I want to display a table from database in phpMyAdmin by putting the following conditions that in every different options in drop down menu it displays different table from database by pressing the button of
search. But it is not doing so.
<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
$con=mysqli_connect("localhost","root","","test");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['from']) and isset($_POST['To'])) {
$from = $_POST['from'] ;
$to = $_POST['To'] ;
$table = array($from, $to);
switch ($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);
?>
