我正在学习一个创建注册和登录系统的教程。我做的每件事都一样，但当我在表格上按提交时，我仍然会遇到一个奇怪的"严格标准"错误。
<?php
require 'pdoconnect.php';
$message = '';
if(!empty($_POST['email']) && !empty($_POST['password'])):
// Enter the new user in the database
$sql = "INSERT INTO users (email, password) VALUES (:email, :password)";
$stmt = $conn->prepare($sql);
$stmt->bindParam(':email', $_POST['email']);
$stmt->bindParam(':password', password_hash($_POST['password'], PASSWORD_BCRYPT));
if( $stmt->execute() ):
$message = 'Successfully created new user';
else:
$message = 'Sorry there must have been an issue creating your account';
endif;
endif;
?>
上面写着
严格的标准：只有变量应该在第14行（带有密码散列的行）通过引用传递。
我到处寻找解决方案，但找不到。
在绑定之前对密码进行哈希：
// Enter the new user in the database
$sql = "INSERT INTO users (email, password) VALUES (:email, :password)";
$stmt = $conn->prepare($sql);
$hashedPass = password_hash($_POST['password'], PASSWORD_BCRYPT);
$stmt->bindParam(':email', $_POST['email']);
$stmt->bindParam(':password', $hashedPass );
不能将函数传递给绑定参数，只能传递变量。