First drop down menu to auto change the options of a second dropdown

See below to see the Working Example without using a Database.

Working Example Using MySQL Database

If you wanna connect it using Database, yeah, it is surely possible. Consider this table:

CREATE TABLE `contents` (
    `id` INT PRIMARY KEY AUTO_INCREMENT,
    `name` VARCHAR (255),
    `parent` INT DEFAULT 0
);

INSERT INTO `contents` (`name`, `parent`) VALUES
    ('Names', 0),
    ('Places', 0),
    ('Animals', 0),
    ('Praveen', 1),
    ('Bill Gates', 1),
    ('Steve Jobs', 1),
    ('India', 2),
    ('New York', 2),
    ('London', 2),
    ('Singapore', 2),
    ('Cat', 3),
    ('Dog', 3),
    ('Tiger', 3),
    ('Deer', 3)

Table Structure

+----+------------+--------+
| id | name       | parent |
+----+------------+--------+
|  1 | Names      |      0 |
|  2 | Places     |      0 |
|  3 | Animals    |      0 |
|  4 | Praveen    |      1 |
|  5 | Bill Gates |      1 |
|  6 | Steve Jobs |      1 |
|  7 | India      |      2 |
|  8 | New York   |      2 |
|  9 | London     |      2 |
| 10 | Singapore  |      2 |
| 11 | Cat        |      3 |
| 12 | Dog        |      3 |
| 13 | Tiger      |      3 |
| 14 | Deer       |      3 |
+----+------------+--------+

Initial HTML & PHP Code

Now, lets use PHP to first populate the initial <select>:

<?php
    mysql_connect();
    mysql_select_db("contents");
    $result = mysql_query("SELECT * FROM `contents` WHERE `parent` = 0");
    while(($data = mysql_fetch_array($result)) !== false)
        echo '<option value="', $data['id'],'">', $data['name'],'</option>'
?>

Now the <select> is ready. With its onchange function, we can fire an AJAX event to get the new <select> with the data provided by the parent <select>.

<select onchange="ajaxfunction(this.value)">
    <!-- Options would have been initially populated here -->
</select>

Now for the jQuery function, you can do this way:

<script type="text/javascript">
    function ajaxfunction(parent)
    {
        $.ajax({
            url: 'process.php?parent=" + parent;
            success: function(data) {
                $("#sub").html(data);
            }
        });
    }
</script>

In the HTML, after the <select>, you need to give another select with an id as sub.

<select onchange="ajaxfunction(this.value)">
    <!-- Options would have been initially populated here -->
</select>
<select id="sub"></select>

Processing PHP Source Code

Finally the source code of process.php:

<?php
    mysql_connect();
    mysql_select_db("contents");
    $result = mysql_query("SELECT * FROM `contents` WHERE `parent` = " . mysql_real_escape_string($_GET["parent"]));
    while(($data = mysql_fetch_array($result)) !== false)
        echo "<option value="', $data['id'],'">', $data['name'],'</option>'
?>

Working Example without using a Database

You just need to replace this in the PHP.

<?php
    $parent = array("Name", "Place", "Animals");
    foreach ($parent as $id => $name)
        echo '<option value="s', $id,'">', $name,'</option>'
?>

And for the process.php:

<?php
    $parent = array("Name", "Place", "Animals");
    $s0 = array("Praveen", "Bill Gates", "Steve Jobs");
    foreach (${$_GET["parent"]} as $id => $name)
        echo '<option value="', $data['id'],'">', $data['name'],'</option>'
?>