警告: mysql_result() [Function.mysql-结果]：无法跳转到MySQL结果索引5的第1行

Question

我正在尝试执行以下代码。但是得到了这个错误："mysql_result() function.mysql- result :无法跳转到MySQL结果索引5的第1行“，第0索引的值被打印出来，但剩下的值给出了上面提到的错误。我的代码是：

$s = mysql_query( "SELECT USER_NAME, LOCATION, LANGUAGE, PHONE FROM USERS WHERE USER_ID = '$userid'" );
    if(mysql_fetch_row( $s ) ){
        $username = mysql_result( $s, 0 );
        $location = mysql_result( $s, 1 );
        $language = mysql_result( $s, 2 );
        $phoneNum = mysql_result( $s, 3 );

    echo $username;
    echo $location;
    echo $language;
    echo $phoneNum;
}

请您解释一下错误。

Answer 1

$s = mysql_query( "SELECT USER_NAME, LOCATION, LANGUAGE, PHONE FROM USERS WHERE USER_ID = '$userid'" );

$row    = mysql_fetch_row($s);
    if($row){
        $username = $row[0];
        $location = $row[1];
        $language = $row[2];
        $phoneNum = $row[3];

    echo $username;
    echo $location;
    echo $language;
    echo $phoneNum;

}