Question

您好我正在尝试创建一个用户类并使用该类来处理登录，但是由于某种原因，我的用户类构造函数正在将我的mysqli对象作为字符串读取

这是我的代码

Class.php

<?php

/**
* 
*/
class user{ 
    protected $email;
    protected $password; 
    private $db;
    protected $user;

    function __construct(mysqli $db, $email, $password){
        $this -> $db = $db;
        $this -> $email = $email;
        $this -> $password = $password;
    }

    protected function check(){
        $query = 'SELECT * FROM user WHERE email =?';
        $handler = $this -> $db -> prepare($query);
        $handler -> execute(array($this->email));
        if($handler -> num_rows() > 0){
            $user = $handler -> fetch();
            $submitpassword = ($this -> $password);
            if ($submitpassword == $user['password']) {
                return $user;
            }
        }
        return false;
    }
    public function login(){
        $user = $this->check();
        if ($user) {
            $this->_user = $user; // store it so it can be accessed later
            $_SESSION['user_id'] = $user['id'];
            return $user['id'];
        }
        return false;
    }

public function getUser(){
    return $this->_user;
}

}

?>

mysql.php

<?php

if (!isset($_POST['submit'])) {
    echo 'Page not found';
    exit();
}

$servername = "localhost";
$username = "root";
$password = "";
$db = "imagesite";

// Create connection
$conn = new mysqli($servername, $username, $password,$db);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

?>

checklogin.php

<?php

include 'mysql.php';
include 'class.php';

// username and password sent from form 
$email=$_POST['email']; 
$pass=$_POST['password']; 



$user = new user($conn, $email, $pass);
$id = $user -> login();
if ($id) {
    echo 'Logged it as user id: '.$user_id;
}else{
    echo 'did not work';
}


?>

它给了我这个错误

Catchable fatal error: Object of class mysqli could not be converted to string in C:\wamp\www\class.php on line 13

我用这个把头发拉出来，有什么想法吗？

Answer 1

要使用实例属性，不需要美元符号。只需更改

function __construct(mysqli $db, $email, $password){
    $this -> $db = $db;
    $this -> $email = $email;
    $this -> $password = $password;
}

对此：

function __construct(mysqli $db, $email, $password){
    $this->db = $db;
    $this->email = $email;
    $this->password = $password;
}

如何将mysqli对象传递给类构造函数

1 个答案: