Question

当我定义了一个函数时，我收到call to undefined function错误。这是代码：

foreach($db->query("SELECT * FROM 'members' WHERE username = '$user'") as $row) {
    echo "Rank: ".get_rank_tag($user)."<br>";
}

以下是get_rank_tag()功能

的课程

include('password.php');

class User extends Password {
  private $_db;

  function __construct($db) {
    parent::__construct();
    $this->_db = $db;
  }

  public function get_rank_tag($user){   
     try {
        $stmt = $this->_db->prepare('SELECT rank FROM members WHERE username = $user ');
        $stmt->execute();
        $row = $stmt->fetch();
        if($row['rank']==0){return "<span class='label label-default'>Default</span>";}
        elseif($row['rank']==1){return "<span class='label label-success'>VIP</span>";}
        elseif($row['rank']==2){return "<span class='label label-success'>MVP</span>";}
        elseif($row['rank']==3){return "<span class='label label-success'>Elite</span>";}
        elseif($row['rank']==4){return "<span class='label label-success'>Master</span>";}
        elseif($row['rank']==5){return "<span class='label label-success'>Champion</span>";}
        elseif($row['rank']==6){return "<span class='label label-primary'>JM</span>";}
        elseif($row['rank']==7){return "<span class='label label-primary'>Trusted</span>";}
        elseif($row['rank']==8){return "<span class='label label-info'>GM</span>";}
        elseif($row['rank']==9){return "<span class='label label-info'>GM+</span>";}
        elseif($row['rank']==10){return "<span class='label label-warning'>Admin</span>";}
        elseif($row['rank']==11){return "<span class='label label-danger'>Owner</span>";}
     } catch(PDOException $e) {
        echo '';
     }
  }

}

我是PHP的初学者，所以请耐心等待。

Answer 1

您确定您拥有User课程的实例吗？确保你已经实例化了这个类，因为PHP无法识别你的函数。

Answer 2

你必须先创建一个新的类实例才能开始调用它中的函数，否则它不知道在哪里查找函数

$newUser = new User($db);
$dataHold = $newUser->get_rank_tag($user);

然后你可以使用$dataHold来回应你的排名。

调用未定义的函数错误

2 个答案: