I am working with ASP.Net Core 2.1, and trying to upload a file while returning it's url, without refreshing the page.
I am trying to write the JavaScript in site.js as the _RenderPartial("scripts") renders all scripts at the end of the page and hence directly using script tag in the razor view is not working. Secondly, adding it to site.js gives me an opportunity to call the script across the site views.
My Controller action looks like :
[HttpPost]
[DisableRequestSizeLimit]
public async Task<IActionResult> Upload()
{
// Read & copy to stream the content of MultiPart-Form
// Return the URL of the uploaded file
return Content(FileName);
}
My view looks like :
<form id="FileUploadForm" action="~/Resources/Upload" method="post" enctype="multipart/form-data">
<input name="uploadfile" type="file" />
<button name="uploadbtn" type="submit" onclick="SubmitForm(this.parentElement, event)">Upload</button>
The site.js currently looks like :
function SubmitForm(form, caller) {
caller.preventDefault();
$.ajax(
{
type: form.method,
url: form.action,
data: form.serialize(),
success: function (data) { alert(data); },
error: function (data) { alert(data); }
})}
Presently, the code bypasses the entire script and the file is uploaded and new view displaying the file name is returned. I need help to create the javascript.
Unfortunately the jQuery serialize() method will not include input file elements. So the file user selected is not going to be included in the serialized value (which is basically a string).
What you may do is, create a FormData object, append the file(s) to that. When making the
ajax call, you need to specify processData and contentType property values to false
<form id="FileUploadForm" asp-action="Upload" asp-controller="Home"
method="post" enctype="multipart/form-data">
<input id="uploadfile" type="file" />
<button name="uploadbtn" type="submit">Upload</button>
</form>
and here in the unobutrusive way to handle the form submit event where we will stop the regular behavior and do an ajax submit instead.
$(function () {
$("#FileUploadForm").submit(function (e) {
e.preventDefault();
console.log('Doing ajax submit');
var formAction = $(this).attr("action");
var fdata = new FormData();
var fileInput = $('#uploadfile')[0];
var file = fileInput.files[0];
fdata.append("file", file);
$.ajax({
type: 'post',
url: formAction,
data: fdata,
processData: false,
contentType: false
}).done(function (result) {
// do something with the result now
console.log(result);
if (result.status === "success") {
alert(result.url);
} else {
alert(result.message);
}
});
});
})
Assuming your server side method has a parameter of with name same as the one we used when we created the FormData object entry(file). Here is a sample where it will upload the image to the uploads directory inside wwwwroot.
The action method returns a JSON object with a status and url/message property and you can use that in the success/done handler of the ajax call to whatever you want to do.
public class HomeController : Controller
{
private readonly IHostingEnvironment hostingEnvironment;
public HomeController(IHostingEnvironment environment)
{
_context = context;
hostingEnvironment = environment;
}
[HttpPost]
public async Task<IActionResult> Upload(IFormFile file)
{
try
{
var uniqueFileName = GetUniqueFileName(file.FileName);
var uploads = Path.Combine(hostingEnvironment.WebRootPath, "uploads");
var filePath = Path.Combine(uploads, uniqueFileName);
file.CopyTo(new FileStream(filePath, FileMode.Create));
var url = Url.Content("~/uploads/" + uniqueFileName);
return Json(new { status = "success", url = url });
}
catch(Exception ex)
{
// to do : log error
return Json(new { status = "error", message = ex.Message });
}
}
private string GetUniqueFileName(string fileName)
{
fileName = Path.GetFileName(fileName);
return Path.GetFileNameWithoutExtension(fileName)
+ "_"
+ Guid.NewGuid().ToString().Substring(0, 4)
+ Path.GetExtension(fileName);
}
}
Sharing the code that worked for me, implementing #Shyju's answer.
