Related
I have a form that I want to send its data to admin-ajax:
<form method="POST" id="form">
<input type="text" name="name">
<input type="number" name="phone">
<input type="email" name="email">
<textarea name="overview"></textarea>
<input type="submit" name="submit" id="submit" value="Submit">
</form>
Javascript/jQuery code to send the data using Ajax:
document.getElementById('submit').addEventListener('click', function(e){
e.preventDefault();
var form_data = $('#form').serialize();
$.ajax('http://mywebsite.com/wordpress/wp-admin/admin-ajax.php', {
method: "POST",
data: form_data + {action: 'my_action'},
success: function (response) {
console.log(response);
},
error: function (err) {
console.log('Error:' + err);
}
});
});
Also tried formData:
var form_data = new FormData();
form_data.append('form', $('#custom').serialize());
form_data.append('action', 'my_action');
How to send the form data and the action my_action?
you need to change this line from data: form_data + {action: 'my_action'}, to data: {action: 'my_action', form_data:form_data},
jQuery(document).on("click","#submit", function(){
e.preventDefault();
var form_data =jQuery('#form').serializeArray();
jQuery.ajax('http://mywebsite.com/wordpress/wp-admin/admin-ajax.php', {
method: "POST",
data: {action: 'my_action', form_data:form_data},
success: function (response) {
console.log(response);
},
error: function (err) {
console.log('Error:' + err);
}
});
});
and change input type submit to button.
In general i prefer to use this way, like in your case you are using submit type button:
$(document).on('click', '#submit', function(){ // the id of your submit button
var form_data = $('#your_form_data_id'); // here is the id of your form
form_data.submit(function (e) {
var my_action = "my_action";
e.preventDefault();
$.ajax({
type: form_data.attr('method'), // use this if you have declared the method in the form like: method="POST"
url: form_data.attr('action'), // here you have declared the url in the form and no need to use it again or you can also use the path like in your code
data: form_data.serialize() +'&'+$.param({ 'my_action': my_action}), // here you are sending all data serialized from the form and appending the action value you assing when declare var my_action
success: function (data) {
// after the success result do your other stuff like show in console, print something or else
},
});
});
});
Hope it helps you. if anything is not clear feel free to ask, i try to explain in the comments.
You should just pass form to new FormData() so in your case when submitting the form just pass new FormData(e.target);
Hope it helps
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();
Good day all,
I have a form wil multiple fields in it. Also, the form is being submitted through form data method using ajax to a php file.
The following is the javascript code submitting the form data.
$(".update").click(function(){
$.ajax({
url: 'post_reply.php',
type: 'POST',
contentType:false,
processData: false,
data: function(){
var data = new FormData();
data.append('image',$('#picture').get(0).files[0]);
data.append('body' , $('#body').val());
data.append('uid', $('#uid').val());
return data;
}(),
success: function(result) {
alert(result);
},
error: function(xhr, result, errorThrown){
alert('Request failed.');
}
});
$('#picture').val('');
$('#body').val('');
});
And, the following is the actual form:
<textarea name=body id=body class=texarea placeholder='type your message here'></textarea>
<input type=file name=image id=picture >
<input name=update value=Send type=submit class=update id=update />
This form and javascript work good as they are. However, I am trying to be able to upload multiple files to the php file using this one single type=file field attribute. As it is now, it can only take one file at a time. How do I adjust both the form and the javascript code to be able to handle multiple files uploads?
Any help would be greatly appreciated.
Thanks!
Here is ajax, html and php global you can access. Let me know if it works for you.
// Updated part
jQuery.each(jQuery('#file')[0].files, function(i, file) {
data.append('file-'+i, file);
});
// Full Ajax request
$(".update").click(function(e) {
// Stops the form from reloading
e.preventDefault();
$.ajax({
url: 'post_reply.php',
type: 'POST',
contentType:false,
processData: false,
data: function(){
var data = new FormData();
jQuery.each(jQuery('#file')[0].files, function(i, file) {
data.append('file-'+i, file);
});
data.append('body' , $('#body').val());
data.append('uid', $('#uid').val());
return data;
}(),
success: function(result) {
alert(result);
},
error: function(xhr, result, errorThrown){
alert('Request failed.');
}
});
$('#picture').val('');
$('#body').val('');
});
Updated HTML:
<form enctype="multipart/form-data" method="post">
<input id="file" name="file[]" type="file" multiple/>
<input class="update" type="submit" />
</form>
Now, in PHP, you should be able to access your files:
// i.e.
$_FILES['file-0']
Here's another way.
