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Croppie Working on Localhost, but not on live shared hosting

I got a problem with no error message. I don't know why this happen. My code working 100% on localhost. But when I upload the code to live shared hosting, there is nothing happen after uploading process. I am using codeigniter 3. This is my html code:
<div class="modal-dialog modal-lg">
<div class="modal-content">
<form method="post" enctype="multipart/form-data">
<div class="modal-header bg-dark">
<h5 class="modal-title">Edit Gambar Profil</h5>
<button type="button" class="close" data-dismiss="modal">×</button>
</div>
<div class="modal-body text-center">
<div class="row">
<div class="col-md-12">
<div class="alert alert-warning">
Sebelum melanjutkan, pastikan Foto Profil yang Anda upload sudah tepat. Hanya berkas <b>JPG</b> dan <b>PNG</b> yang diperbolehkan. Maksimal ukuran berkas <b>2 MB</b>.
</div>
</div>
<div class="col-md-4">
<div class="form-group">
<input type="file" class="form-control-uniform-custom" name="gambar" accept="image/*" id="upload" required>
</div>
</div>
<div class="col-md-8">
<center>
<div id="upload-demo" style="width:350px"></div>
<div id="upload-demo-i"></div>
</center>
</div>
</div>
</div>
<div class="modal-footer" id="upload_footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button class="btn bg-dark upload-result"><i class="icon-check pr-2"></i> Set Gambar Profil</button>
</div>
</form>
</div>
</div>
</div>
And this is my javascript code:
<script>
$("#upload-demo").hide();
$("#upload_footer").hide();
//Mulai crop Image
$uploadCrop = $('#upload-demo').croppie({
enableExif: true,
viewport: {
width: 320,
height: 320,
//type: 'circle'
},
boundary: {
width: 350,
height: 350
}
});
$('#upload').on('change', function () {
$("#upload-demo").show();
$("#upload_footer").show();
var reader = new FileReader();
reader.onload = function (e) {
$uploadCrop.croppie('bind', {
url: e.target.result
}).then(function(){
console.log('jQuery bind complete');
});
}
reader.readAsDataURL(this.files[0]);
});
$('.upload-result').on('click', function (ev) {
$uploadCrop.croppie('result', {
type: 'canvas',
size: 'viewport'
}).then(function (resp) {
$.ajax({
url: "<?php echo base_url('profil/postfotoprofil');?>",
type: "POST",
data: {"image":resp, "no_register" : "<?php echo sesuser('no_register');?>"},
success: function (data) {
console.log(data);
html = '<img src="' + resp + '" />';
$("#upload-demo-i").html(html);
}
});
});
});
//Akhir Crop image
</script>
And this is my post ajax url controller:
public function postfotoprofil()
{
header("Access-Control-Allow-Headers: Authorization, Content-Type");
header("Access-Control-Allow-Origin: *");
$post = $this->input->post();
$data = $post['image'];
list($type, $data) = explode(';', $data);
list(, $data) = explode(',', $data);
$data = base64_decode($data);
$imageName = $post['no_register'].'.png';
unlink('sekolah_berkas/user/'.$imageName);
if(file_put_contents('sekolah_berkas/user/'.$imageName, $data)){
$post_data = array(
"gambar" => $imageName
);
$hajar = $this->db->update("pengguna", $post_data, array("no_register" => $post['no_register']));
if($hajar){
$this->session->set_flashdata('pesen', '<script>sukses("proses");</script>');
}else{
$this->session->set_flashdata('pesen', '<script>gagal("proses");</script>');
}
}else{
$this->session->set_flashdata('pesen', '<script>gagal("proses");</script>');
}
redirect(base_url("profil"));
}
Anyone knows about this, please help.
Problem Solved
Alright guys... thank you very much for reply to my thread. The problem is only on the submit button. I change my code from :
<button class="btn bg-dark upload-result"><i class="icon-check pr-2"></i> Set Gambar Profil</button>
become like this :
<a class="btn bg-dark upload-result"><i class="icon-check pr-2"></i> Set Gambar Profil</a>
and the problem solved. With (button) syntax you will submit immediately to server. Therefore you don't have much time while submitting. Then I change it to syntax (a) and really solved my problem. Look like small problem, with no error message you will get headache.