View ( Razor Page ):
<form name="UploadForm" action="~/Resources/Upload" method="post" enctype="multipart/form-data">
<input name="uploadfile" type="file" />
<button name="uploadbtn" type="submit" onclick="SubmitForm(this.parentElement, event)">Upload</button>
AJAX code added in Site.js (to make it a reusable):
// The function takes Form and the event object as parameter
function SubmitForm(frm, caller) {
caller.preventDefault();
var fdata = new FormData();
var file = $(frm).find('input:file[name="uploadfile"]')[0].files[0];
fdata.append("file", file);
$.ajax(
{
type: frm.method,
url: frm.action,
data: fdata,
processData: false,
contentType: false,
success: function (data) {
alert(data);
},
error: function (data) {
alert(data);
}
})
};
if you want to submit the form without using ajax request
var form = document.getElementById('formId');
form.submit();
When I do this way I am getting base64 encoded image. I need to just upload the file. How can I change the code
<script>
submitBox = new Vue({
el: "#submitBox",
data: {
username: '',
category: '',
subcategory: [],
image: '',
},
methods: {
onFileChange(e) {
var files = e.target.files || e.dataTransfer.files;
if (!files.length)
return;
this.createImage(files[0]);
},
createImage(file) {
var image = new Image();
var reader = new FileReader();
var vm = this;
reader.onload = (e) => {
vm.image = e.target.result;
};
reader.readAsDataURL(file);
},
handelSubmit: function(e) {
var vm = this;
data = {};
data['username'] = this.username;
data['category'] = this.category;
data['subcategory'] = this.subcategory;
data['image'] = this.image;
$.ajax({
url: 'http://127.0.0.1:8000/api/add/post/',
data: data,
type: "POST",
dataType: 'json',
success: function(e) {
if (e.status) {
alert("Registration Success")
window.location.href = "https://localhost/n2s/registersuccess.html";
} else {
vm.response = e;
alert("Registration Failed")
}
}
});
return false;
}
},
});
</script>
My html code is
<div id="submitBox">
<form method="POST" onSubmit="return false;" data-parsley-validate="true" v-on:submit="handelSubmit($event);">
<input name="username" type="text" class="form-control" id="name" placeholder="Name" required="required" v-model="username" data-parsley-minlength="4"/>
<select title="Select" v-model="category" name="category" ref="category">
<option v-for="post in articles" v-bind:value="post.name" >{{post.name}}</option>
</select>
<input class="form-control" type="file" id="property-images" #change="onFileChange">
</form>
</div>
How can I able to upload images without base64 encoding? When I am doing this way I am only able to upload image in base64 format. I need just file upload?
Remove onSubmit attribute (that should really be spelled onsubmit (lowercased)
Use vues owns event modifiers v-on:submit.prevent="handelSubmit"
Remove return false; in handelSubmit function
You are say "images" like it means plural?
So maybe you wanna add the multiple attribute to the file input?
<input type="file" multiple>
To me this all really seems like you want to turn a post request of a from into a ajax request. All the information is already in the form so you could just easily use the FormData constructor and select the form element. You just need to set the name attribute to the file input as well.
<input type="file" name="image" multiple>
But I don't see the subcategory which you might need to add manually.
So here is what i would have done:
handelSubmit(e) {
var form = e.target; // Get hold of the form element from the event
var fd = new FormData(form); // create a FormData
// Add the subcategory
fd.append('subcategory', this.subcategory.join(', '));
// or do this to get more of a array like:
// (depends upon server logic)
//
// this.subcategory.forEach(function(category){
// fd.append('subcategory', category);
// })
// to inspect what the formdata contains
// better to remove in production
// spread operator isn't cross browser compitable
console.log(...fd);
$.ajax({
url: 'http://127.0.0.1:8000/api/add/post/',
data: fd, // data is a FormData instance
type: "POST",
processData: false, // stop jQuery's transformation
contentType: false, // stop jQuery's from setting wrong content-type header
success(e) {
if (e.status) {
alert("Registration Success")
window.location.href = "https://localhost/n2s/registersuccess.html";
} else {
vm.response = e;
alert("Registration Failed")
}
}
});
// For persornal reason i don't like how
// jQuery are unhandel to accept a FormData
//
// So here are some insporation if you can use the new fetch instead
/*
fetch('http://127.0.0.1:8000/api/add/post/', {method: 'post', body: fd})
.then(function(res){
// Here you have only recived the headers
console.log(res.ok) // true for 2xx responses, false otherwise
return res.text() // or .json()
})
.then(function(text){
// Here you have recived the hole response
console.log(text)
})
*/
}
I fail to see the reason why you would need the createImage and the onFileChange functions.