Assuming your HTML is like this:
<form id="theform">
<textarea name="body" id="body" class="texarea" placeholder="type your message here"></textarea>
<!-- note the use of [] and multiple -->
<input type="file" name="image[]" id="picture" multiple>
<input name="update" value="Send" type="submit" class="update" id="update">
</form>
You could simply do
$("#theform").submit(function(e){
// prevent the form from submitting
e.preventDefault();
$.ajax({
url: 'post_reply.php',
type: 'POST',
contentType:false,
processData: false,
// pass the form in the FormData constructor to send all the data inside the form
data: new FormData(this),
success: function(result) {
alert(result);
},
error: function(xhr, result, errorThrown){
alert('Request failed.');
}
});
$('#picture').val('');
$('#body').val('');
});
Because we used [], you would be accessing the files as an array in the PHP.
<?php
print_r($_POST);
print_r($_FILES['image']); // should be an array i.e. $_FILES['image'][0] is 1st image, $_FILES['image'][1] is the 2nd, etc
?>
More information:
FormData constructor
Multiple file input
I'm using jQuery and Ajax for my forms to submit data and files but I'm not sure how to send both data and files in one form?
I currently do almost the same with both methods but the way in which the data is gathered into an array is different, the data uses .serialize(); but the files use = new FormData($(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);
?>
So I have a table that gets transformed to an array using:
var result = $("#enrolledStudents").sortable('toArray');
But when I go a head an pass that into my controller like so:
$("#update-enroll").click(function () {
var result = $("#enrolledStudents").sortable('toArray');
$.ajax({
url: '#Url.Action("Enrollment", "Classroom")',
data: { students: result },
type: 'POST',
traditional: true
});
});
My debugging breakpoint gets set off twice, causing issues to arise. What is the proper way to submit data to my controller on POST?
Per my comments, there are a couple things that could be causing this.
You have have the unobtrusive file(s) loaded multiple times
Your form has an action method defined, and your button is inside the form tag as a submit button. This will submit the form, and then the click will also submit the form - see example
Example
<form action="/somerowout/someaction">
<input type="text" id="text1"/>
<input type="text" id="text1"/>
<input type="submit" />
</form>
If you need to validate a value on your form before posting, don't hook up an additional Ajax call. Your javascript will look something like:
$(document).ready(function () {
$("form").submit(function(){
var result = $("#enrolledStudents").sortable('toArray');
if(result == null){
//do something to show validation failed
return false;
}
return true;
});
});
And your form code would then look something like:
#using (#Ajax.BeginForm(new AjaxOptions { })) {
<input type="text" id="text1"/>
<input type="text" id="text1"/>
<input type="submit" />
}
If you want to use Ajax rather than the Html Helpers, use a div instead of a form, and you won't get a duplicate post. Here's how you could achieve this:
<div id="enrolledStudents">
<--! your elements -->
<button id="saveStudents">Save</button>
</div>
JavaScript
$(document).ready(function () {
$("saveStudents").click(function(){
var result = $("#enrolledStudents").sortable('toArray');
if(result !== null){ /* do some kind of check here. */
$.ajax({
url: '#Url.Action("Enrollment", "Classroom")',
data: { students: result },
type: 'POST',
traditional: true,
success : function(data) {
if (data.status) {
window.location = data.route;
}
}
})
} else {
/* notify ui that save didn't happpen */
}
});
});
Example Controller Action
When posting your data using Ajax, here's an example of how to pass the route
[HttpPost]
public ActionResult SomethingPost(SomeModel model) {
if (Request.IsAjaxRequest()) {
var json = new {
status = true,
route = #Url.RouteUrl("MyRouteName", new { /* route values */ })
};
return Json(json, JsonRequestBehavior.AllowGet);
}
}
Are you sure you are preventing the default behaviour (form POSTING) of the submit button ? use preventDefault to do so.
$("#update-enroll").click(function (e) {
e.preventDefault();
//rest of the code
});
EDIT : As per the comment
To do the redirect in the ajax handler, you need to return the URL to be redirected in a JSON Response back to the calle.
[HttpPost]
public ActionResult Classroom(string students)
{
//do some operaton
if(Request.IsAjax())
{
//This is an Ajax call
return Json(new
{
Status="Success",
NewUrl = Url.Action("Index","Home")
});
}
else
{
//Normal request. Use RedirectToActiom
return RedirectToAction("Index","Home");
}
}
Now in your ajax call check the JSON result and do the redirect.
$.ajax({
url: '#Url.Action("Enrollment", "Classroom")',
data: { students: result },
type: 'POST',
dataType: "json",
contentType: "application/json; charset=utf-8",
success: function (data) {
if(data.Status=="Success")
{
window.location.href = data.Newrl;
}
else
{
alert("some error");
}
}
});
Check if you don't have the jquery files loaded twice. I had this behavior and the problem was files loaded twice.