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POST output into a modal from any form class on page

Sorry I am a beginner with jQuery and Javascript. I want to be able to get the results into my modal from any form on the page that has class ajax. My code is below but not working correctly. Currently it opens the post result in a new page and not in the modal. Can anyone shed any light on my code?
Many thanks
$(document).ready(function() {
$('.ajax').click(function() {
var that = $(this),
url = that.attr('action'),
type = that.attr('method'),
data = {};
that.find('name').each(function(index, value) {
var that = $(this),
name = that.attr('name'),
value = that.val();
data[name] = value;
});
console.log(value);
// AJAX request
$.ajax({
url: url,
type: type,
data: data,
success: function(response){
// Add response in Modal body
$('.modal-body').html(response);
// Display Modal
$('#aaModal').modal('show');
}
});
});
});

This probably happens because your browser submits the form by default. It doesnt know youre doing AJAX stuff. To prevent this, use preventDefault().
In addition to that, jQuery has a built in function for serializing (1 and 2) form data.
$(document).ready(function() {
$('form.ajax').click(function(event) {
event.preventDefault(); // prevents opening the form action url
var $form = $(this),
url = $form.attr('action'),
type = $form.attr('method'),
data = $form.serialize();
// console.log(value); // value doesnt exist outside of your loop btw
// AJAX request
$.ajax({
url: url,
type: type,
data: data,
success: function(response){
// Add response in Modal body
$('.modal-body').html(response);
// Display Modal
$('#aaModal').modal('show');
}
});
});
});
Also, its not quite clear if you bind the click event handler to a form or a button, I guess the first one. You should change the handler to the following:
$(document).ready(function() {
$('form.ajax').on('submit', function(event) {

How to Replace Form Without Triggering the Change Event?

I would like to validate a form with an AJAX request to the server and then swap the form html in the web browser with the form html from the server because this would be an easy implementation in theory. It is proving a nightmare though because the change event is triggered without the user interacting further after the first interaction which triggered the first change event. Consequently an infinite loop of AJAX requests to the server is happening.
The html form sits inside a div which has classes 'container mb-4'. This is the JS code -
var _cont = $('.container.mb-4')
var _form = $('.custom-form')
function ajax_validation(form) {
form.on('change', 'input, select, textarea', function() {
form_data = form.serialize()
$.ajax({
url: "/form/6/",
type: "POST",
data: form_data,
success: function(data) {
if(!(data['success'])) {
_cont.empty()
_cont.append(data['form_html'])
form = _cont.find('form')
ajax_validation(form)
}
},
error: function () {
form.find('.error-message').show()
}
});
})
}
ajax_validation(_form)
The change event I am assuming is triggered because the server returns a form input field with a different csrf token as the value to the previous input field - all other fields are the same. So an obvious solution would be to keep the same csrf token. But I want to understand why the JS code isn't working. I thought destroying the form would destroy the change event bound to it. So am at a loss to explain this infinite loop. How do I change this so I can just swap the form and not trigger another change event until the user really does change something?

It's not a good thing to use events in function no need to do that
Also your event here for input , select , textarea for serialize you need to select the closest() form
Try the next code
var _cont = $('.container.mb-4');
var _form = $('.custom-form');
_cont.on('change', 'form input,form select,form textarea', function() {
var ThisForm = $(this).closest('form');
var form_data = ThisForm.serialize();
$.ajax({
url: "/form/6/",
type: "POST",
data: form_data,
success: function(data) {
if(!(data['success'])) {
_cont.html(data['form_html']);
}
},
error: function () {
ThisForm.find('.error-message').show()
}
});
});
And logically if(!(data['success'])) { should be if(data['success']) {

First let's understand the issue that you have. You have a function called ajax_validation that is defining a change event on the form's elements which, on response will call ajax_validation. So, if any change happens on your elements, then a new request is sent to the server. So, if any value is changed, like a token, the request will be sent again. You could use a semaphore, like this:
var semaphore = true;
function ajax_validation(form) {
form.on('change', 'input, select, textarea', function() {
if (!semaphore) return;
semaphore = false;
form_data = form.serialize()
$.ajax({
url: "/form/6/",
type: "POST",
data: form_data,
success: function(data) {
if(!(data['success'])) {
_cont.empty()
_cont.append(data['form_html'])
form = _cont.find('form')
ajax_validation(form)
}
semaphore = true;
},
error: function () {
form.find('.error-message').show()
}
});
})
}
Something like this should solve your issue for the time being, but you should consider refactoring your code, because what you experience is well-known and is called callback hell.

Turns out the password field was coming back blank from the server - this django must do out of the box if the PasswordInput widget is used. So the form is replaced with a new form which lacks the password input from the before. The browser was then applying the autofill password value to the form which was triggering the change event.
This is my code now. It checks that the form_data about to be sent for validation really is different to before minus the csrf token which will be different.
It is based on Mohamed's answer -
var _cont = $('.container.mb-4');
var _form = $('.custom-form');
var prev_data = undefined
_cont.on('change', 'form input,form select,form textarea', function() {
var ThisForm = $(this).closest('form');
var form_data_wo_csrf = ThisForm.find("input, textarea, select").not("input[type='hidden']").serialize()
if(form_data_wo_csrf == prev_data) {
return
}
var form_data = ThisForm.serialize()
$.ajax({
url: "/form/6/",
type: "POST",
data: form_data,
success: function(data) {
if(!(data['success'])) {
_cont.html(data['form_html']);
prev_data = form_data_wo_csrf
}
},
error: function () {
ThisForm.find('.error-message').show()
}
});
});

Set interval to ajax form

So I'm trying to send the form info to php with ajax form every second, but for some reason it doesn't want to.
Here is my latest attempt, I tried every other similar combination(like put everything in the function or just put everything into the setInterval).
$(document).ready(function() {
var ajaxCall=function() {
$("#myForm").ajaxForm(function(e) {
$.ajax({
type:'post',
url:'php1.php',
data:$("#myForm").serialize(),
success:function(data) {
document.getElementById("result").innerHTML=data;
}
});
});
}
setInterval(ajaxCall,1000);
});
EDIT
Solved with M.M answer, thank you for the help!

