Submitting multipart/form-data via Ajax breaks HttpPostedFileBase submission [duplicate] - javascript

How do I pass a whole set model object through formdata and convert it to model type in the controller?
Below is what I've tried!
JavaScript part:
model = {
EventFromDate: fromDate,
EventToDate: toDate,
ImageUrl: imgUrl,
HotNewsDesc: $("#txthtDescription").val().trim(),
};
formdata.append("model",model);
then pass it through AJAX, it will be a string, and if I check the value of Request.Form["model"] the result will be same, that is it will be received as string and value will be "[object object]"
Is there any way to pass model through formdata and receive it in the controller?

If your view is based on a model and you have generated the controls inside <form> tags, then you can serialize the model to FormData using
var formdata = new FormData($('form').get(0));
This will also include any files generated with <input type="file" name="myImage" .../>
and post it back using
$.ajax({
url: '#Url.Action("YourActionName", "YourControllerName")',
type: 'POST',
data: formdata,
processData: false,
contentType: false,
});
and in your controller
[HttpPost]
public ActionResult YourActionName(YourModelType model)
{
}
or (if your model does not include a property for HttpPostedFileBase)
[HttpPost]
public ActionResult YourActionName(YourModelType model, HttpPostedFileBase myImage)
{
}
If you want to add additional information that is not in the form, then you can append it using
formdata.append('someProperty', 'SomeValue');

If you want to send Form data using Ajax.This is the way to send
var formData = new FormData();
//File Upload
var totalFiles = document.getElementById("Iupload").files.length;
for (var i = 0; i < totalFiles; i++) {
var file = document.getElementById("Iupload").files[i];
formData.append("Document", file);
}
formData.append("NameCode", $('#SelecterID').val());
formData.append("AirLineCode", $('#SelecterID').val());
$.ajax({
url: "/Controller/ActionName",
type: "POST",
dataType: "JSON",
data: formData,
contentType: false,
processData: false,
success: function (result) {
}
})

Using Pure Javascript, considering you have
<form id="FileUploadForm">
<input id="textInput" type="text" />
<input id="fileInput" type="file" name="fileInput" multiple>
<input type="submit" value="Upload file" />
</form>
JS
document.getElementById('FileUploadForm').onsubmit = function () {
var formdata = new FormData(); //FormData object
var fileInput = document.getElementById('fileInput');
//Iterating through each files selected in fileInput
for (i = 0; i < fileInput.files.length; i++) {
//Appending each file to FormData object
formdata.append(fileInput.files[i].name, fileInput.files[i]);
}
//text value
formdata.append("textvalue",document.getElementById("textInput").value);
//Creating an XMLHttpRequest and sending
var xhr = new XMLHttpRequest();
xhr.open('POST', '/Home/UploadFiles');
xhr.send(formdata); // se
xhr.onreadystatechange = function () {
if (xhr.readyState == 4 && xhr.status == 200) {
//on success alert response
alert(xhr.responseText);
}
}
return false;
}
in your C# controller you can get values it as below
[HttpPost]
public ActionResult UploadFiles(YourModelType model, HttpPostedFileBase fileInput)
{
//save data in db
}
Reference : File Uploading using jQuery Ajax or Javascript in MVC

In view side ,if you are using ajax then,
$('#button_Id').on('click', function(){
var Datas=JSON.stringify($('form').serialize());
$.ajax({
type: "POST",
contentType: "application/x-www-form-urlencoded; charset=utf-8",
url: '#Url.Action("ActionName","ControllerName")',
data:Datas,
cache: false,
dataType: 'JSON',
async: true,
success: function (data) {
},
});
});
In Controller side,
[HttpPost]
public ActionResult ActionName(ModelName modelObj)
{
//Some code here
}

