How to make multiple MySQL connections from PHP

itemprop="text">

The code below is working for one if
statement but not giving results for another else if and it is showing the 'flights'
table in first query but after another condition not display the other table named 'isb
to
muree'

$from
            = isset($_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");
$result = mysql_query("SELECT * FROM
            flights");
if (!$result) {

 die("Query to show fields
            from table failed");
}



$fields_num
            = mysql_num_fields($result);

echo "
Table:
            'flights'
";
echo "            border='1'>";



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

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



            echo "\n";
}
}

else if( $from
            =='Islamabad'){
 if($To == 'murree'){
if (!mysql_connect($db_host,
            $db_user))
 die("Can't connect to database");

if
            (!mysql_select_db($database))

 die("Can't select
            database");

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

echo "
Table: 'isb to
            murree'
";
echo "            border='1'>";



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

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



            echo
            "\n";

}
}
}
}


mysqli_close($con);

?>

Answer

You should move the database connection
variables to the top of your code (so they are outside of the if
statement)

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