1094

I have a form with name orderproductForm and an undefined number of inputs.

I want to do some kind of jQuery.get or ajax or anything like that that would call a page through Ajax, and send along all the inputs of the form orderproductForm.

I suppose one way would be to do something like

jQuery.get("myurl",
 {action : document.orderproductForm.action.value,
 cartproductid : document.orderproductForm.cartproductid.value,
 productid : document.orderproductForm.productid.value,
 ...

However I do not know exactly all the form inputs. Is there a feature, function or something that would just send ALL the form inputs?

Kamil Kiełczewski
93.5k34 gold badges401 silver badges374 bronze badges
asked Dec 25, 2009 at 1:31
0

22 Answers 22

1695

This is a simple reference:

// this is the id of the form
$("#idForm").submit(function(e) {
 e.preventDefault(); // avoid to execute the actual submit of the form.
 var form = $(this);
 var actionUrl = form.attr('action');
 
 $.ajax({
 type: "POST",
 url: actionUrl,
 data: form.serialize(), // serializes the form's elements.
 success: function(data)
 {
 alert(data); // show response from the php script.
 }
 });
 
});
Kamlesh
6,2511 gold badge51 silver badges55 bronze badges
answered Aug 5, 2011 at 17:57
Sign up to request clarification or add additional context in comments.

5 Comments

But form.serialize() can't post <input type="file"> in IE
can the url called come from an "http" or does it have to be local? your example: var url = "path/to/your/script.php";
I found this answer while looking for the same solution and it looks like form.serializeArray() is another good option to look into. api.jquery.com/serializeArray/#serializeArray
Do we need to have a action attribute in the form?
It's worth noting that it won't serialize the "submit" field too eg. <input type="submit" name="wont-be-serialized" value="Sadly" />
870

You can use the ajaxForm/ajaxSubmit functions from Ajax Form Plugin or the jQuery serialize function.

AjaxForm:

$("#theForm").ajaxForm({url: 'server.php', type: 'post'})

or

$("#theForm").ajaxSubmit({url: 'server.php', type: 'post'})

ajaxForm will send when the submit button is pressed. ajaxSubmit sends immediately.

Serialize:

$.get('server.php?' + $('#theForm').serialize())
$.post('server.php', $('#theForm').serialize())

AJAX serialization documentation is here.

Deelaka
13.8k9 gold badges38 silver badges67 bronze badges
answered Dec 25, 2009 at 1:36

2 Comments

.ajaxSubmit()/.ajaxForm() are not core jQuery functions- you need the jQuery Form Plugin
This doesn't seem to be a core jquery function either: malsup.github.io/jquery.form.js
494

Another similar solution using attributes defined on the form element:

<form id="contactForm1" action="/your_url" method="post">
 <!-- Form input fields here (do not forget your name attributes). -->
</form>
<script type="text/javascript">
 var frm = $('#contactForm1');
 frm.submit(function (e) {
 e.preventDefault();
 $.ajax({
 type: frm.attr('method'),
 url: frm.attr('action'),
 data: frm.serialize(),
 success: function (data) {
 console.log('Submission was successful.');
 console.log(data);
 },
 error: function (data) {
 console.log('An error occurred.');
 console.log(data);
 },
 });
 });
</script>

Comments

189

There are a few things you need to bear in mind.

1. There are several ways to submit a form

  • using the submit button
  • by pressing enter
  • by triggering a submit event in JavaScript
  • possibly more depending on the device or future device.

We should therefore bind to the form submit event, not the button click event. This will ensure our code works on all devices and assistive technologies now and in the future.

2. Hijax

The user may not have JavaScript enabled. A hijax pattern is good here, where we gently take control of the form using JavaScript, but leave it submittable if JavaScript fails.

We should pull the URL and method from the form, so if the HTML changes, we don't need to update the JavaScript.

3. Unobtrusive JavaScript

Using event.preventDefault() instead of return false is good practice as it allows the event to bubble up. This lets other scripts tie into the event, for example analytics scripts which may be monitoring user interactions.

Speed

We should ideally use an external script, rather than inserting our script inline. We can link to this in the head section of the page using a script tag, or link to it at the bottom of the page for speed. The script should quietly enhance the user experience, not get in the way.

