PHP未定义的mysqli_result


PHP undefined mysqli_result

我正在检查数据库是否使用了用户名,但我一直收到错误"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.';
    }
  }
}
?>

也许是因为我使用jQuery来处理来自不同文件的输入,并用它来检查它,这意味着它是未定义的。这段代码在我正在观看的教程中运行得很好,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.';
    }

mysqli_result类还有其他应该使用的函数,例如mysqli_fetch_assoc

也就是说,最好使用准备好的语句来绑定和检索数据,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);