Codeigniter & AJAX - Reset/clear the modal form when close the edit modal

I have a problem on my modal form when I open the edit modal ,the modal pop up and it fetched my data and that is WORKING WELL but when I close the modal and click the add new user, the data is automatically fetched why is it fetched when I close the modal it should be reset to blank?
how can I reset or clear my form modal after i close the modal and not doing anything?
here is my Edit Javascript/Ajax code
$('#btnClose').click(function(){
$('#myForm')[0].reset();
}); //I Tried this block of code, but it didnt work
//Edit/Show
$('#showdata').on('click','.item-edit', function(){
var id = $(this).attr('data');
$('#myModal').modal('show');
$('#myModal').find('.modal-title').text('Edit Employee');
$('#myForm').attr('action', '<?php echo base_url() ?>employees/updateEmployee');
$.ajax({
type: 'ajax',
method: 'get',
url: '<?php echo base_url() ?>employees/editEmployee',
data: {id:id},
async: false,
dataType:'json',
success: function(data){
$('input[name=txtEmployeeName]').val(data.employee_name);
$('textarea[name=txtAddress]').val(data.address);
$('input[name=txtId]').val(data.id);
},
error: function(){
aler('Could not edit Data');
}
});
});
and here is my modal (I use this modal to create and edit my data so it is just a one modal)
<!--MODAL-->
<div id="myModal" class="modal fade" tabindex="-1" role="dialog">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 class="modal-title">Modal title</h4>
</div>
<div class="modal-body">
<form id="myForm" action="" method="post" class="form-horizontal">
<input type="hidden" name="txtId" value="0">
<div class="form-group">
<label for="name" class="label-control col-md-4">Employee Name</label>
<div class="col-md-8">
<input type="text" name="txtEmployeeName" class="form-control">
</div>
</div>
<div class="form-group">
<label for="address" class="label-control col-md-4">Address</label>
<div class="col-md-8">
<textarea class="form-control" name="txtAddress"></textarea>
</div>
</div>
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="button" id="btnSave" class="btn btn-primary">Save changes</button>
</div>
</div><!-- /.modal-content -->
</div><!-- /.modal-dialog -->
</div><!-- /.modal -->
<!--END MODAL-->
IN CASE you need to see my add/create javascript code here it is
$('#btnSave').click(function(){
var url = $('#myForm').attr('action');
var data = $('#myForm').serialize();
//form validation
var employeeName = $('input[name=txtEmployeeName]');
var address = $('textarea[name=txtAddress]');
var formValid = true;
if (employeeName.val() == '') {
employeeName.parent().parent().addClass('has-error');
}else{
employeeName.parent().parent().removeClass('has-error');
formValid = false;
}
if (address.val()=='') {
address.parent().parent().addClass('has-error');
}else{
address.parent().parent().removeClass('has-error');
formValid = false;
}
if (!formValid) {
$.ajax({
type: 'ajax',
method: 'post',
url: url,
data: data,
async: false,
dataType: 'json',
success: function(response){
if (response.success) {
$('#myModal').modal('hide');
$('#myForm')[0].reset();
if (response.type=='add') {
var type = 'added'
}else if(response.type=='update'){
var type = 'updated'
}
$('.alert-success').html('Employee '+type+' successfully').fadeIn().delay(3000).fadeOut('slow');
showAllEmployees();
}else{
alert('Error');
}
},
error: function(){
alert('Data is not added');
}
});
}
});

Replace
$('#btnClose').click(function(){
$('#myForm')[0].reset();
});
with the following code,
$("#myModal").on("hidden.bs.modal", function () {
$('#myForm')[0].reset();
});

