mysql_fetch_array() retourneert 'meegeleverd argument is geen geldige MySQL-resultaatbron'

U heeft geen database geselecteerd - gebruik mysql_select_db()

Dat zou zoiets zijn als:

<?php
    $link = mysql_connect('localhost', 'root', 'geheim');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    echo 'Connected successfully';

    $db_selected = mysql_select_db('foo', $link);
    if (!$db_selected) {
        die ('Error selecting database: '. mysql_error());
    }
    echo 'Using database successfully';

    $query = "SELECT * FROM Auctions";
    $result = mysql_query($query);
    while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        foreach($row as $field=>$value) {
            echo "$field: {$value} <br />";
        }
    }
    mysql_close($link);
?>