PHP 未定义的 mysqli_result

Question

如果用户名被占用，我正在检查数据库，但我不断收到此错误“在第 9 行调用未定义的函数 mysqli_result()。”

<?
$connect = mysqli_connect('localhost', 'root', '', 'mydb');
if(isset($_POST['username'])){
  $username = mysqli_real_escape_string($connect, $_POST['username']);
  if (!empty($username)){
    $username_query = mysqli_query($connect, "SELECT COUNT ('id') FROM 'users' WHERE 'username' = '$username'");
    $username_result = mysqli_result($username_query, 0);

    if ($username_result == 0){
        echo 'Available.';
        } else if($username_result == 1){
        echo 'Username unavailable.';
    }
  }
}
?>

可能因为我正在使用 jQuery 处理来自不同文件的输入并用它检查它，它说它是未定义的。同样的代码在我正在观看的教程中运行良好，但他使用的是 mysql 而不是 mysqli。编辑:固定代码看起来像是替换了 $username_query 和 $username_rsult -

    $data = mysqli_query($connect, "SELECT COUNT(`username`) AS num FROM `users` WHERE `username`='".$username."'") or die(mysqli_error());
    $row = mysqli_fetch_assoc($data);
    $numUsers = $row['num'];
if ($numUsers >= 1){
        echo 'That user already exists.';
    }else{
        echo 'Username is available.';
    }

Answer 1

mysqli_result还有其他功能你应该使用的类，例如mysqli_fetch_assoc.

也就是说，最好使用准备好的语句来绑定(bind)和检索数据，例如

$_stmt = $_mysqli->prepare('SELECT `permissies` FROM `gebruiker` WHERE (`id`=? AND `pwdenc`=?)');
$_stmt->bind_param('is', $_cred[0], $_cred[1]);
$_stmt->execute();
$_stmt->bind_result($perm);