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Data is not display after insertion through Bootstrap modal with jquery ajax in php Mysqli

Data is not display after insert through bootstrap modal with ajax in php mysqli.
jQuery Ajax code:
$(document).ready(function() {
fetch_data();
function fetch_data() {
var action = "fetch";
$.ajax({
url: 'action.php',
method: 'post',
data: {action: action},
success: function (data) {
$("#all_table_data").html(data);
}
});
}
});
function addData() {
var formData = new FormData($('#employee_insert_form')[0]);
formData.append('action', 'add');
$.ajax({
method: 'post',
url: 'action.php',
data: formData,
processData: false,
contentType: false,
cache: false,
success: function (response) {
$("#all_table_data").html(response);
$('#empInsert').modal('hide');
fetch_data();
}
});
}

Your PHP code never return anything to your view. So try to echo result of select * from tbl_employee after the insertion

Getting array data from PHP to jQuery for outputting

Here is the partial code from the remove PHP file:
if($action == 'trackings_get') {
$result = $trackings->get(getCourierSlugByID($GLOBALS['tracking_id']), $GLOBALS['tracking_id']);
$result_history = $result['data']['tracking']['checkpoints'];
echo json_encode($result_history);
// debugging
//pretty_print($result_history);
}
Here is the JS from the remote site i am trying to call the data for:
$.ajax({
url: '/login/tracking.php',
type: 'POST',
dataType: "json",
data: {
action: action,
tracking_id: tracking_id
},
success: function(json){
//debug
alert(JSON.stringify(json));
}
});

try this
function test(){
$.ajax({
url: 'url',
type: 'POST',
dataType: "json",
data: {
action: action,
tracking_id: tracking_id
},
success: function(json){
}
});
}

i try this code in inspect element in this page https://tracking.ambientlounge.com/
function test(){
$.ajax({
url: 'your url',
type: 'POST',
dataType: "json",
data: {
action: "action",
tracking_id: "tracking_id"
},
success: function(json){
//debug
console.log(json);
}
});
}
result is array . dont need JSON.stringify .

How prevent that an user change the input value?

I've a website with the post, and when an user comment its I send an Ajax request.
$(document).on('submit', '.theFormComment', function(e){
var data_to_send = $(this).serializeArray(); // convert form to array
data_to_send.push({name: "cs_spotale", value: $.cookie("c_spotale") });
$.ajax({
type: "POST",
url: '/post/addComment',
data: $.param(data_to_send),
dataType: 'json',
success: function(data){
console.log(data);
if(data.user_id !== data.user_to_id) {
$.ajax({
type: "POST",
url: "/notification/addNotification",
data: {
post_id: data.the_post,
user_from_id: data.user_id,
user_to_id: data.user_to_id,
action: data.action,
cs_spotale: $.cookie("c_spotale")
},
dataType: 'json',
success: function(x){
socket.emit('sendNotification', {data: data, type: x.type, image: x.image});
},
error: function(){}
});
}
$('#comment-box-'+ data.the_post).val('');
$("input[id='csrf_prot']").val(data.c);
$(data.html).prependTo("#comment-" + data.the_post);
if(data.own !== data.username){
socket.emit('notify', data.own, data.username);
}
socket.emit('updateComment', {permaRoom: data.the_post, html: data.html});
},
error: function(data){
console.log(data);
}
});
e.preventDefault();
return false;
});
I saw that if I change the input value of my hidden field "post_id", the comment goes to another "post_id". There is a way to prevent this problem? Or any other idea?

Post array from javascript to php function using Ajax

i want to post an array from java script to php by ajax. But i don't know how do that, especially send it to php function like controller class. Correct me if i'm wrong, this is my java script source, as a function to send an array :
<script>
function send(){
var obj = JSON.stringify(array);
window.location.href = "post.php?q=" + obj;
}
</script>
i was try, but still fail. really need help..

As described in the JQuery API documentation, you can use
var rootPath="http://example.com/"
var jsonData = $.toJSON({ q: array });
var urlWS = rootPath + "post.php";
$.ajax({
url: urlWS,
type: "POST",
dataType: "json",
contentType: "application/json; charset=utf-8",
data: jsonData,
success: function(result) {
// do something here with returned result
}
});

var array= [];
array[0] = 'hi';
array[1] = 'hello';
$.ajax({
url: 'http://something.com/post.php',
data: {array: array},
type: 'POST'
});

try like this,
var data_to_send = $.serialize(array);
$.ajax({
type: "POST",
url: 'post.php',
data: data_to_send,
success: function(msg){
}
});
or
you can pass as json like below,
$.ajax({
type: "POST",
url: 'post.php',
dataType: "json",
data: {result:JSON.stringify(array)},
success: function(msg){
}
});

var arr = <?php echo json_encode($postdata); ?>;
ajax: {
url:"post.php"
type: "POST",
data: {dataarr: arr},
complete: function (jqXHR, textStatus) {
}
You can try this .this will work

example
ajax code:
$.ajax({
url: 'save.php',
data: {data: yourdata},
type: 'POST',
dataType: 'json', // you will get return json data
success:function(result){
// to do result from php file
}
});
PHP Code:
$data['something'] = "value";
echo json_encode($data);

jQuery var doesn't change

I have an AJAX registration form:
var id = null;
$.ajax({
type: 'POST',
url: requestUrl,
data: $(".defaultRequest").serialize(),
dataType: 'json',
success: function(data) {
if(data.response){
$('div.errormsg').remove();
if(data.step){
openStep(data.step);
}else{
openStep('next');
}
}else{
$('div.errormsg').remove();
$('<div class="errormsg">'+data.message+"</div>").insertBefore(form);
}
}
});
When a user is successful registered, I want to show him his unique ID, but it stays NULL. How can I solve it?
<script type="text/javascript">
$('#linkkk').text('Your id is: '+id+'');
</script>

you need to set id first in success callback.
$.ajax({
type: 'POST',
url: requestUrl,
data: $(".defaultRequest").serialize(),
dataType: 'json',
success: function(data) {
if(data.response){
$('div.errormsg').remove();
if (data.step) {
openStep(data.step);
} else {
openStep('next');
}
$('#linkkk').text('Your id is: ' + data.id);
} else {
$('div.errormsg').remove();
$('<div class="errormsg">'+data.message+"</div>").insertBefore(form);
}
}
});