PHP致命錯誤：調用pdo登錄中的非對象的成員函數fetchAll（）

Question

我一直在爲客戶端開發Web應用程序，並在構建數據庫登錄代碼時遇到此錯誤。

的config.php文件代碼

php try { 
    // Create connection 
    $dbh = new PDO("mysql:host=" . $localhost . ";dbname=" . $database, $username, $dbpassword); 
} catch (PDOException $e) { 
    // Echo error message if the connection fails 
    echo 'Connection failed: ' . $e->getMessage(); 
    exit; 
} 

而且login.php中碼

require 'config.php'; 
$sql = "SELECT * FROM `Users` WHERE `email` = ?"; 
$stmt = $dbh->prepare($sql); 
$result = $stmt->execute([$_POST['email']]); 
$users = $result->fetchAll(); 

if (isset($users[0])) { 
    if (password_verify($_POST['password'], $users[0]->password)) { 
     // valid login 
     include 'dashboard.php'; 
    } 
    else { 
     // invalid password 
     echo "<p class=''>The password you entered is Incorrect. Please try again.</p>"; 
    } 
} 
else { 
    // invalid email 
    echo "<p class=''>We're sorry, That email is not in our Records. Please Check the spelling and try again.</p>"; 
} 

的Register.php工作，並將該領域的數據庫，但我會添加使當然，我不會錯過任何東西，因爲它試圖訪問數據庫來驗證記錄。

require 'config.php'; 
require 'lib/password.php'; 

$company = $_POST["company"]; 
$email = $_POST["email"]; 
$password = $_POST["password"]; 
$hash = password_hash($password, PASSWORD_BCRYPT); 

$stmt = $dbh->prepare("insert into Users(id,company,email,password) VALUES (?,?,?,?)"); 
$stmt->execute([$id, $company, $email, $hash]); 

Answer 1

您使用了錯誤的對象（$結果）來獲取

您需要使用$語句 做這樣

$users = $stmt->fetchAll(); 

也，你需要的變量傳遞爲在這之前

$stmt->execute(array($_POST['email'])); 

http://php.net/manual/en/pdostatement.fetchall.php

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