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reloading a function in javascript

I'm initializing three functions on loading the page and then performing some events on click so I want that when a click event is pressed I just reload those functions completely in the given code it loads the function again but the previous one remains there so I want that when I load the function again it should remove the previously loaded one.
$('#locationdeleteconfirm').click(function(e) {
e.preventDefault();
$.ajax({
url: "php/deleteLocationByID.php",
type: 'POST',
dataType: 'json',
data: {
name: $('#dname').val(),
locationID: $('#dlocation').val(),
id: $('#editlocationid').val(),
},
//if user enters correct data and api gives back the data then we run success funtion
success: function(result) {
console.log(result);
//if status of data is ok we fetch the data from fields and put them in results table in html
if (result.status.name == "ok") {
//fetching fields from api and showing them in html table
Swal.fire({
title: "Deleted",
text: "Deleted",
type: "success"
}).then(function() {
$("#locdeleteconfirm").modal("hide");
$("#editlocationmodal2").modal("hide");
$('.modal-backdrop').removeClass('modal-backdrop');
$('#locationtable').load(location.href + " #locationtable>*")
getlocations();
getdepartments();
getallemployees();
});
} else {
//if the upper condition fails
console.log("error");
}
},
//if there is no success in retrieving data from api
error: function(jqXHR, textStatus, errorThrown) {
console.log("error");
$('#txterror').html("Enter Valid Id");
}
});
})
getloactions();
getdepartments();
getallemployees();
these are the three functions that are loading on the page load event I want them to reload once the click event is performed

JQuery confirmation dialog after AJAX request

I need to validate, on server side, if a person with a given registration number is already on the database. If this person is already registered, then I proceed with the program flow normally. But, if the number is not already registered, then I'd like to show a confirmation dialog asking if the operator wants to register a new person with the number entered and, if the operator answers yes, then the person will be registered with the number informed on the form on it's submission.
I've tried
Server side(PHP):
if (!$exists_person) {
$resp['success'] = false;
$resp['msg'] = 'Do you want to register a new person?';
echo json_encode($resp);
}
Client side:
function submit(){
var data = $('#myForm').serialize();
$.ajax({
type: 'POST'
,dataType: 'json'
,url: 'myPHP.php'
,async: 'true'
,data: data
,error: function(response){
alert('response');
}
});
return false;
}
I can't even see the alert, that's where I wanted to put my confirmation dialog, with the message written on server side. Other problem, how do I resubmit the entire form appended with the operator's answer, so the server can check if the answer was yes to register this new person?
EDIT
I was able to solve the problem this way:
Server side(PHP):
$person = find($_POST['regNo']);
if ($_POST['register_new'] === 'false' && !$person) {
$resp['exists'] = false;
$resp['msg'] = 'Do you want to register a new person?';
die(json_encode($resp)); //send response to AJAX request on the client side
} else if ($_POST['register_new'] === 'true' && !$person) {
//register new person
$person = find($_POST['regNo']);
}
if($person){
//proceed normal program flow
}
Client side:
function submit(e) {
e.preventDefault();
var data = $('#myForm').serialize();
var ajax1 = $.ajax({
type: 'POST'
, dataType: 'json'
, async: 'true'
, url: 'myPHP.php'
, data: data
, success: function (response) {
if (!response.exists && confirm(response.msg)) {
document.getElementById('register_new').value = 'true'; //hidden input
dados = $('#myForm').serialize(); //reserialize with new data
var ajax2 = $.ajax({
type: 'POST'
, dataType: 'json'
, async: 'true'
, url: 'myPHP.php'
, data: data
, success: function () {
document.getElementById('register_new').value = 'false';
$('#myForm').unbind('submit').submit();
}
});
} else if (response.success) {
alert(response.msg);
$('#myForm').unbind('submit').submit();
}
}
});
}

