Submit form twice display error of blank input - php

Question

I made a form for my website using php, Ajax & jquery. When i submit the form with some inputs 1st time it works fine and but the second time without reloading the page it shows errors. The html input tags are simple with using class id and type attr.
What i want here is the code also runs for the second time if user submit form.
php and jquery:
if(isset($_POST['submit'])){
$name = $_POST['name'];
$website = $_POST['website'];
// $category = $_POST['category'];
$toscheckbox = $_POST['toscheckbox'];
$errorEmpty = false;
$website_error = false;
$name_error = false;
$toscheckbox_error = false;
if(empty($name) || empty($website) || empty($toscheckbox)){
echo "<span class='alert alert-danger' role='alert'>Please fill in all fields!</span>";
$errorEmpty = true;
}
elseif (!filter_var($website, FILTER_VALIDATE_URL)){
echo "<span class='alert alert-danger' role='alert'>Write a valid URL!</span>";
$website_error = true;
}
elseif (!preg_match("/^[a-zA-Z0-9][a-zA-Z0-9 \t]*$/", $name)) {
echo "<span class='alert alert-danger' role='alert'>Please check your Website Name</span>";
$name_error = true;
}
else {
echo "<span class='alert alert-success'>Thanks your submission listed in our Database!</span>";
}
}
else {
echo "<span class='alert alert-danger'>There was something wrong with your form</span>";
}
<script>
// $("#name, #website, #toscheckbox").removeClass("border border-danger");
var errorEmpty = "<?php echo $errorEmpty; ?>";
var website_error = "<?php echo $website_error; ?>";
var name_error = "<?php echo $name_error; ?>";
var toscheckbox_error = "<?php echo $toscheckbox_error; ?>";
if(errorEmpty == true){
$("#name, #website, #toscheckbox").addClass("border border-danger");
}
if(website_error == true) {
$("#website").addClass("border border-danger");
}
if(name_error == true) {
$("#name").addClass("border border-danger");
}
if (errorEmpty == false && name_error == false && website_error == false && toscheckbox_error == false) {
$("#name, #website, #toscheckbox").val("");
}
</script>
<script>
$(document).ready(function() {
$("form").submit(function(event){
event.preventDefault();
var name = $("#name").val();
var website = $("#website").val();
var toscheckbox = $("#toscheckbox").val();
var submit = $("#submit").val();
$(".text-danger").load("ajaxreturn.php", {
name: name,
website: website,
toscheckbox: toscheckbox,
submit: submit
});
});
});
</script>
EDIT html:
<input class="form-control" id="name" type="text" name="name">
<input name="toscheckbox" class="form-check-input" id="toscheckbox" type="checkbox" value="ToS">
<input class="form-control form-control" id="website" type="text" name="website">