Mysql select box with jquery post

Question

i have a country state city database that displaying in select boxes and everything is working fine. What i want when state select box selected to display city results in a div or another element. When select box defined for city results there is no problem but if i use div gives the error message

Notice: Undefined index: country in C:\wamp\www\list\post.php on line 3.

here is my codes, thanks.

post.php

include("connect.php");
$country = $_POST['country'];
$query = mysqli_query($connect, "select * from state where country_id='$country'");

while( $list = mysqli_fetch_array( $query ) ) {
    echo '<option value=' . $list["id"]. '>' . $list["state_name"] . '</option>';
}

 $state = $_POST['state'];
$query = mysqli_query($connect, "select * from city where state_id='$state'");

while( $list = mysqli_fetch_array( $query ) ) {
    echo $list["city_name"];
}

jquery codes

$( "#countries" ).change( function(){
    $("#states").empty();
    var val = $(this).val();
    $.post("post.php", {country:val}, function(a) {
        $("#states").append(a);
        });
});

$( "#states" ).change( function(){
    $("#cities").empty();
    var val = $(this).val();
    $.post("post.php", {state:val}, function(a) {
        $("#cities").append(a);
        });
});

index.php

<?php include("connect.php"); ?>
<select id="countries">
<option>select</option>
<?php
$query = mysqli_query($connect, 'select * from country');
while( $list = mysqli_fetch_array( $query ) ) {
    echo '<option value=' . $list["id"]. '>' . $list["country_name"] . '</option>';
}
?>
</select>

<select id="states">
<option>select</option>
</select>

<div id="cities"></div>

Answer 1

The script needs to cheeck whether it was called with the country or state parameter, and perform the appropriate query. And the code for returning the cities needs to put them in <option>, just like returning states does.

include("connect.php");
if (isset($_POST['country'])) {
    $country = mysqli_real_escape_string($connect, $_POST['country']);
    $query = mysqli_query($connect, "select * from state where country_id='$country'");

    while( $list = mysqli_fetch_array( $query ) ) {
        echo '<option value=' . $list["id"]. '>' . $list["state_name"] . '</option>';
    }
elseif (isset($_POST['state'])) {

    $state = mysqli_real_escape_string($connect, $_POST['state']);
    $query = mysqli_query($connect, "select * from city where state_id='$state'");

    while( $list = mysqli_fetch_array( $query ) ) {
        '<option value=' . $list["id"]. '>' . $list["city_name"] . '</option>';
    }
}

You should also learn to use parametrized queries instead of substituting variables, to prevent SQL injection. Until then, you should at least escape the parameters.