The thing you also have to notices is that when using jQuery's ajax and and Formdata is that you need to set both contentType and processData to false otherwise jQuery will do incorrect things to the request
This will change the post from a json payload to a multipart payload which is the best choice for uploading files... you save more bandwidth and need to do less on the server
I have a working modal window that saves an entity to the database, my question is how to send files from the modal?
Here as my current JavaScript function:
$('#btn-save').on('click', function () {
$.ajax({
url: '#Url.Action("CreateModal", "Suggestions")',
type: 'POST',
data: $('#modal-form').serialize(),
success: function (data) {
if (data.success == true) {
$('#suggestion-modal').modal('hide');
location.reload(true);
} else {
$('.modal-content').html(data);
$('#suggestion-modal').modal('show');
}
}
});
});
This part does not send data, but works fine when not using modal and using standard MVC Create template:
<form id="modal-form" method="post" enctype="multipart/form-data">
#Html.AntiForgeryToken()
#Html.ValidationSummary(true, "", new { #class = "text-danger" })
#Html.HiddenFor(model => model.UserId)
<div>
#Html.LabelFor(model => model.Title)
#Html.EditorFor(model => model.Title)
#Html.ValidationMessageFor(model => model.Title, "")
</div>
<div>
#Html.LabelFor(model => model.Images, "Images")
<input type="file" name="upload" multiple />
</div>
<div>
<input id="btn-save" type="button" value="" />
</div>
</form>
I've left the rest of the partial view out as that all works correctly.
EDIT: Just added where my button was in the form. It was there, I just removed much of the code not relevant to the question - should have left the button in. Also added extra model properties - These must be sent with the file, including validation token.
EDIT: Many thanks to Jasen. I've included the JavaScript below for anyone else struggling with this.
$('#btn-save').on('click', function () {
var formData = new FormData($('form').get(0));
$.ajax({
url: '#Url.Action("CreateModal", "Suggestions")',
type: 'POST',
processData: false,
contentType: false,
data: formData,
success: function (data) {
if (data.success == true) {
$('#suggestion-modal').modal('hide');
location.reload(true);
} else {
$('.modal-content').html(data);
$('#suggestion-modal').modal('show');
}
}
});
});
Im trying to post data with ajax but in console i recive something like.. Uncaught ReferenceError: file is not defined
This is my input with my button to upload data:
<div id='file_browse_wrapper'>
<input class="iconos" type="file" name="File Upload" id='file_browse' accept=".csv" />
</div>
<label id="url-archivo"><b>Selecciona tu archivo...</b></label>
<button class="btn btn-success" id="carga-archivo">Subir</button>
This is my jquery/ajax:
$("#carga-archivo").click(function () {
var fileUpload = document.getElementById("file_browse");
if (fileUpload .value != null) {
var uploadFile = new FormData();
var files = $("#file_browse").get(0).files;
// Add the uploaded file content to the form data collection
if (files.length > 0) {
uploadFile.append('file-'+i, file);
$.ajax({
url: "/2/delivery/client/massive",
contentType: false,
processData: false,
data: uploadFile,
type: 'POST',
success: function () {
alert("file upload successfuly");
}
});
}
}
});
But the data is not POST i think is something missing
You have created a variable called files, but are using a variable called file. Try changing this portion of your code:
var file = $("#file_browse").get(0).files;
// Add the uploaded file content to the form data collection
if (file.length > 0) {
uploadFile.append('file-'+i, file);
$.ajax({
url: "/2/delivery/client/massive",
contentType: false,
processData: false,
data: uploadFile,
type: 'POST',
success: function () {
alert("file upload successfuly");
}
});
}
This just renames the files variable to file and changes the one other place that used files.
I'm using jQuery and Ajax for my forms to submit data and files but I'm not sure how to send both data and files in one form?
I currently do almost the same with both methods but the way in which the data is gathered into an array is different, the data uses .serialize(); but the files use = new FormData($(this)[0]);
Is it possible to combine both methods to be able to upload files and data in one form through Ajax?
