as the code below is accesing the database table directly but i want it to display the table content on giving conditions in drop down menu like when i select islamabad in one drop down menu and lahore in other as given in code and press search buttonn then it display the table flights.but it is displaying it directly

<p class="h2">Quick Search</p>
    <div class="sb2_opts">
     <p>
   </p>
<form method="post" action="">
 <p>Enter your source and destination.</p>
<p>
    From:</p>
<select name="from">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<p>
    To:</p>
   <select name="To">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<input type="submit" value="search" /> 
</form>
</form> </table>

<?php


 $from =  isset($_POST['from'])?$_POST['from']:'';
 $to = isset($_POST['to'])?$_POST['to']:'';

 if( $from =='Islamabad'){
     if($to == 'Lahore'){


$db_host = 'localhost';
$db_user = 'root';


$database = 'homedb';
$table = 'flights';

if (!mysql_connect($db_host, $db_user))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
    die("Query to show fields from table failed");
}



$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
    die("Query to show fields from table failed");
}

$fields_num = mysql_num_fields($result);

echo "<h1>Table: {$table}</h1>";
echo "<table border='1'><tr>";


while($row = mysql_fetch_row($result))
{
    echo "<tr>";

    // $row is array... foreach( .. ) puts every element
    // of $row to $cell variable
    foreach($row as $cell)
        echo "<td>$cell</td>";

    echo "</tr>\n";
}
}



}
mysqli_close($con);
?>