resume[]" class="inputFileHidden" multiple>
<input type="submit">
</form>
I am uploading files using ajax using formdata. The thing is that I don't want to send all files in one go using ajax. Instead I want to send single file per ajax request.
To upload files I am using following jQuery code
$('#submit-form').submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
var url = "upload-files.php";
$.ajax({
url: url,
type: 'post',
data: formData,
success: function(response) {
alert(response);
},
cache: false,
contentType: false,
processData: false
})
})
You can just use forEach method of FormData:
$('#submit-form').submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
var url = "upload-files.php";
formData.forEach(function(entry) {
if (entry instanceof File) {
var fileForm = new FormData()
fileForm.append('resume', entry)
$.ajax({
url: url,
type: 'post',
data: fileForm,
success: function(response) {
alert(response);
},
cache: false,
contentType: false,
processData: false
})
}
})
})
below you can see the html code, javascript/jquery and PHP. I'm trying to send a request using AJAX PHP, to send multiple variable to PHP, one of them is FormData and the other is some text or values from inputs etc.
Here's html code:
<div>
<p>Name:</p>
<input id="newP_name"></input>
<p>Product Image</p>
<input type="file" id="img"></input>
<p>Description:</p>
<textarea id="newP_desc"></textarea>
<button name="submitProductBtn" id="submitProductBtn">Submit</button>
Here's my javascript:
$("#submitProductBtn").click(function(){
var file_data = $('#img').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
$.ajax({
url: 'imgParser.php',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data, name: "John",
type: 'POST',
success: function (response) {
alert(response);
},
error: function (response) {
alert("ERROR OCCURED");
}
});
});
As you can see I'm trying to pass to 'data:' the FormData (image) and a second variable name.
Here's PHP code:
<?php
if(isset($_FILES["file"]["name"])){
if($_FILES["file"]["error"] == 0){
move_uploaded_file($_FILES["file"]["tmp_name"], "images/".$_FILES["file"]["name"]);
echo $_POST["name"];
}else{
echo $_POST["name"];
}
}
?>
I would expect it to print "John", but the alert is blank, image showing ->
Any help appreciated,
thanks in advance.
EDIT:
Changed slightly the jQuery code, still outputs empty, but when I had code commented out in PHP (below closing tag), it printed it.
here's my new jquery code:
$("#submitProductBtn").click(function(){
var file_data = $('#img').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('name', "John");
$.ajax({
url: 'imgParser.php',
dataType: 'text',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'POST',
success: function (response) {
alert(response);
},
error: function (response) {
alert("ERROR OCCURED");
}
});
});
You append all data fields to the form data object just like you did with the file
form_data.append('file', file_data);
form_data.append('name', 'John');
I'm not sure what is form_data maybe you can try this,
form_data.append('file', file_data);
form_data.append('name', 'john');
and simply pass form_data in data property
$.ajax({
...
...
data: form_data
I'm trying to upload files with datas using ajax.
here is my html form :
<body>
<input type="text" id="name" value="test" />
<input type="file" id="pic" accept="image/*" />
<input id = "submit" type="submit" />
</body>
when I send the uploaded file alone with ajax it is working using new FormData();
var file_data = $('#pic').prop('files');
var form_data = new FormData();
form_data.append('file', file_data);
alert(form_data);
$.ajax({
url: 'test.php', // point to server-side PHP script
dataType: 'text', // what to expect back from the PHP script, if anything
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
alert(php_script_response);
}
});
but Idon't know how to send the input 'name' with Data
var DATA = 'name='+name;
$.ajax({
url: "test.php",
type: "post",
data: DATA,
success: function (response) {
console.log($response);
},
});
Thanks
Here is mycode
function addPackage(elem)
{
var dataimg = new FormData();
dataimg.append('', $("#browseimg"+elem).prop('files')[0]);
var startdate=$("#from_date"+elem).val();
var enddate=$("#to_date"+elem).val();
$.ajax({
url: "addpackage/",
type:"post",
contentType:false,
data:{startdate:startdate,enddate:enddate,packageid:elem,img:dataimg},
success: function(data) {
}
});
}
I tried post method ajax to upload image and input field data without form. In ajax call it showing [object object]. How to post image with input field without form in jquery ajax?
You can do it like following. Hope this will help you.
function addPackage(elem)
{
var dataimg = new FormData();
dataimg.append('startdate', $("#from_date"+elem).val());
dataimg.append('enddate', $("#to_date"+elem).val());
dataimg.append('packageid', elem);
dataimg.append('img', $("#browseimg"+elem)[0].files[0]);
$.ajax({
url: "addpackage/",
type:"post",
cache : false,
contentType : false,
processType : false,
data: dataimg,
success: function(data) {
}
});
}
You can try this:
Your JS Code:
<script type="text/javascript">
var data = new FormData(document.getElementById("yourFormID")); //your form ID
var url = $("#yourFormID").attr("action"); // action that you mention in form action.
$.ajax({
type: "POST",
url: url,
data: data,
enctype: 'multipart/form-data',
processData: false, // tell jQuery not to process the data
contentType: false, // tell jQuery not to set contentType
dataType: "json",
success: function(response)
{
// some code after succes from php
},
beforeSend: function()
{
// some code before request send if required like LOADING....
}
});
</script>
Dummy HTML:
<form method="post" action="addpackage/" id="yourFormID">
<input type="text" name="firstvalue" value="some value">
<input type="text" name="secondvalue" value="some value">
<input type="file" name="imagevalue">
</form>
in addpackage php file:
print_r($_POST);
print_r($_FILES);