小编S. *_*tha的帖子

以下是我的PHP代码.

<?php
session_start();

if(isset($_SESSION['user_id'])) {
    header("Location: /");
}

require 'database.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 an account.';
        endif;
    endif;

它显示错误说:

注意:在第17行的C:\ xampp\htdocs\auth\register.php中只能通过引用传递变量

在第17行,以下代码位于:

$stmt->bindParam(':password',password_hash($_POST['password'], PASSWORD_BCRYPT));

知道问题是什么,只有变量应该通过引用传递？