How to insert data from a modal PHP Codeigniter

I have a modal form (simple) that I want to insert in my BD, but I'm not getting the expected result
modalview
this is the button that calls my modal:
<button type='button' class='btn btn-info' data-toggle="modal" data-target="#modal-default-cliente"><span class='fa fa-plus'></span></button>
this is my modal:
<div class="modal fade" id="modal-default-cliente">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span></button>
<h4 class="modal-title">Agregar Usuario</h4>
</div>
<div class="modal-body">
<form method="POST" id="clienteform">
<div class="form-group">
<label for="nombrecompleto">Nombre Completo:</label>
<input type="text" class="form-control" name="nombremodal" id="nombremodal" required="required">
</div>
<div class="form-group">
<label for="telefono">Teléfono:</label>
<input type="text" class="form-control" name="telefonomodal" id="telefonomodal">
</div>
<div class="form-group">
<label for="direccion">Dirección:</label>
<input type="text" class="form-control" name="direccionmodal" id="direccionmodal">
</div>
<div class="form-group">
<input type="submit" name="action" class="btn btn-success" value="Guardar">
</div>
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-primary pull-right" data-dismiss="modal">Cerrar</button>
</div>
</div>
</div>
</div>
This is my Javascript code, which by the way the parameters of the form arrive
$(document).on("submit", "#clienteform", function(event){
event.preventDefault();
var nombre = $("#nombremodal").val();
var telefono = $("#telefonomodal").val();
var direccion = $("#direccionmodal").val();
$.ajax({
url: base_url+"mantenimiento/Ventas/agregarClientemodal",
method:'POST',
success: function(data){
alert(data);
$("#modal-default-cliente").modal("hide");
}
});
});
This is my action in the "Ventas" Controller:
public function agregarClientemodal(){
$data = array(
'Nombre' => $this->input->post("nombremodal") ,
'Telefono' => $this->input->post("telefonomodal"),
'Direccion' => $this->input->post("direccionmodal"),
'Estado' => "1"
);
$this->Ventas_model->agregarClientemodal($data);
}
and finally my function in the Sales model:
public function agregarUsuariomodal($data){
return $this->db->insert("Clientes",$data);
}
I'm new to Codeigniter, and when I click on my save button, my modal window does nothing
Expected behavior: save the record and hide the modal
behavior obtained: clicking on the submit does nothing
what am I doing wrong? What do I need to validate? any help for me?

Hope this will help you :
Pass form post values with ajax's data using var formdata = $(this).serialize();, and use site_url() for the URL
Your ajax should be like this :
$(document).ready(function(){
$("#clienteform").on("submit", function(event){
event.preventDefault();
var formdata = $(this).serialize();
$.ajax({
url: '<?=site_url("mantenimiento/Ventas/agregarClientemodal");?>',
method:'POST',
data : formdata,
success: function(data)
{
alert(data);
$("#modal-default-cliente").modal("hide");
}
});
});
});
Your agregarClientemodal method should be like this :
public function agregarClientemodal()
{
$data = array(
'Nombre' => $this->input->post("nombremodal") ,
'Telefono' => $this->input->post("telefonomodal"),
'Direccion' => $this->input->post("direccionmodal"),
'Estado' => "1"
);
$this->Ventas_model->agregarClientemodal($data);
echo "success";
exit;
}

Your base_url variable is not defined in javascript/jquery.
So you need to change that line into:
$.ajax({
url: '<?php echo base_url("mantenimiento/Ventas/agregarClientemodal");?>',
method:'POST',
success: function(data){
alert(data);
$("#modal-default-cliente").modal("hide");
}
});
it will generate correct url.
Further you can check console for error logs.

PHP redirect after form POST

i'm making a download button on my website that opens a modal with a request for password and then if the password is right should take the user to the real download. This is my javascript:
function submit() {
var postData = $("#passwd").serialize();
//alert(postData);
$.ajax({
type: "POST",
url: "http://www.redcraft.it/submit.php",
data: postData,
success: function(redirect) {
alert('Submitted')
}/*,
error: function() {
alert('Failure');
}*/
});
}
and this is my php:
<?php
function redirect() {
$data = $_POST["postData"];
if($data == is_null()) {
echo "Error";
} else {
echo "<script>window.open('http://www.google.com/" + $data + "', '_self')</script>";
}
}
?>
I'm using echo with a js script in php to try different methods to redirect since header() doesn't work. I'm sure the php is correctly called because i've added a row to create a file when the function is called, and the file is correctly generated.
Please, i need your help and don't kill me if i've made some "noobie" errors.
Edit:
This is the html of the modal:
<div class="modal fade bs-modal-sm" tabindex="-1" role="dialog" aria-labelledby="loginModalLabel" aria-hidden="true">
<div class="modal-dialog modal-mediom">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Download mondo RedCraft</h4>
</div>
<div class="modal-body">
<p>Per scaricare il mondo della redcraft inserisci la password che trovi nel video di presentazione del download!</p>
<div class="control-group">
<label class="control-label" for="userid">Password:</label>
<div class="controls">
<input id="passwd" required="" name="passwordinput" type="input" class="form-control input-medium">
</div>
</div>
<div class="control-group">
<label class="control-label" for="confirmsignup"></label>
<div class="controls">
<button id="confirm" onClick="submit()" name="confirmsignup" class="btn btn-primary" data-dismiss="modal">Conferma</button>
</div>
</div>
</div>
</div>
</div>
</div>