Code

Assuming you agree with all the above, and you want to catch the submit event, and handle it via AJAX (a hijax pattern), you could do something like this:

$(function() {
 $('form.my_form').submit(function(event) {
 event.preventDefault(); // Prevent the form from submitting via the browser
 var form = $(this);
 $.ajax({
 type: form.attr('method'),
 url: form.attr('action'),
 data: form.serialize()
 }).done(function(data) {
 // Optionally alert the user of success here...
 }).fail(function(data) {
 // Optionally alert the user of an error here...
 });
 });
});

You can manually trigger a form submission whenever you like via JavaScript using something like:

$(function() {
 $('form.my_form').trigger('submit');
});

Edit:

I recently had to do this and ended up writing a plugin.

(function($) {
 $.fn.autosubmit = function() {
 this.submit(function(event) {
 event.preventDefault();
 var form = $(this);
 $.ajax({
 type: form.attr('method'),
 url: form.attr('action'),
 data: form.serialize()
 }).done(function(data) {
 // Optionally alert the user of success here...
 }).fail(function(data) {
 // Optionally alert the user of an error here...
 });
 });
 return this;
 }
})(jQuery)

Add a data-autosubmit attribute to your form tag and you can then do this:

HTML

<form action="/blah" method="post" data-autosubmit>
 <!-- Form goes here -->
</form>

JS

$(function() {
 $('form[data-autosubmit]').autosubmit();
});
answered May 7, 2013 at 12:44

Comments

56

You can also use FormData (But not available in IE):

var formData = new FormData(document.getElementsByName('yourForm')[0]);// yourForm: form selector 
$.ajax({
 type: "POST",
 url: "yourURL",// where you wanna post
 data: formData,
 processData: false,
 contentType: false,
 error: function(jqXHR, textStatus, errorMessage) {
 console.log(errorMessage); // Optional
 },
 success: function(data) {console.log(data)} 
});

This is how you use FormData.

Mikael Dúi Bolinder
2,2872 gold badges26 silver badges53 bronze badges
answered Jan 19, 2013 at 18:52

1 Comment

+1 This is the best solution. IE was considered in the past decade. Now FormData is the standard solution. metamug.com/article/html5/ajax-form-submit.html
31

Simple version (does not send images)

<form action="/my/ajax/url" class="my-form">
...
</form>
<script>
 (function($){
 $("body").on("submit", ".my-form", function(e){
 e.preventDefault();
 var form = $(e.target);
 $.post( form.attr("action"), form.serialize(), function(res){
 console.log(res);
 });
 });
 )(jQuery);
</script>

Copy and paste ajaxification of a form or all forms on a page

It is a modified version of Alfrekjv's answer

  • It will work with jQuery>= 1.3.2
  • You can run this before the document is ready
  • You can remove and re-add the form and it will still work
  • It will post to the same location as the normal form, specified in the form's "action" attribute

JavaScript

jQuery(document).submit(function(e){
 var form = jQuery(e.target);
 if(form.is("#form-id")){ // check if this is the form that you want (delete this check to apply this to all forms)
 e.preventDefault();
 jQuery.ajax({
 type: "POST",
 url: form.attr("action"), 
 data: form.serialize(), // serializes the form's elements.
 success: function(data) {
 console.log(data); // show response from the php script. (use the developer toolbar console, firefox firebug or chrome inspector console)
 }
 });
 }
});

I wanted to edit Alfrekjv's answer but deviated too much from it so decided to post this as a separate answer.

Does not send files, does not support buttons, for example clicking a button (including a submit button) sends its value as form data, but because this is an ajax request the button click will not be sent.

To support buttons you can capture the actual button click instead of the submit.

jQuery(document).click(function(e){
 var self = jQuery(e.target);
 if(self.is("#form-id input[type=submit], #form-id input[type=button], #form-id button")){
 e.preventDefault();
 var form = self.closest('form'), formdata = form.serialize();
 //add the clicked button to the form data
 if(self.attr('name')){
 formdata += (formdata!=='')? '&':'';
 formdata += self.attr('name') + '=' + ((self.is('button'))? self.html(): self.val());
 }
 jQuery.ajax({
 type: "POST",
 url: form.attr("action"), 
 data: formdata, 
 success: function(data) {
 console.log(data);
 }
 });
 }
});

On the server side you can detect an ajax request with this header that jquery sets HTTP_X_REQUESTED_WITH for php

PHP

if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
 //is ajax
}
answered Jul 10, 2013 at 9:35

