更新表使用PDO：語法錯誤

Question

林做一個函數來改變從表中的用戶密碼，但即時得到一個語法錯誤

下面是函數：

public function changepassword($password, $newpassword) { 
     $user_id = $_SESSION["userSession"]; 
     $stmt = $this->db->prepare("SELECT * FROM user WHERE user_id=:user_id"); 
     $stmt->execute(array(":user_id" => $user_id)); 
     $userRow = $stmt->fetch(PDO::FETCH_ASSOC); 
     if ($password = $userRow['password']) { 
      $sql = "UPDATE user set password=:password WHERE user_id=:user_id"; 
      $stmt2 = $this->db->query($sql); 
      $stmt->execute(array(":user_id" => $user_id, ":password" => $password)); 
      $stmt2->execute(); 
      return true; 
     } else { 
      return false; 
     } 
    } 

這是函數調用

if (isset($_POST['btn-save'])) { 
    $password = $_POST['password']; 
    $newpassword = $_POST['newpassword']; 
    $newpassword2 = $_POST['newpassword2']; 

    if ($newpassword == $newpassword2) { 
     if ($user->changepassword($password, $newpassword)) { 
      header("Location: selfedit.php?inserted"); 
     } else { 
      header("Location: selfedit.php?failure"); 
     } 
    } else { 
     header("Location: selfedit.php?failurematch"); 
    } 
} 

這是錯誤我得到：

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ':password WHERE user_id=:user_id' at line 1' in C:\xampp\htdocs\aDatabase2\class.user.php:37 Stack trace: #0 C:\xampp\htdocs\aDatabase2\class.user.php(37): PDO->query('UPDATE user set...') #1 C:\xampp\htdocs\aDatabase2\selfedit.php(8): USER->changepassword('fereira', 'umdois') #2 {main} thrown in C:\xampp\htdocs\aDatabase2\class.user.php on line 37

編輯1

我根據這個問題的意見改變了功能，讓不同的錯誤

public function changepassword($password, $newpassword) { 
    $user_id = $_SESSION["userSession"]; 
    $stmt = $this->db->prepare("SELECT * FROM user WHERE user_id=:user_id"); 
    $stmt->execute(array(":user_id" => $user_id)); 
    $userRow = $stmt->fetch(PDO::FETCH_ASSOC); 
    if ($password == $userRow['password']) { 
     $sql = "UPDATE user set password=:newpassword WHERE user_id=:user_id"; 
     $stmt2 = $this->db->prepare($sql); 
     $stmt2->execute(array(":user_id" => $user_id, ":password" => $newpassword)); 
     return true; 
    } else { 
     return false; 
    } 
} 

新的錯誤：

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: parameter was not defined' in C:\xampp\htdocs\aDatabase2\class.user.php:37 Stack trace: #0 C:\xampp\htdocs\aDatabase2\class.user.php(37): PDOStatement->execute(Array) #1 C:\xampp\htdocs\aDatabase2\selfedit.php(8): USER->changepassword('ferreira', 'anotherpass') #2 {main} thrown in C:\xampp\htdocs\aDatabase2\class.user.php on line 37

Answer 1

它的工作方式發生變化，從編輯1，this line：

$sql = "UPDATE user set password=:newpassword WHERE user_id=:user_id"; 

要這樣：

$sql = "UPDATE user set password=:newpassword WHERE user_id=:user_id"; 

最終功能：

public function changepassword($password, $newpassword) { 
    $user_id = $_SESSION["userSession"]; 
    $stmt = $this->db->prepare("SELECT * FROM user WHERE user_id=:user_id"); 
    $stmt->execute(array(":user_id" => $user_id)); 
    $userRow = $stmt->fetch(PDO::FETCH_ASSOC); 
    if ($password == $userRow['password']) { 
     $sql = "UPDATE user set password=:password WHERE user_id=:user_id"; 
     $stmt2 = $this->db->prepare($sql); 
     $stmt2->execute(array(":user_id" => $user_id, ":password" => $newpassword)); 
     return true; 
    } else { 
     return false; 
    } 
}