just add one div in your html
<div id="outputredirect"></div>
and add below line in your ajax success or replace with your alert
if(redirect == "Error"){
alert(redirect);
} else {
jQuery('#outputredirect').html(redirect);
}

Your javascript code should redirect to the url returned by the php script :
function submit() {
var postData = $("#passwd").serialize();
//alert(postData);
$.ajax({
type: "POST",
url: "http://www.redcraft.it/submit.php",
data: postData,
success: function(redirect) {
location.href(redirect);
}/*,
error: function() {
alert('Failure');
}*/
});
}

How to use error messages from PHP file into AJAX response?

I've following code for file upload :
HTML code :
<a href="#" id="promotion_status_1">
<button type="button" class="btn btn-default brmodalbtn" data-toggle="modal" data-target="#BrandImageModal" id="1">On</button>
</a>
<div class="container">
<div class="modal fade" id="BrandImgeModaal">
<div class="modal-dialog">
<div class="modal-content">
<form id="form" enctype="multipart/form-data" role="form">
<input type="text" class="form-control" name="brand_id" id="brand_id" value="{$data.id}">
<input type="text" name="admin_url" id="admin_url" value="http://localhost/abc.com">
<input type="text" name="op" value="upload_brand_image">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">×</span><span class="sr-only">Close</span></button>
<h4 class="modal-title">Brand Image</h4>
</div>
<div class="modal-body">
<div id="messages"></div>
<input type="file" name="file" id="file">
</div>
<div class="modal-footer">
<button type="submit" class="btn btn-primary">Save</button>
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
</form>
</div>
</div>
</div>
</div>
jQuery AJAX Code :
$('#form').submit(function(e) {
var form = $(this);
var formdata = false;
if(window.FormData) {
formdata = new FormData(form[0]);
}
var formAction = form.attr('action');
$.ajax({
type : 'POST',
url : 'manufacturers.php',
cache : false,
data : formdata ? formdata : form.serialize(),
//data : formdata ? formdata : form.serialize() + '&' + $.param({'op':'upload_brand_image'}),
contentType : false,
processData : false,
success: function(response) {
if(response != 'error') {
//$('#messages').addClass('alert alert-success').text(response);
// OP requested to close the modal
$('#BrandImgeModaal').modal('hide');
} else {
$('#messages').addClass('alert alert-danger').text(response);
}
}
});
e.preventDefault();
});
Now the PHP code is as follows :
$request = $_REQUEST ;
switch( $op ) {
case "add":
case "upload_brand_image":
//print_d($request);
//here if error comes in validation I've an array of errors which needs to be printed in AJAX response. Following is the array
$error_messages = array(
"email" => "Email Id can't be blank",
"email_invalid" => "Email Id is not valid"
);
die;
break;
}
My issue is how should I use the array $error_messages in ajax response and if the error doesn't come then the success message should be passed to ajax function instead of this $errror_messages array.
How to achieve this? Can someone please help me in this regard?
Thanks.

why not use json format?
$.ajax({
dataType : 'json',
........
........
success: function(response) {
if(response.email_invalid){
//do something
}else{
//do something
}
}
});
then in your php echo the error
$error_messages = array(
"email" => "Email Id can't be blank",
"email_invalid" => "Email Id is not valid"
);
echo json_encode($error_messages);