Related

POST file list to controller in ajax post

I tried to upload documents but I couldnt pass to list my controllers parameter. My scenario is:
user click to "choose file" button and pick the files and press done
then my some functions get file list and pass to controller for save locally via POST merhod like below:
view side: (get file list)
function saveDocuments(documentList) {
if (documentList.length > 0)
{
var formList = new Array;
for (var i = 0; i < documentList.length; i++) {
var form = new FormData();
var file = documentList[i];
form.append('FormFile', file);
formList.push(form);
}
savePhysicalFile(formList);
}
}
view side: (post file list)
function savePhysicalFile(formData)
{
if (formData != null)
{
$.ajax({
url: "Installation/SavePhysicalPath",
type: 'POST',
dataType: "json",
contentType: "multipart/form-data",
data:formData,
processData: false,
contentType: false,
success: function (result) {
console.log("Success", result);
},
error: function (data) {
console.log(data);
}
});
}
}
In my controller side; the parameter "model" is always null. I couldnt pass view side list here. How can I figure out ?
controller side
public JsonResult SavePhysicalPath([FromForm] List<FileModel> model)
{
var savingRootPath = #"C:\MyDocuments";
//I'm doing save locally
return Json(savingRootPath);
}
model side
public class FileModel
{
public string Files { get; set; }
public IFormFile FormFile { get; set; }
}
From your code,you may pay attention to two things here:
1.For each property of the complex type, model binding looks through the sources for the name pattern prefix.property_name. If nothing is found, it looks for just property_name without the prefix.For model you receive in backend is a List,you need give the name like:[index].FormFile or model[index].FormFile.
2.Your model has a IFormFile and your action receives a list model,if you only pass the IFormFile you need remove FromForm attribute and be sure do not have [ApiController].It is a known github issue and this has been moved to Next sprint planning milestone.
Here is a whole working demo:
View:
<input type="file" multiple onchange="saveDocuments(this.files)"/>
<div class="form-group">
<input type="button" value="Submit" id="submit" class="btn btn-primary" />
</div>
#section Scripts
{
<script>
function saveDocuments(documentList) {
if (documentList.length > 0) {
var form = new FormData();
for (var i = 0; i < documentList.length; i++) {
var file = documentList[i];
//change here
form.append('model['+i+'].FormFile', file);
}
savePhysicalFile(form);
}
}
function savePhysicalFile(formData) {
if (formData != null) {
$.ajax({
url: "/Installation/SavePhysicalPath",
type: 'POST',
dataType: "json",
contentType: "multipart/form-data",
data: formData,
processData: false,
contentType: false,
success: function (result) {
console.log("Success", result);
},
error: function (data) {
console.log(data);
}
});
}
}
</script>
}
Controller:
[HttpPost]
public JsonResult SavePhysicalPath(List<FileModel> model)
{
var savingRootPath = #"C:\MyDocuments";
//I'm doing save locally
return Json(savingRootPath);
}
Result:

How can I send a viewmodel to controller through Ajax

The user is able to write text and upload a file.
Using Javascript, I create an object with the file and text as properties, then send it to the controller using Ajax.
Using debugging, it seems that the problem is in the Ajax part. Everything before that(file upload, object creation) seems to be fine.
Here is the Javascript/JQuery:
function AddPost() {
var fileInput = document.getElementById('file').files;
var textInput = document.getElementById('addPostText').value;
if (fileInput != null || textInput != null) {
file = null;
if (fileInput.length > 0) {
file = fileInput[length-1];
}
var AddPostViewModel =
{
"Text": textInput,
"File": file
};
$.ajax(
{
url: "#Url.Action('AddPost')",
type: "POST",
dataType: "json",
data: JSON.stringify(AddPostViewModel),
contentType: 'application/json; charset=utf-8'
}
);
}
}
Here is the method in the controller:
[HttpPost]
public void AddPost(AddPostViewModel viewModel)
{
}
And here is the object:
public class AddPostViewModel
{
[DataType(DataType.MultilineText)]
public string Text { get; set; }
public IFormFile File { get; set; }
}
The HTML:
<textarea id="addPostText"></textarea>
<div class="upload-button">
<div class="label">Add image</div>
<input id="file" type="file" size="1" />
</div>
You can't send the File inside a json. Also, you've a bug in line file = fileInput[length-1]; (length is wrong, should be 0 or fileInput.length-1.
Here is a working function. This function should be inside a Razor Page because it needs to build the form action url #Url.Action("AddPost").
function AddPost() {
var fileInput = document.getElementById('file').files;
var textInput = document.getElementById('addPostText').value;
if (fileInput != null || textInput != null) {
var file = null;
if (fileInput.length > 0) {
file = fileInput[0];
}
var form = new FormData();
form.append("Text", textInput);
form.append("File", file);
$.ajax(
{
url: '#Url.Action("AddPost")',
type: "POST",
data: form,
processData: false,
contentType: false
}
);
}
}
EDIT Here is a full demo using the tag helpers asp-action, asp-for and a small JS/jQuery code to intercept the form submission and use Ajax.
1: Controller.
public class FileController : Controller
{
public IActionResult AddPost()
{
return View();
}
[HttpPost]
public void AddPost(AddPostViewModel viewModel)
{
//process the viewModel here
}
}
2: View
#model AddPostViewModel
<form asp-action="AddPost" id="theForm">
<textarea asp-for="Text"></textarea>
<input asp-for="File" size="1" />
<button type="submit">Send</button>
</form>
#section Scripts{
<script>
$("#theForm").submit(function(event) {
event.preventDefault();
$.ajax(
{
url: $(this).prop("action"),
type: "POST",
data: new FormData(this),
processData: false,
contentType: false,
success: function() {
console.log("form sent!");
}
}
);
});
</script>
}
You cant stringify a file data and you cant send file as JSON. You should use FormData to send file with ajax request.
Try to send request as follow
function AddPost() {
var fileInput = document.getElementById('file').files;
var textInput = document.getElementById('addPostText').value;
if (fileInput != null || textInput != null) {
file = null;
if (fileInput.length > 0) {
file = fileInput[length-1];
}
var AddPostViewModel =
{
"Text": textInput,
"File": file
};
var formData = new FormData();
Object.getOwnPropertyNames(AddPostViewModel).forEach(function ( attr, index ) {
var value = AddPostViewModel[attr];
formData.append(attr, value)
});
$.ajax(
{
url: "#Url.Action('AddPost')",
type: "POST",
contentType: false,
processData: false,
data: formData,
}
);
}
}

