drop downs from table data

worldop · 12-04-2008, 11:02 PM

Hello, I'm developing a booking system. I've got a database, with records such as( colon representing column break ):
Bristol : Newcastle
Bristol : Manchester
Manchester : Bristol
and so on, I'm trying to make drop down menu's for Departing City, and Destination City using <select><option> html. I'm getting duplicate drop-down options and im not sure how to display one of each of the departing/destination records, Also another thing I don't know how to do is, only display the corresponding Destination: e.g.
if the Departing City is Bristol, then only the destinations from Bristol should appear in the second drop-down.

Here is my code so far, if anyone could help it would be appreciated, thanks.

php Code:

$Query = "SELECT DISTINCT DepartingFrom FROM flights";
$RunQuery = mysqli_query($dbc, $Query) or trigger_error("Query: $Query\n<br />MySQL Error: " . mysqli_error($dbc));


if(mysqli_num_rows($RunQuery) > 0){
    echo '<h1>Book your flight here</h1><br />
            <b>Airport to fly from:</b>';
    echo '<select>';
    while($Row = mysqli_fetch_array($RunQuery)){
            echo '<option>' . $Row['DepartingFrom'] . '</option>';
        } /* end of while loop */         
            echo '</select><br />';
            echo '<b>Airport to Fly to:</b>
            <select>';
            $Query = "SELECT ArrivalAt FROM flights";
            $RunQuery = mysqli_query($dbc, $Query) or trigger_error("Query: $Query\n<br />MySQL Error: " . mysqli_error($dbc)); 
            while($Row = mysqli_fetch_array($RunQuery)){
                echo'<option>' . $Row['ArrivalAt'] . '</option>';
            } /* end of while loop */
            echo'</select>';
        echo '<br />';
    
} /* end of if statement - add error if there were no rows in flights */