Simply change ajaxForm to ajaxSubmit
See this (question) and this (documentation) for more information on AjaxForm vs AjaxSubmit
Essentially AjaxForm submits when the user clicks the button and AjaxSubmit does it immediately so your code should be:
$(document).ready(function()
{
var ajaxCall=function()
{
$("#myForm").ajaxSubmit(function(e)
{
$.ajax(
{
type:'post',
url:'php1.php',
data:$("#myForm").serialize(),
success:function(data)
{
document.getElementById("result").innerHTML=data;
}
});
});
}
setInterval(ajaxCall,1000);
});
Update after comment explanation
$(document).ready(function(){
//live feed
var ajaxCall=function(){
$("#myForm").ajaxSubmit(function(e){
ajax_submit();
});
}
setInterval(ajaxCall,1000);
//real submit
$("#myForm").ajaxForm(function(e){
ajax_submit();
});
function ajax_submit(){//ajax_code
$.ajax({
type:'post',
url:'php1.php',
data:$("#myForm").serialize(),
success:function(data) {
document.getElementById("result").innerHTML=data;
}
});
}
});
If you wish to differentiate the feed from the submit you can pass a parameter to the ajax_submit function

Getting rid of the ajaxForm() call seems to accomplish what you are trying to do:
$(document).ready(function() {
var ajaxCall = function() {
$.ajax({
type: 'post',
url: 'php1.php',
data: $("#myForm").serialize(),
success: function(data) {
document.getElementById("result").innerHTML = data;
}
});
}
setInterval(ajaxCall, 1000);
});

Multiple form submit with only one jQuery Ajax

I have a page where is 3-4 form in divs, and I want to submit it with only one script and I want to refresh the actual div content (where is the form). But I don't know how to specify the actual div or form for the ajax post.
$(document).ready(function(){
$(".ajaxform").submit(function() {
var id = $(this).attr('id');
var dataString = $(this).serialize();
$.ajax({
url:'../tools/tools.php',
data: dataString + '&form=' + id,
type: 'POST',
success: function(html)
{
$('#actualdiv').load(document.URL + ' #actualdiv');
}
});
});
});
The forms or divs have unique id for the tools.php where I can run the actual form's mysql.

Since you are posting data and then receiving new forms back as a response, I would make the function more dynamic.
function loadNextForm(form, id){
var dataString = form.serialize();
$.ajax({
url:'../tools/tools.php',
data: dataString + '&form=' + id,
type: 'POST',
success: function(newForm) {
$('#actualdiv').html(newForm);
}
});
}
$(document).ready(function(){
$(".ajaxform").on("submit", function(){
loadNextForm($(this), $(this).attr('id'));
});
});
This should allow you to POST the data, get the new form, place it into the DIV, and allow it to be a functional form that can process the next form.

Ajax Form Cannot Prevent Page Reload with event.preventDefault();

I have the following code which is supposed submit a form via Ajax without having to reload the page:
$( document ).on('submit', '.login_form', function( event ){
event.preventDefault();
var $this = $(this);
$.ajax({
data: "action=login_submit&" + $this.serialize(),
type: "POST",
url: _ajax_login_settings.ajaxurl,
success: function( msg ){
ajax_login_register_show_message( $this, msg );
}
});
});
However for some reason, despite the event.preventDefault(); function which is supposed to prevent the form from actually firing, it actually does fire.
My question is, how do I prevent the above form from reloading the page?
Thanks

don't attach a listener on document instead use a on click handler on the submit button and change the type to button.
<button id="form1SubmitBtn">Submit</button>
$('#form1SubmitBtn').click(function(){
//do ajax here
});
Happy Coding !!!

for instance you can write like this
$(".login_form").on('submit', function( event ){
var $this = $(this);
$.ajax({
data: "action=login_submit&" + $this.serialize(),
type: "POST",
url: _ajax_login_settings.ajaxurl,
success: function( msg ){
ajax_login_register_show_message( $this, msg );
}
});
event.preventDefault();
});

You can use jquery and ajax to do that. Here is a nice piece code below that doesn't refresh the page but instead on submit the form gets hidden and gets replaced by a thank you message. The form data is sent to an email address using sendmail.php script.
Assuming your form has 3 input fields - name, email and message.
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script>
<script type="text/javascript">
jQuery(function() {
jQuery("#button").click(function() {
var name=jQuery('#name').val();
var email=jQuery('#email').val();
var message=jQuery('#message').val();
var dataString = 'name='+ name + '&email=' + email + '&message=' + message;
jQuery.ajax({
type: "POST",
url: "sendmail.php",
data: dataString,
success: function() {
jQuery('#contact_form').html("<div id='message'></div>");
jQuery('#contactForm').hide();
jQuery('#message').html("<h2>Contact Form Submitted!</h2>")
.append("<p>Thank you for your submission. We will be in touch shortly.</p>").hide()
.fadeIn(1500, function() {
});
}
});
return false;
});
});
</script>
On top of your form tag just add this to display the thank you message.
<div id='message'></div>
Enjoy coding!!!!!!!