Submit a Form using AJAX in ASP.Net Core MVC

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();

Upload file using MVC Razor from Ajax - Beginner logic issue

I am trying to upload a file using AJAX using C#-Razor. When I submit by clicking on the button the controller method is not being executed. How can I solve this ?
My code is as follows:
View
<div class="form-group">
#Html.TextBoxFor(model => model.IMG, new { #class = "control-label col-md-12", type = "file", placeholder = "Industry", name = "files[]", id="FileUpload" })
#Html.LabelFor(model => model.IMG, new { #class = "col-md-12 " })
#Html.ValidationMessageFor(model => model.IMG)
</div>
<div class="form-group">
<div class="col-md-offset-2 col-md-10">
<input type="button" value="Create" class="btn btn-default" id="UseShipAddr" />
</div>
</div>
AJAX
$('#UseShipAddr').click(function () {
var formData = new FormData();
var totalFiles = document.getElementById("FileUpload").files.length;
for (var i = 0; i < totalFiles; i++) {
var file = document.getElementById("FileUpload").files[i];
formData.append("IMG", file);
alert("h" + file);
}
formData.append("name", "James");
formData.append("age", "1");
$.ajax({
url: "/Post/New",
type: "POST",
data: formData,
cache: false,
async: true,
success: function (data) {
alert(data);
}
});
});
Controller
[HttpPost]
//[ValidateAntiForgeryToken]
public async Task<ActionResult> New([Bind(Include="age","name","IMG")] POST rec)
{
if (ModelState.IsValid)
{
db.REC.Add(rec);
await db.SaveChangesAsync();
return RedirectToAction("My", "Post");
}
return View(rec);
}
Send the extra parameters in a querystring. Here is the AJAX code:
$.ajax({
url: "/Post/New?name=James&age=1",
type: "POST",
data: formData,
cache: false,
async: true,
contentType: false, // Not to set any content header
processData: false, // Not to process data
success: function (data) {
alert(data);
}
});
And your controller should be similar to below:
public async Task<ActionResult> New(string name, int age)
{
try
{
foreach (string file in Request.Files)
{
var fileContent = Request.Files[file];
if (fileContent != null && fileContent.ContentLength > 0)
{
// get a stream
var stream = fileContent.InputStream;
// and optionally write the file to disk
var fileName = Path.GetFileName(file);
using (var fileStream = new MemoryStream())
{
stream.CopyTo(fileStream);
}
// File is in the memory stream, do whatever you need to do
}
}
}
catch (Exception)
{
// whatever you want to do
}
}
You are not specifying any form while creating the form in object in your form submit #UseShipAddr click event. please specify your form while creating object as:
var formData = new FormData($(#formID)[0]);
or create form constructor
var dataString = new FormData();
append file to the form
dataString.append("UploadedFile", selectedFile);
var form = $('#formID')[0];
var dataString = new FormData(form);
now send that string to the action in controller.
There are lots of problem in your request. First remove this then you'll be able to call that action of the controller.

XMLHTTPREQUEST send file and parameters [duplicate]

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);
?>

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