Data jQuery, Ajax and html
$("form#data").submit(function(){
var formData = $(this).serialize();
$.ajax({
url: window.location.pathname,
type: 'POST',
data: formData,
async: false,
success: function (data) {
alert(data)
},
cache: false,
contentType: false,
processData: false
});
return false;
});
<form id="data" method="post">
<input type="text" name="first" value="Bob" />
<input type="text" name="middle" value="James" />
<input type="text" name="last" value="Smith" />
<button>Submit</button>
</form>
Files jQuery, Ajax and html
$("form#files").submit(function(){
var formData = new FormData($(this)[0]);
$.ajax({
url: window.location.pathname,
type: 'POST',
data: formData,
async: false,
success: function (data) {
alert(data)
},
cache: false,
contentType: false,
processData: false
});
return false;
});
<form id="files" method="post" enctype="multipart/form-data">
<input name="image" type="file" />
<button>Submit</button>
</form>
How can I combine the above so that I can send data and files in one form via Ajax?
My aim is to be able to send all of this form in one post with Ajax, is it possible?
<form id="datafiles" method="post" enctype="multipart/form-data">
<input type="text" name="first" value="Bob" />
<input type="text" name="middle" value="James" />
<input type="text" name="last" value="Smith" />
<input name="image" type="file" />
<button>Submit</button>
</form>
The problem I had was using the wrong jQuery identifier.
You can upload data and files with one form using ajax.
PHP + HTML
<?php
print_r($_POST);
print_r($_FILES);
?>
<form id="data" method="post" enctype="multipart/form-data">
<input type="text" name="first" value="Bob" />
<input type="text" name="middle" value="James" />
<input type="text" name="last" value="Smith" />
<input name="image" type="file" />
<button>Submit</button>
</form>
jQuery + Ajax
$("form#data").submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
url: window.location.pathname,
type: 'POST',
data: formData,
success: function (data) {
alert(data)
},
cache: false,
contentType: false,
processData: false
});
});
Short Version
$("form#data").submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.post($(this).attr("action"), formData, function(data) {
alert(data);
});
});
another option is to use an iframe and set the form's target to it.
you may try this (it uses jQuery):
function ajax_form($form, on_complete)
{
var iframe;
if (!$form.attr('target'))
{
//create a unique iframe for the form
iframe = $("<iframe></iframe>").attr('name', 'ajax_form_' + Math.floor(Math.random() * 999999)).hide().appendTo($('body'));
$form.attr('target', iframe.attr('name'));
}
if (on_complete)
{
iframe = iframe || $('iframe[name="' + $form.attr('target') + '"]');
iframe.load(function ()
{
//get the server response
var response = iframe.contents().find('body').text();
on_complete(response);
});
}
}
it works well with all browsers, you don't need to serialize or prepare the data.
one down side is that you can't monitor the progress.
also, at least for chrome, the request will not appear in the "xhr" tab of the developer tools but under "doc"
I was having this same issue in ASP.Net MVC with HttpPostedFilebase and instead of using form on Submit I needed to use button on click where I needed to do some stuff and then if all OK the submit form so here is how I got it working
$(".submitbtn").on("click", function(e) {
var form = $("#Form");
// you can't pass Jquery form it has to be javascript form object
var formData = new FormData(form[0]);
//if you only need to upload files then
//Grab the File upload control and append each file manually to FormData
//var files = form.find("#fileupload")[0].files;
//$.each(files, function() {
// var file = $(this);
// formData.append(file[0].name, file[0]);
//});
if ($(form).valid()) {
$.ajax({
type: "POST",
url: $(form).prop("action"),
//dataType: 'json', //not sure but works for me without this
data: formData,
contentType: false, //this is requireded please see answers above
processData: false, //this is requireded please see answers above
//cache: false, //not sure but works for me without this
error : ErrorHandler,
success : successHandler
});
}
});
this will than correctly populate your MVC model, please make sure in your Model, The Property for HttpPostedFileBase[] has the same name as the Name of the input control in html i.e.
<input id="fileupload" type="file" name="UploadedFiles" multiple>
public class MyViewModel
{
public HttpPostedFileBase[] UploadedFiles { get; set; }
}
Or shorter:
$("form#data").submit(function() {
var formData = new FormData(this);
$.post($(this).attr("action"), formData, function() {
// success
});
return false;
});
EDIT: with the new version of JQuery (3.6), you could also try using contentType function argument instead of enctype. Try contentType: multipart/form-data.
For me, it didn't work without enctype: 'multipart/form-data' field in the Ajax request. I hope it helps someone who is stuck in a similar problem.
Even though the enctype was already set in the form attribute, for some reason, the Ajax request didn't automatically identify the enctype without explicit declaration (jQuery 3.3.1).
