php - How can i populate a dropdown list by selecting the value from another dropdown list?

Question

I have build two drop downs (like state and city) by fetching the records of both drop downs from mysql database and am trying to build the tool in which, while selecting any value (i.e. any state) from first drop down, at that time in second drop down (in city) only those values (cities) under that value (state) selected in first drop down should be visible.

Here's my code:

<tr>    
        <td id='hed'><span style="font-family:verdana,geneva,sans-  serif">State</state></td>
        <td>
        <?php 
        $dbcon = mysql_connect("@ip","@username","@password");

        if($dbcon)
        {
            mysql_select_db("@database", $dbcon);
        }
        else
        {
            die('error connecting to the database');
        }

        $qry = "select @value(state) from @tablename  ";
        $result = mysql_query($qry) or die(mysql_error());

        $dropdown = "<select name='@valuename' id='officeItemList' style='cursor:pointer;cursor:hand;'>";
        while($row = mysql_fetch_array($result))
        {           
            $dropdown .= "
<option value='{$row['@value']}' > {$row['@value']} </option>";
        }
        $dropdown .= "
</select>"; 
        echo $dropdown;
        mysql_close($dbcon);
        ?>
        </td> 
    </tr>

        <tr>
        <td id='hed'><span style="font-family:verdana,geneva,sans-serif">City</span></td>
        <td colspan="1"> 
        <?php 
        $dbcon = mysql_connect("@ip","@username","@password");

        if($dbcon)  
        {
            mysql_select_db("@database", $dbcon);
        }  
        else
        {
            die('error connecting to the database');
        }  

        $qry = "select value2(city) from @tablename where ";
        $result = mysql_query($qry) or die(mysql_error()); 

        $dropdown = "<select name='@value2' id='officeItemList' style='cursor:pointer;cursor:hand;'>";
        while($row = mysql_fetch_array($result)) 
        {

            $dropdown .= "
<option value='{$row['@value2']}' > {$row['@value2']} </option>";
        }
        $dropdown .= "
</select>"; 
        echo $dropdown;
        mysql_close($dbcon);
        ?>      


        </td>
    </tr>

Answer 1

That is the wrong way. Your PHP code is fully executed before showing the page to user. So second query can never know that user choses something.

Right way #1: Do it in two pages. First page contains first combo and when it is submitted second page is generated and shows the second combo.

Right way #2 although not optimal: Do it in one page. Load all possible records for second combo to some JS array. Place listener to first array. When user choses something fill second combo with right records from JS-array.

Right way #3 (most right of them): Do it in a page with AJAX-request in it. User selects a value in the first combo. Its listener sends a request to some server script which returns JSON-object with records for second combo.