Zebra Stripe PHP MYSQL-tabel

Dit zou moeten werken, de syntaxis kan onjuist zijn omdat ik het niet heb getest. Maar de logica is er.

$currentState = 'odd';
$html = '';
while($row = mysql_fetch_array($result)){
    $currentState = ($currentState == 'odd' ? 'even' : 'odd');
    $html .= '<tr class="'.$currentState.'">';
    $html .= '<td><a href="page.php?id=' . $row['ID'] . '">'. $row['Short Name of Fund'] .'</a></td>';
    $html .= '<td>' . $row['I/C'] . '</td>';
    $html .= '<td>' . $row['Fund Manager Company'] . '</td>';
    $html .= '<td>' . $row['Class'] . '</td>';
    $html .= '<td>' . $row['Special Class'] . '</td>';
    $html .= '<td>' . $row['TTR year-to-date %'] . '</td>';
    $html .= '<td>' . $row['Mgmt Fee Effectively Charged'] . '</td>';
    $html .= '<td>' . $row['Total Expenses %'] . '</td>';
    $html .= '<td>' . $row['Fund Size'] . '</td>';
    $html .= '</tr>';
}
echo $html;

BEWERKEN :Code bijgewerkt zodat het werkt.