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);
?>