// Tested, this works for me (jQuery 3.3.1)
fileUploadForm.submit(function (e) {
e.preventDefault();
$.ajax({
type: 'POST',
url: $(this).attr('action'),
enctype: 'multipart/form-data',
data: new FormData(this),
processData: false,
contentType: false,
success: function (data) {
console.log('Thank God it worked!');
}
}
);
});
// enctype field was set in the form but Ajax request didn't set it by default.
<form action="process/file-upload" enctype="multipart/form-data" method="post" >
<input type="file" name="input-file" accept="text/plain" required>
...
</form>
As others mentioned above, please also pay special attention to the contentType and processData fields.
A Simple but more effective way:
new FormData() is itself like a container (or a bag). You can put everything attr or file in itself.
The only thing you'll need to append the attribute, file, fileName eg:
let formData = new FormData()
formData.append('input', input.files[0], input.files[0].name)
and just pass it in AJAX request. Eg:
let formData = new FormData()
var d = $('#fileid')[0].files[0]
formData.append('fileid', d);
formData.append('inputname', value);
$.ajax({
url: '/yourroute',
method: 'POST',
contentType: false,
processData: false,
data: formData,
success: function(res){
console.log('successfully')
},
error: function(){
console.log('error')
}
})
You can append n number of files or data with FormData.
and if you're making AJAX Request from Script.js file to Route file in Node.js beware of using
req.body to access data (ie text)
req.files to access file (ie image, video etc)
The code below works for me
$(function () {
debugger;
document.getElementById("FormId").addEventListener("submit", function (e) {
debugger;
if (ValidDateFrom()) { // Check Validation
var form = e.target;
if (form.getAttribute("enctype") === "multipart/form-data") {
debugger;
if (form.dataset.ajax) {
e.preventDefault();
e.stopImmediatePropagation();
var xhr = new XMLHttpRequest();
xhr.open(form.method, form.action);
xhr.onreadystatechange = function (result) {
debugger;
if (xhr.readyState == 4 && xhr.status == 200) {
debugger;
var responseData = JSON.parse(xhr.responseText);
SuccessMethod(responseData); // Redirect to your Success method
}
};
xhr.send(new FormData(form));
}
}
}
}, true);
});
In your Action Post Method, pass parameter as HttpPostedFileBase UploadFile and make sure your file input has same as mentioned in your parameter of the Action Method.
It should work with AJAX Begin form as well.
Remember over here that your AJAX BEGIN Form will not work over here since you make your post call defined in the code mentioned above and you can reference your method in the code as per the Requirement
I know I am answering late but this is what worked for me
Just to remind, in 2022 you don't need to use jquery. Try js standard Fetch API
var formData = new FormData(this);
fetch(url, {
method: 'POST',
body: formData
})
.then(response => {
if(response.ok) {
//success
alert(response);
} else {
throw Error('Server error');
}
})
.catch(error => {
console.log('fail', error);
});
This is a solution that I implemented
var formData = new FormData();
var files = $('input[type=file]');
for (var i = 0; i < files.length; i++) {
if (files[i].value == "" || files[i].value == null) {
return false;
}
else {
formData.append(files[i].name, files[i].files[0]);
}
}
var formSerializeArray = $("#Form").serializeArray();
for (var i = 0; i < formSerializeArray.length; i++) {
formData.append(formSerializeArray[i].name, formSerializeArray[i].value)
}
$.ajax({
type: 'POST',
data: formData,
contentType: false,
processData: false,
cache: false,
url: '/Controller/Action',
success: function (response) {
if (response.Success == true) {
return true;
}
else {
return false;
}
},
error: function () {
return false;
},
failure: function () {
return false;
}
});
---Solution for DOT NET CORE MVC Implementation---
While looking at this question I though I should right .NET CORE implementation for this because the question is not specific to any backend language.
So guys here is the standalone implementation example.