There doesn't appear to be anything wrong with your PHP.
The problem is (1) You are doing the alert inside of an error callback, and your request isn't failing, so you don't see the alert. (2) You are alerting the string 'response' instead of the variable response.
It is also worth noting that you should be using the .done() and .fail() promise methods (http://api.jquery.com/jquery.ajax/#jqXHR).
Here is the fixed JS:
function submit() {
var data = $('#myForm').serialize();
// Same as before, with the error callback removed
var myAjaxRequest = $.ajax({
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
});
// The request was successful (200)
myAjaxRequest.done(function(data, textStatus, jqXHR) {
// The data variable will contain your JSON from the server
console.log(data);
// Use a confirmation dialog to ask the user your question
// sent from the server
if (confirm(data.msg)) {
// Perform another AJAX request
}
});
// The request failed (40X)
myAjaxRequest.fail(function(jqXHR, textStatus, errorThrown) {
console.log(errorThrown);
});
return false;
}
Also, you are setting a 'status' in PHP and checking that in the JS (I presume). What you want to be doing is setting a HTTP status code from the server, as below:
if (!$exists_person)
{
$resp['msg'] = 'Do you want to register a new person?';
// 400 - Bad Request
http_response_code(400);
echo json_enconde($resp);
}
Then, jQuery will determine whether the request failed based on the status code you respond with. 200 is a successful request, and 400 numbers are fail.
Check out this page for a full list: https://httpstatuses.com/

Okay so this is a two part question; I'll try my best to answer both parts:
Part 1: How to detect if success is false and trigger the confirmation popup?
In jQuery.ajax the error handler is triggered based on response code. This is probably not what you want. You can use your success handler and test the value res.success to see if it's true or false. It would be something along the lines of:
function submit(e) {
e.preventDefault();
var data = $('#myForm').serialize();
$.ajax({
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
}).done(function(res) {
if (!res.success) {
alert(res.msg);
}
});
}
Part 2: How do I resubmit with a confirmation?
Working off of our previous code we will make some changes that allow for submit() to be passed an argument registerNew. If registerNew is true we will pass it as a param to the ajax handler in the PHP so it knows we want to register a new person. The Javascript will look something like this:
function submit(e, registerNew) {
if (e) e.preventDefault();
var data = $('#myForm').serialize();
var ajax_options = {
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
};
ajax_options.data.register_new = !!registerNew;
$.ajax(ajax_options).done(function(res) {
if (!res.success && confirm(res.msg)) {
submit(null, true);
}
});
}
As you can see here, we are passing a new register_new param in the data in our ajax options. Now we need to detect this on the PHP side, which is easy enough and looks like this (this goes in your php ajax handler):
if ($_POST["register_new"]) {
// new user registration code goes here
} else {
// your existing ajax handler code
}

Add confirm inside submit function
function submit(){
var data = $('#myForm').serialize();
if (confirm('Are you ready?')) {
$.ajax({
type: 'POST'
,dataType: 'json'
,url: 'myPHP.php'
,async: 'true'
,data: data
,error: function(response){
alert('response');
}
});
}
return false;
}

Stop Symfony form from redirecting after submit

I have a form that is rendered in a pop-up. When the form is submitted and valid, the data is saved and I'm redirected to the route of the form. How could I stop that redirection? I don't want to render another page, my goal is closing the pop-up after a valid submit.
public function addFeedbackAction(Request $request)
{
$view = View::create();
$feedback = new Feedback();
$feedbackService = $this->get('main.feedback.service');
$form = $this->createForm(new FeedbackType(), null, ['action' => 'feedback']);
$form->handleRequest($request);
if ($form->isValid()) {
$formData = $form->getData();
$feedbackService->create($formData, $feedback);
return null;
}
$view
->setData($form)
->setTemplateData($form)
->setTemplate('MainBundle:Modals:feedback.html.twig');
return $view;

You can catch submit event with Javascript.
You can find an exemple using JQuery here:
https://api.jquery.com/submit/

Send data using AJAX
$.ajax({
type: "POST",
url: "yourControllerHere",
data: form.serialize(),
beforeSend: function() {
//TODO: Loading
},
error: function( xhr){
},
success: function( data ){
}
});
Then you can process what you gonna do in success: function (data)

How to get a value from Ajax call to be used in a Bootbox button's function?

