求个搜索引擎来路判断代码进行跳转解决方法

$RUrl = $_SERVER ['HTTP_REFERER'];if (! empty ( $RUrl )) {    $UArray = parse_url ( $RUrl );    $UHost = $UArray ['host'];    switch ($UHost) {        case 'www.baidu.com' :            header ( "Location: http://aaa.com" );            break;        case 'www.google.com' :            header ( "Location: http://bbb.com" );            break;        case 'www.sogou.com' :            header ( "Location: http://ccc.com" );            break;        default :            echo $UHost;            break;    }}echo '直接輸入';