Populate a second dropdown accoriding to the values returned by Jquery Ajax

I scrapped this code to check if the mechanism works. Actually what I needed to achieve was to populate a "city dropdown" as per the selection made with the "state" dropdown. The HTML code is as follows.

     <select id="id"></select>
     <select id="state"></select>
     <select id="city"></select>

The Ajax call is as follows.

$(document).on("change", '#state', function(e) {
            var state = $(this).val();


        $.ajax({
            type: "POST",
            data: {state: state},
            url: 'get_name_list.php',
            dataType: 'json',
            success: function(json) {

                var $el = $("#city");
                $el.empty(); // remove old options
                $el.append($("<option></option>")
                        .attr("value", '').text('Please Select'));
                $.each(json, function(value, key) {
                    $el.append($("<option></option>")
                            .attr("value", value).text(key));
                });                                                     




            }
        });

    });

The PHP code encodes the list of values into a JSON with the following structure.

As a test, I inserted the JSON into a Database table and it is working until this point. So it is not a theoretical process. It works until this point in the code.

The JSON structure is as follows.

[{"name":"Alappuzha"},{"name":"Ernakulam"},{"name":"Idukki"},{"name":"Kannur"},{"name":"Kasaragod"},{"name":"Kollam"},{"name":"Kottayam"},{"name":"Kozhikode"},{"name":"Malappuram"},{"name":"Palakkad"},{"name":"Pathanamthitta"},{"name":"Thiruvananthapuram"},{"name":"Thrissur"},{"name":"Wayanad"}]

Then I echo the JSON.

But the dropdown back at the first page is not getting populated. The PHP file is as follows.

$json= array();
$id=$_POST["id"];
//echo  $sample;

 //mysqli_query($mysqli,"INSERT into test_tablr values('','$id')") or die(mysqli_error($mysqli)); 


if ($result = $mysqli->query("SELECT name FROM geo_locations  WHERE parent_id =$id ")) {

    while ($row = mysqli_fetch_assoc($result)) {
            $json1[] = $row;
    }
    $json=json_encode($json1);
}


mysqli_query($mysqli,"INSERT into test_tablr values('','$json')") or die(mysqli_error($mysqli)); 
echo $json;

But unfortunately, the dropdown is not getting populated. Any lights?