Comments

27

This code works even with file input

$(document).on("submit", "form", function(event)
{
 event.preventDefault(); 
 $.ajax({
 url: $(this).attr("action"),
 type: $(this).attr("method"),
 dataType: "JSON",
 data: new FormData(this),
 processData: false,
 contentType: false,
 success: function (data, status)
 {
 },
 error: function (xhr, desc, err)
 {
 }
 }); 
});
answered Feb 7, 2015 at 19:35

1 Comment

Just to note formData() is IE10+ developer.mozilla.org/en-US/docs/Web/API/FormData
13

I really liked this answer by superluminary and especially the way he wrapped is solution in a jQuery plugin. So thanks to superluminary for a very useful answer. In my case, though, I wanted a plugin that would allow me to define the success and error event handlers by means of options when the plugin is initialized.

So here is what I came up with:

;(function(defaults, ,ドル undefined) {
 var getSubmitHandler = function(onsubmit, success, error) {
 return function(event) {
 if (typeof onsubmit === 'function') {
 onsubmit.call(this, event);
 }
 var form = $(this);
 $.ajax({
 type: form.attr('method'),
 url: form.attr('action'),
 data: form.serialize()
 }).done(function() {
 if (typeof success === 'function') {
 success.apply(this, arguments);
 }
 }).fail(function() {
 if (typeof error === 'function') {
 error.apply(this, arguments);
 }
 });
 event.preventDefault();
 };
 };
 $.fn.extend({
 // Usage:
 // jQuery(selector).ajaxForm({ 
 // onsubmit:function() {},
 // success:function() {}, 
 // error: function() {} 
 // });
 ajaxForm : function(options) {
 options = $.extend({}, defaults, options);
 return $(this).each(function() {
 $(this).submit(getSubmitHandler(options['onsubmit'], options['success'], options['error']));
 });
 }
 });
})({}, jQuery);

This plugin allows me to very easily "ajaxify" html forms on the page and provide onsubmit, success and error event handlers for implementing feedback to the user of the status of the form submit. This allowed the plugin to be used as follows:

 $('form').ajaxForm({
 onsubmit: function(event) {
 // User submitted the form
 },
 success: function(data, textStatus, jqXHR) {
 // The form was successfully submitted
 },
 error: function(jqXHR, textStatus, errorThrown) {
 // The submit action failed
 }
 });

Note that the success and error event handlers receive the same arguments that you would receive from the corresponding events of the jQuery ajax method.

answered Nov 12, 2013 at 23:06

1 Comment

It's beforeSubmit in current versions of jquery form plugin jquery-form.github.io/form/options/#beforesubmit
9

I got the following for me:

formSubmit('#login-form', '/api/user/login', '/members/');

where

function formSubmit(form, url, target) {
 $(form).submit(function(event) {
 $.post(url, $(form).serialize())
 .done(function(res) {
 if (res.success) {
 window.location = target;
 }
 else {
 alert(res.error);
 }
 })
 .fail(function(res) {
 alert("Server Error: " + res.status + " " + res.statusText);
 })
 event.preventDefault();
 });
}

This assumes the post to 'url' returns an ajax in the form of {success: false, error:'my Error to display'}

You can vary this as you like. Feel free to use that snippet.

answered Jan 21, 2014 at 20:47

1 Comment

What is target for here? Why would you submit a form with AJAX, only to redirect to a new page?
9

jQuery AJAX submit form, is nothing but submit a form using form ID when you click on a button

Please follow steps

Step 1 - Form tag must have an ID field

<form method="post" class="form-horizontal" action="test/user/add" id="submitForm">
.....
</form>

Button which you are going to click

<button>Save</button>

Step 2 - submit event is in jQuery which helps to submit a form. in below code we are preparing JSON request from HTML element name.

$("#submitForm").submit(function(e) {
 e.preventDefault();
 var frm = $("#submitForm");
 var data = {};
 $.each(this, function(i, v){
 var input = $(v);
 data[input.attr("name")] = input.val();
 delete data["undefined"];
 });
 $.ajax({
 contentType:"application/json; charset=utf-8",
 type:frm.attr("method"),
 url:frm.attr("action"),
 dataType:'json',
 data:JSON.stringify(data),
 success:function(data) {
 alert(data.message);
 }
 });
});

for live demo click on below link

How to submit a Form using jQuery AJAX?

answered Jan 11, 2018 at 8:05

Comments

9

I know this is a jQuery related question, but now days with JS ES6 things are much easier. Since there is no pure javascript answer, I thought I could add a simple pure javascript solution to this, which in my opinion is much cleaner, by using the fetch() API. This a modern way to implements network requests. In your case, since you already have a form element we can simply use it to build our request.

