如何使用此php代码从wamp数据库访问mysql表？

Question

如何通过在Wamp Server中使用php从数据库访问表。我已经完成了以下代码，但由于某些原因而无法正常工作。是否有任何内容要放入“ action =“” page.i想在下拉菜单中的任何其他条目上并按搜索按钮显示数据库中的表。

<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


if(isset($_POST['from']) and isset($_POST['To'])) {
$from = $_POST['from'] ;
$to = $_POST['To'] ;
$table = array($from, $to);
$con=mysqli_connect("localhost");
$mydb=mysql_select_db("homedb"); 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}


switch ($table) {
  case array ("Islamabad", "Lahore") :
$result = mysqli_query($con,"SELECT * FROM flights");
echo "</flights>";                                    //table name is flights


 break;
  case array ("Islamabad", "Murree") :

$result = mysqli_query($con,"SELECT * FROM isb to murree");
echo "</isb to murree>";                                     //table name isb to murree
  ;
 break;
  case array ("Islamabad", "Muzaffarabad") :

$result = mysqli_query($con,"SELECT * FROM isb to muzz");
echo "</isb to muzz>";
 break;
//.....
//......
default:
echo "Your choice is nor valid !!";
}

}
mysqli_close($con);
?>