Objective :- To submit form fields including files and how we can get data in a single model at backend
HTML Code / View Code - Views/Home/Index.cshtml
#{
ViewData["Title"] = "Home Page";
}
<input type="file" id="FileUpload1" multiple />
<div>
<label>Enter First Name :</label>
<input type="text" id="nameText" maxlength="50" />
</div>
<input type="button" id="btnUpload" value="Submit Form with Files" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
$(document).ready(function () {
$('#btnUpload').click(function () {
// Checking whether FormData is available in browser
if (window.FormData !== undefined) {
var fileUpload = $("#FileUpload1").get(0);
var files = fileUpload.files;
// Create FormData object
var fileData = new FormData();
// Looping over all files and add it to FormData object
for (var i = 0; i < files.length; i++) {
fileData.append("files", files[i]);
}
// Adding one more key to FormData object
fileData.append('FirstName', $("#nameText").val());
$.ajax({
url: '/Home/UploadFiles',
type: "POST",
contentType: false, // Not to set any content header
processData: false, // Not to process data
data: fileData,
success: function (result) {
alert(result);
},
error: function (err) {
alert(err.statusText);
}
});
} else {
alert("FormData is not supported.");
}
});
});
</script>
Backend Code / Controller action method Controllers/HomeController.cs
public class HomeController : Controller
{
private readonly ILogger<HomeController> _logger;
private readonly IWebHostEnvironment _environment;
public HomeController(ILogger<HomeController> logger, IWebHostEnvironment environment)
{
_logger = logger;
_environment = environment;
}
public IActionResult Index()
{
return View();
}
public IActionResult Privacy()
{
return View();
}
[HttpPost]
public async Task<IActionResult> UploadFiles(MyForm myForm)
{
var files = myForm.Files;
// First Name
string name = myForm.FirstName;
// check All files
foreach (IFormFile source in files)
{
string filename = ContentDispositionHeaderValue.Parse(source.ContentDisposition).FileName.Trim('"');
filename = this.EnsureCorrectFilename(filename);
string fileWithPath = this.GetPathAndFilename(filename);
// Create directory if not exist
Directory.CreateDirectory(Path.GetDirectoryName(fileWithPath));
using (FileStream output = System.IO.File.Create(fileWithPath))
await source.CopyToAsync(output);
}
return Ok("Success");
}
[ResponseCache(Duration = 0, Location = ResponseCacheLocation.None, NoStore = true)]
public IActionResult Error()
{
return View(new ErrorViewModel { RequestId = Activity.Current?.Id ?? HttpContext.TraceIdentifier });
}
public class MyForm
{
public string FirstName { get; set; }
public IList<IFormFile> Files { get; set; }
}
private string EnsureCorrectFilename(string filename)
{
if (filename.Contains("\\"))
filename = filename.Substring(filename.LastIndexOf("\\") + 1);
return filename;
}
private string GetPathAndFilename(string filename)
{
return Path.Combine(_environment.ContentRootPath, "uploadedFiles", filename);
}
}
Full Source Code Repo: https://github.com/rj-learning/DotNetCoreFileUpload
In my case I had to make a POST request, which had information sent through the header, and also a file sent using a FormData object.
I made it work using a combination of some of the answers here, so basically what ended up working was having this five lines in my Ajax request:
contentType: "application/octet-stream",
enctype: 'multipart/form-data',
contentType: false,
processData: false,
data: formData,
Where formData was a variable created like this:
var file = document.getElementById('uploadedFile').files[0];
var form = $('form')[0];
var formData = new FormData(form);
formData.append("File", file);
you can just append them on your formdata, add your files and datas in it.you can read this..
https://developer.mozilla.org/en-US/docs/Web/API/FormData/append
for better understanding. you can separately retrieve them $_FILES for your files and $_POST for your data.
<form id="form" method="post" action="otherpage.php" enctype="multipart/form-data">
<input type="text" name="first" value="Bob" />
<input type="text" name="middle" value="James" />
<input type="text" name="last" value="Smith" />
<input name="image" type="file" />
<button type='button' id='submit_btn'>Submit</button>
</form>
<script>
$(document).on("click", "#submit_btn", function (e) {
//Prevent Instant Click
e.preventDefault();
// Create an FormData object
var formData = $("#form").submit(function (e) {
return;
});
//formData[0] contain form data only
// You can directly make object via using form id but it require all ajax operation inside $("form").submit(<!-- Ajax Here -->)
var formData = new FormData(formData[0]);
$.ajax({
url: $('#form').attr('action'),
type: 'POST',
data: formData,
success: function (response) {
console.log(response);
},
contentType: false,
processData: false,
cache: false
});
return false;
});
</script>
///// otherpage.php
<?php
print_r($_FILES);
?>