const form = document.forms["orderproductForm"];
const formInputs = form.getElementsByTagName("input"); 
let formData = new FormData(); 
for (let input of formInputs) {
 formData.append(input.name, input.value); 
}
fetch(form.action,
 {
 method: form.method,
 body: formData
 })
 .then(response => response.json())
 .then(data => console.log(data))
 .catch(error => console.log(error.message))
 .finally(() => console.log("Done"));
answered Oct 16, 2019 at 19:28

Comments

8

Try

fetch(form.action,{method:'post', body: new FormData(form)});

function send(e,form) {
 fetch(form.action,{method:'post', body: new FormData(form)});
 console.log('We submit form asynchronously (AJAX)');
 e.preventDefault();
}
<form method="POST" action="myapi/send" onsubmit="send(event,this)" name="orderproductForm">
 <input hidden name="csrfToken" value="0ドルmeh@$h">
 <input name="email" value="[email protected]">
 <input name="phone" value="123-456-666">
 <input type="submit"> 
</form>
Look on Chrome Console > Network after/before 'submit'

answered Jan 21, 2020 at 18:48

Comments

5

consider using closest

$('table+table form').closest('tr').filter(':not(:last-child)').submit(function (ev, frm) {
 frm = $(ev.target).closest('form');
 $.ajax({
 type: frm.attr('method'),
 url: frm.attr('action'),
 data: frm.serialize(),
 success: function (data) {
 alert(data);
 }
 })
 ev.preventDefault();
 });
answered Oct 26, 2016 at 10:49

Comments

5

You may use this on submit function like below.

HTML Form

<form class="form" action="" method="post">
 <input type="text" name="name" id="name" >
 <textarea name="text" id="message" placeholder="Write something to us"> </textarea>
 <input type="button" onclick="return formSubmit();" value="Send">
</form>

jQuery function:

<script>
 function formSubmit(){
 var name = document.getElementById("name").value;
 var message = document.getElementById("message").value;
 var dataString = 'name='+ name + '&message=' + message;
 jQuery.ajax({
 url: "submit.php",
 data: dataString,
 type: "POST",
 success: function(data){
 $("#myForm").html(data);
 },
 error: function (){}
 });
 return true;
 }
</script>

For more details and sample Visit: http://www.spiderscode.com/simple-ajax-contact-form/

roberrrt-s
8,2302 gold badges49 silver badges61 bronze badges
answered Dec 20, 2016 at 7:48

Comments

4

To avoid multiple formdata sends:

Don't forget to unbind submit event, before the form submited again, User can call sumbit function more than one time, maybe he forgot something, or was a validation error.

 $("#idForm").unbind().submit( function(e) {
 ....
answered Nov 13, 2014 at 18:43

Comments

4

If you're using form.serialize() - you need to give each form element a name like this:

<input id="firstName" name="firstName" ...

And the form gets serialized like this:

firstName=Chris&lastName=Halcrow ...
answered Apr 26, 2016 at 23:48

Comments

4

I find it surprising that no one mentions data as an object. For me it's the cleanest and easiest way to pass data:

$('form#foo').submit(function () {
 $.ajax({
 url: 'http://foo.bar/some-ajax-script',
 type: 'POST',
 dataType: 'json',
 data: {
 'foo': 'some-foo-value',
 'bar': $('#bar').val()
 }
 }).always(function (response) {
 console.log(response);
 });
 return false;
});

Then, in the backend:

// Example in PHP
$_POST['foo'] // some-foo-value
$_POST['bar'] // value in #bar
answered Apr 30, 2019 at 21:32

1 Comment

OP said he doesnt know the field names / undetermined so one can't use data in this instance as one doesnt know the field names.
2

This is not the answer to OP's question,
but in case if you can't use static form DOM, you can also try like this.

var $form = $('<form/>').append(
 $('<input/>', {name: 'username'}).val('John Doe'),
 $('<input/>', {name: 'user_id'}).val('john.1234')
);
$.ajax({
 url: 'api/user/search',
 type: 'POST',
 contentType: 'application/x-www-form-urlencoded',
 data: $form.serialize(),
 success: function(data, textStatus, jqXHR) {
 console.info(data);
 },
 error: function(jqXHR, textStatus, errorThrown) {
 var errorMessage = jqXHR.responseText;
 if (errorMessage.length > 0) {
 alert(errorMessage);
 }
 }
});
answered Apr 30, 2019 at 3:48

