Mysql_fetch_data意外打印

Question

我想用MySQL做密碼驗證，但它不起作用。相反，如果子句中比較它打印MySQL_fetch_object

public function __construct($userId, $userPassword) { 
    $this->userId = $userId; 
    $this->userPassword = $userPassword; 
    $connect = mysql_connect('localhost', 'php', 'geheim') or exit(); 
    $users = mysql_select_db("chat", $connect) or exit(); 
    $request = "SELECT userPassword FROM user WHERE userId=$userId"; 
    $result = mysql_query($result); 
    if(mysql_fetch_object($result)->userPassword === $userPassword){ 
     echo "Right Password"; 
    } 

，但只打印這個雖然我從來沒有說過要呼應它：

verificateUser對象（[用戶名：verificateUser：私人] = > 1的userPassword：verificateUser：私人] => 12345）

我在做什麼錯？ Thx的幫助！

Answer 1

試試這個

public function __construct($userId, $userPassword) { 
$this->userId = $userId; 
$this->userPassword = $userPassword; 
$connect = mysql_connect('localhost', 'php', 'geheim') or exit(); 
$users = mysql_select_db("chat", $connect) or exit(); 
$request = "SELECT userPassword FROM user WHERE userId=$userId"; 
$result = mysql_query($result); 
$userpass =mysql_fetch_assoc($result); 
if($userpass['userPassword'] == $userPassword){ 
    echo "Right Password"; 
} 
} 

編輯：，如果你感興趣下面關於

public function __construct($userId, $userPassword) { 
$this->userId = $userId; 
$this->userPassword = $userPassword; 
$connect = mysql_connect('localhost', 'php', 'geheim') or exit(); 
$users = mysql_select_db("chat", $connect) or exit(); 
$request = "SELECT userPassword FROM user WHERE userId=$userId"; 
$result = mysql_query($result); 
$userpass =mysql_fetch_assoc($result); 
if($userpass['userPassword'] == $userPassword){ 
    return true: 
} 
return false; 
} 

然後，此方法可以檢查像

if(__construct($userId, $userPassword) == true) { you welcome} 
    if(__construct($userId, $userPassword) == false) {do your pass again}