遇到PDO查詢問題

Question

有人可以看一下這段代碼嗎？ 我對PDO方法非常陌生，由於某種原因，每當我提交時，它都會一直導致500錯誤。

我將其范圍縮小到：

可能是這部分嗎？ $hash = $stmt['hash'];

if(empty($response['error'])){
    $stmt = $db->prepare("SELECT * FROM Login WHERE username= :username"); // Prepare the query
 // Bind the parameters to the query
    $stmt->bindParam(':username', $username);
    //Carry out the query
    $stmt->execute();
    $hash = $stmt['hash'];

    $affectedRows = $stmt->rowCount(); // Getting affected rows count
    if($affectedRows != 1){
        $response['error'][] = "No User is related to the Username";
    }
    if(password_verify($password, $hash))
    {
      $_SESSION['username'] = $_POST['username'];
            $_SESSION['userid'] = $stmt['ID'];
    }
    else
    {
      $response['error'][] = "Your password is invalid.";
    }
}

如果您需要更多信息，請詢問我將竭盡所能。

Answer 1

您需要獲取查詢結果才能訪問它。 我不確定這是否是您的問題，我認為$hash只會設置為Resource Id＃x，而不是您想要的值，而不是500。這是獲取方法（ http://php.net/manual/ zh_cn / pdostatement.fetch.php ）

$stmt = $db->prepare("SELECT * FROM Login WHERE username= :username"); // Prepare the query
 // Bind the parameters to the query
    $stmt->bindParam(':username', $username);
    //Carry out the query
    $stmt->execute();
  //if you will only be getting back one result you dont need the while or hashes as an array
   while($result = $stmt->fetch(PDO::FETCH_ASSOC)){
    $hashes[] = $result['hash'];
   }

這是啟用錯誤報告PHP生產服務器的線程-打開錯誤消息

另外，您不必綁定即可通過PDO傳遞值。 你也可以做

$stmt = $db->prepare("SELECT * FROM Login WHERE username= ?"); // Prepare the query
$stmt->execute(array($username));

Answer 2

您的代碼確實很亂。 只是為了幫助您入門：

if (empty($response['error'])) {
    if (isset($_POST['username'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];
        $stmt = $db->prepare("SELECT * FROM Login WHERE username= :username"); 
        $stmt->bindParam(':username', $username);
        $stmt->execute();
        if ($row  = $stmt->fetch(PDO::FETCH_ASSOC)) {
           $hash = $row['hash'];
           if(password_verify($password, $hash)) {
              $_SESSION['username'] = $username;
              $_SESSION['userid'] = $stmt['ID'];
           } else {
              $response['error'][] = "Your password is invalid.";
           }
        } else {
           $response['error'][] = "No User is related to the Username";
        }
    } else {
      $response['error'][] = "Username is not set!";
    }
}