Comments

2

JavaScript

(function ($) {
 var form= $('#add-form'),
 input = $('#exampleFormControlTextarea1');
 form.submit(function(event) {
 event.preventDefault(); 
 var req = $.ajax({
 url: form.attr('action'),
 type: 'POST',
 data: form.serialize()
 });
 req.done(function(data) {
 if (data === 'success') {
 var li = $('<li class="list-group-item">'+ input.val() +'</li>');
 li.hide()
 .appendTo('.list-group')
 .fadeIn();
 $('input[type="text"],textarea').val('');
 }
 });
});
}(jQuery));

HTML

 <ul class="list-group col-sm-6 float-left">
 <?php
 foreach ($data as $item) {
 echo '<li class="list-group-item">'.$item.'</li>';
 }
 ?>
 </ul>
 <form id="add-form" class="col-sm-6 float-right" action="_inc/add-new.php" method="post">
 <p class="form-group">
 <textarea class="form-control" name="message" id="exampleFormControlTextarea1" rows="3" placeholder="Is there something new?"></textarea>
 </p>
 <button type="submit" class="btn btn-danger">Add new item</button>
 </form>
answered Nov 26, 2019 at 13:05

Comments

0

It's a great and easy-to-use function I made to handle any forms, which you can use by just calling handleForm() function. Of course, you can send your own selector to specify the form, even your own special function to handle the success, error, etc.: (I hope it's useful)

const handleForm = function(
 selector = 'form',
 successFunc = function(data,selector){
 $(selector + ' .result').remove();
 const resultMsg = 'The form submitted successfully!';
 $(selector).append("<div class='result success'>"+resultMsg+"</div>");
 },
 errorfunc = function(data,selector){
 $(selector + ' .result').remove();
 const resultMsg = data??'There a problem occured!';
 $(selector).append("<div class='result error'>"+resultMsg+"</div>");
 },
 beforeFunc = function(data,selector){
 $(selector + ' .result').remove();
 },
 processHandler = function(data){},
 timeout = 600000
){
 $(selector).submit(function(e) {
 e.preventDefault();
 const form = $(this);
 const actionPath = form.attr('action');
 const methodName = form.attr('method');
 const requestData = form.serialize();
 const btns = selector+' :is(button:is([type=submit], [type=reset]), input:is([type=button], [type=submit], [type=reset]))';
 $.ajax({
 type: methodName,
 url: actionPath,
 xhr: function () {
 var myXhr = $.ajaxSettings.xhr();
 if (myXhr.upload) myXhr.upload.addEventListener('progress', processHandler, false);
 return myXhr;
 },
 success: function (data) {
 successFunc(data,selector);
 $(btns).removeClass('hide');
 $(selector).css('opacity','1');
 },
 error: function (data) {
 errorfunc(data,selector);
 $(btns).removeClass('hide');
 $(selector).css('opacity','1');
 },
 beforeSend: function (data) {
 beforeFunc(data,selector);
 $(btns).addClass('hide');
 $(selector).css('opacity','.5');
 },
 async: true,
 data: requestData,
 cache: false,
 contentType: (typeof requestData === 'string')?'application/json; charset=utf-8':false,
 processData: false,
 timeout: timeout
 });
 });
}

Enjoy...

answered May 6, 2023 at 13:01

Comments

0

Alternatively, form can also be sent after auto converted to object.

$("#form").submit( function (e) {
 
 e.preventDefault(); //prevent sending via browser, navigating or refreshing. 
 let formData = new FormData(this); 
 let formObj = Object.fromEntries(formData);
 
 $.ajax(uri, {
 type: "post", 
 data: formObj
 });
}
answered Oct 5, 2024 at 23:42

Comments

-24

There's also the submit event, which can be triggered like this $("#form_id").submit(). You'd use this method if the form is well represented in HTML already. You'd just read in the page, populate the form inputs with stuff, then call .submit(). It'll use the method and action defined in the form's declaration, so you don't need to copy it into your javascript.

examples

Souad
5,08216 gold badges92 silver badges148 bronze badges
answered Dec 25, 2009 at 4:50

3 Comments

if you delete your post, you get the penalty points that have been taken away from you back.
This will submit the form, but not via AJAX, so doesn't answer the question.
what @spider said, plus you get a nice badge called "peer pressure"

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.