Select row table using ajax and php - php

Question

I have a drop menu that includes some category every category has their own subcategory i want to show them buy selecting category name
but it's not working, did i miss something or am i doing it completely wrong?
<script type="text/javascript">
$(function() {
$("#error").hide();
$("#category").change(function(){
$("#error").hide();
var category = $("#category").val();
if (category == "") {
$("#error").show();
return false;
}
var data = $("#form").serialize();
$.ajax({
type:"POST",
url:"index.php",
data:data,
success: function(){
}
});
return false;
});
});
</script>
<form id="form" name="form">
<label for="category" id="error">Empty</label>
<select name="category" id="category">
<option></option>
<option value="News">News</option>
<option value="Items">Items</option>
<option value="Updates">Updates</option>
</select>
</form>
<?php
include("connect.php");
if(!empty($_POST['category'])){
$sql=$con->prepare("SELECT * FROM categorys WHERE category=:category ");
$sql->bindparam(":category",$_POST['category']);
$sql->execute();
while($r=$sql->fetch()){
echo $r['subcategory'];
}
}
?>

Answer 1

SomePage.php
<form id="form" name="form">
<div id='category'>
<label for="category" id="error">Empty</label>
<select name="category" id="category">
<option></option>
<option value="News">News</option>
<option value="Items">Items</option>
<option value="Updates">Updates</option>
</select>
</div>
<div id='subcategory'>
</div>
</form>
<script>
$('#category').change(function(){
var CatName= $('#category').val();
$.ajax({url:"AjaxSelectCategory.php?CatName="+CatName,cache:false,success:function(result){
$('#subcategory').html(result);
}});
});
</script>
Create New Page AjaxSelectCategory.php
[NOTE: If you want to change this page name. Change in <script></script> tag too. Both are related.]
<?php
include("connect.php");
if(!empty($_GET['CatName']))
{
$sql=$con->prepare("SELECT * FROM categorys WHERE category=:category ");
$sql->bindparam(":category",$_GET['CatName']);
$sql->execute();
?>
<select name='subcategory'>
<?php
while($r=$sql->fetch())
{?>
<option value="<?php echo $r['subcategory'];?>"><?php echo $r['subcategory'];?></option>
</select>
<?php }
}?>

Answer 2

Try like this,
$.ajax({
type:"POST",
url:"get_subcategory.php",
data:data,
success: function(data){
alert(data)// this will have the second dropdown. add to desired place in your view.
}
});
In get_subcategory.php
$sql = "Select * from table_name where catregory = ".$_POST'category'];
// execute the query.
$sub = "<select name='sub-category'>";
foreach(result from database as $row) {
$sub .= "<option value='$row->id'>$row->name</option>";
}
$sub .= "</select>";
echo $sub;