成員函數錯誤。任何解釋如何解決

Question

我想如果用戶沒有登錄，顯示的數據，我發現了以下錯誤：

(Fatal error: Call to a member function Fetch() on a non-object in /home/a6150953/public_html/php/classes/class.user.php on line 58) Any help

class.users.php

<?php 
class User 
{ 
    public $db; 
    public $error; 

    public function __construct($con){ 
      $this->db = $con; 
    } 

    /*** for login process ***/ 
    public function check_login($username='', $password=''){ 
      // Validate that your email is a real one 
      if(filter_var($username,FILTER_VALIDATE_EMAIL) !== false) { 
        $password = md5($password); 
        $sql  = "SELECT uid from users WHERE (uemail='$username' or uname='$username') and upass = '$password'"; 
        $result  = $this->db->Fetch($sql); 

         if ($result !== 0) { 
           // this login var will use for the session thing 
           $_SESSION['emailusername'] = $result[0]['uemail']; 
           $_SESSION['uid']   = $result[0]['uid']; 
           $_SESSION['user']   = $this->get_fullname($result[0]['uid'],0); 
           $_SESSION['login']   = true; 
          } 
         else 
          $this->error['account'] = 'Invalid Username/Password'; 
       } 
      else 
       $this->error['email'] = 'Invalid Email Address'; 

      return (!isset($_SESSION['emailusername']))? false:true; 
     } 

    /*** for showing the username or fullname ***/ 
    public function get_fullname($uid, $write = 1){ 

      // --> You can prepare, bind, and execute your values here replacing what you have now....<-- 
      $sql    = "SELECT * FROM users WHERE uid = $uid"; 
      $user_data   = $this->db->Fetch($sql); 

      if($user_data !== 0) { 
        $user['fullname'] = $user_data['fullname']; 
        $user['uemail']  = $user_data['uemail']; 
        $user['uid']  = $user_data['uid']; 

        // This gives the option of returning an array (setting session array) or echoing 
        if($write == 1) 
         echo implode("<br />",$user); 
        else 
         return $user; 
       } 
     } 

    public function check_user($uid) 
     { 
      $sql  = "SELECT * from users WHERE uid='$uid'"; 
      $result  = $this->db->Fetch($sql); 
      $count_row = ($result !== 0)? count($result): 0; 

      return ($count_row == 1); 
     } 

    /*** starting the session ***/ 
    public function get_session() 
     { 
      return $_SESSION['login']; 
     } 

    public function user_logout() 
     { 
      $_SESSION['login'] = FALSE; 
      session_destroy(); 
     } 
} 

？ >

profile.php

<?php 
session_start(); 
//session_destroy(); 

include_once('php/classes/db_config.php'); 
include_once('php/classes/class.user.php'); 

$user1 = new User(""); 

if(isset($_POST['logout'])) { 
     session_destroy(); 
     header('Location: index.php'); 
} 

include_once('php/common/head.php'); 

$all = $con->Fetch("select * from users"); 
?> 
<div class="wrapper"> 
<h1> 
<?php 
if(isset($_SESSION['uid'])){ 
    echo "Profile Page for ". $all[0]['fullname'] ." "; 
}else{ 
    echo "Welcome", "<br/><a href='index.php'>Login</a>"; 
} 
?> 
</h1> 
<pre> 
<div id="profile"> 
<?php 
if(isset($_GET['uid']) || isset($_SESSION['uid'])) { 

    if($_GET['uid'] === $_SESSION['uid']) { 

     echo '<p>'.$all[0]['fullname'].'</p>'; 
     echo '<p>'.$all[0]['uemail'].'</p>'; 

     echo "<form action='' method='post'> 
      <input type='hidden' name='logout' value='true' /> 
      <input type='submit' name='submit' value='Logout'> 
     </form>"; 

    }else if($user1->check_user($uid)){ 

     echo '<p>'.$all[0]['fullname'].'</p>'; 
     echo '<p>'.$all[0]['uemail'].'</p>'; 

    }else if(!$user1->check_user($uid)){ 
     echo "Invalid User"; 
    } 

}else{ 
    echo "Incorrect"; 
} 
?> 
</pre> 
</div> 
</div> 
<?php include_once('php/common/foot.php'); ?> 

Answer 1

你構建User沒有有效數據庫的方法Fetch（8號線profile.php）

$user1 = new User(""); 

我會假設類似（沒有看到你的數據庫類）

$user1 = new User($con);