I am using Bootbox for my modals and I am having trouble in showing the form validation errors from an Ajax call to the modal. The callback function for the submit button on my modal calls the add_college function to submit the form via Ajax.
When there are validation errors, the modal is populated with validation errors. The problem is that the modal closes regardless if there are validation errors or not. I want the modal to not close only when there are no validation errors.
I know I can just return false in the callback function on my button when there are validation errors to not close it but I have no way of knowing if there are validation errors since I cannot return a value in the Ajax call since it is asynchronous. What is the proper way of doing it?
Here is my code:
$('#new-college-btn').click(function () {
bootbox.dialog({
title: "New College",
message:
''// <Insert long HTML form here>
,
buttons: {
add: {
label: "Add",
className: "btn btn-primary",
callback: function () {
var form_data = {
college_name: $('#college-name').val(),
college_initials: $('#college-initials').val(),
username: $('#username').val(),
password: $('#password').val(),
confirmation_password: $('#confirmation-password').val()
};
add_college(form_data);
}
},
cancel: {
label: "Cancel",
className: "btn btn-default"
}
}
}); // end bootbox dialog
});
function add_college(form_data) {
console.log(form_data);
$.ajax({
url: 'admin/add_new_college',
type: 'POST',
data: form_data,
dataType: 'JSON',
success: function (response)
{
if (response.error) { // there are form validation errors
// populate modal with validation errors here
} else {
// other data processing here
Result.success('College Successfully Added!');
}
},
error: function () {
console.log("fail");
}
});
}

If you want to control when the dialog closes, make sure the callback for your "submit" button always returns false. Then, in the done() (and probably fail()) callbacks for the ajax function, call bootbox.hideAll() to close the dialog (along with any other dialogs you may have opened).
If you want to only close the current dialog, do something along this line:
var dialog = bootbox.dialog({
/* rest of your options... */,
buttons: {
submit: {
label: "Submit",
callback: function() {
var data = [];
$.post('/url', data)
.done(function(result, status, jqxhr){
// if everything went well...
dialog.modal('hide');
})
.fail(function(jqxhr, status, error){
// etc.
});
return false;
}
}
}
});
Basically, create a reference to the dialog, which you can then use inside of the ajax callback.

Handle different type of PHP responses with AJAX

I am designing some PHP pages to process forms. In these pages I want to redirect if result is successful or print a message if there was an error. Structure of pages is like this:
$arg = $_POST["arg"];
if (isset($arg)) {
if (function($arg)) {
header ("Location: page2.php");
}
else {
echo "Response was not successfully";
}
}
else {
echo "$arg parameter was not defined";
}
When I need to print messages I use cftoast for JQuery (http://www.jqueryscript.net/other/Android-Style-jQuery-Toaster-Messages-Plugin-cftoaster.html)
To handle all forms I am using this Javascript function:
$(document).ready(function() {
$("#asynchronousForm").submit(function() {
$.ajax({
type: $(this).attr('method'),
url: $(this).attr('action'),
data: $(this).serialize(),
success: function(data) {
$("body").cftoaster({content: data});
window.location.href = ""; //Refresh page to clear possible errors
}
})
return false;
});
My problem is, when form redirects, sometimes appears problems like no redirection and shows empty toast, refreshing page with duplicated input fields... How can I solve this problem? I am using JQueryMobile as skeleton of my webpage.

A good way to handle AJAX responses is to use JSON.
It will allow you to send multiples data and do a redirect or show message depending of AJAX result.
In PHP you can use json_encode() to convert and array to JSON.
$arg = $_POST["arg"];
if (isset($arg)) {
if (function($arg)) {
exit(json_encode(array('redirect' => 'page2.php')));
}
else {
exit(json_encode(array('message' => 'Response was not successfully')));
}
}
else {
exit(json_encode(array('message' => $arg.' parameter was not defined')));
}
AJAX:
You just have to add dataType: 'json'.
You can also use $.getJSON()
$.ajax({
type: $(this).attr('method'),
url: $(this).attr('action'),
data: $(this).serialize(),
dataType: 'json',
success: function(json) {
if ( json.redirect )
window.location.href = json.redirect;
else
$("body").cftoaster({content: json.message});
}
})