我正在做一个学校项目，我正在制作一个LMS（学习管理系统）。我遇到了这个奇怪的错误关于mysql数据库如何关闭。然而，我的代码中没有一部分告诉它关闭。
下面是运行的代码：

class user_class {
  // Properties
  private $user_sc;
  public $user_name;
  public $user_pw;
  public $user_fn;
  public $user_mn;
  public $user_ln;
  public $new_data;
  private $query_table;
  private $query_col;
  private $USER_ID;
  private $dbconn;
 
  function __construct($dbconn){
    $this->db = $dbconn;

  }
  
  function login($user_name,$user_pw) {
    $login_info = "SELECT email, pw FROM user_info where email = '$user_name' and pw = '$user_pw'";
    $result = mysqli_query($this->db, $login_info);  
    $count = mysqli_num_rows($result);  
    if($count == 1){ 
      return $this->get_name($user_name);
    }else{  
      return false;  
    }      
  }
}

下面是代码：

class course_class {
    private $dbconn;
 
 function __construct($dbconn){
   $this->db = $dbconn;
}
function dlist($num) {
  $grabdata = "SELECT * FROM couse_def";

  
  $result = mysqli_query($this->db,$grabdata) or die("Couldn't execute query.") ;

  
$tablecode = "<select name='CatID'>";

 //Loop through the results to manage each row
 while ($row = mysqli_fetch_assoc($result))
 {
       $isSelected ="";
       $tablecode .= "<option value='".$row['CID']."'".$isSelected.">".$row['cn']."</option>\n";
 }

 $tablecode .= "</select>";

 return ($tablecode);
 }

}

它们看起来是一样的，但是它们是不同的类，使用相同的sql数据库和相同的数据库连接器代码。
this is the coding error I have been getting
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