Codeigniter，PHP 构造函数 - 缺少参数，即使它存在

Question

我收到这样的警告：

A PHP ERROR WAS ENCOUNTERED
Severity: Warning
Message: Missing argument 1 for User_model::__construct(), called in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\CI_PP\system\core\Loader.php on line 303 and defined
Filename: models/user_model.php
Line Number: 20

在 CodeIgniter、NetBeans、PHP 5.4 中运行应用程序时。

这是models/user_model.php的代码（玩玩，请忽略避免OOP原则）：

<?php

  class User_model extends MY_Model {

    public $_table = 'pp_user';
    public $primary_key = 'u_id';
    public $firstname;
    public $lastname;
    public $emailAddress;
    public $password;
    public $gender;
    public $deliveryAddress;
    public $address;
    public $city;
    public $zip;
    public $country;
    public $isAdmin;

//this is line nr.20:
    public function __construct($firstname, $lastname, $emailAddress, $password, $gender, $address, $deliveryAddress, $city, $zip, $country, $isAdmin) {
        parent::__construct();

        $this->firstname = $firstname;
        $this->lastname = $lastname;
        $this->emailAddress = $emailAddress;
        $this->password = $password; //TODO!
        if ($gender == 'male') {
            $this->gender = 0;
        } else if ($gender == 'female') {
            $this->gender = 1;
        } else {
            $this->gender = -1;
        }
        $this->deliveryAddress = $deliveryAddress;
        $this->address = $address;
        $this->city = $city;
        $this->zip = $zip;
        $this->country = $country;
        $this->isAdmin = $isAdmin;
    }
}

我在registration.php（控制器）中调用构造函数：

            $firstname = $this->input->post('tf_first_name');
        $lastname = $this->input->post('tf_last_name');
        $emailAddress = $this->input->post('tf_email_address');
        $password = $this->input->post('tf_password_base');
        $gender = $this->input->post('tf_gender');
        $address = $this->input->post('tf_address');
        $deliveryAddress = $this->input->post('tf_delivery_addres');
        $city = $this->input->post('tf_city');
        $zip = $this->input->post('tf_zip');
        $country = $this->input->post('tf_country');
        $isAdmin = FALSE;


        $user_instance = new User_model(
                        $firstname,
                        $lastname,
                        $emailAddress,
                        $password,
                        $gender,
                        $address,
                        $deliveryAddress,
                        $city,
                        $zip,
                        $country,
                        $isAdmin);

如果我从以下位置更改 contor 参数：

public function __construct($firstname, $lastname, $emailAddress, $password, $gender, $address, $deliveryAddress, $city, $zip, $country, $isAdmin) 

到：

public function __construct($firstname="", $lastname="", ...) {

然后它可以工作，但我不喜欢这样的解决方案。我一直在网上搜索提示，但根据 PHP OOP 教程和 PHP 手册，它看起来不错。

303 行的 Loader.php 正在这样做：

$CI->$name = new $model();

我在实例化对象时尝试将参数更改为直接参数，我尝试删除与父类的关系，但问题仍然存在。

我真的很好奇可能是什么问题，有什么想法吗？

Answer 1

这是因为 CodeIgniter 在加载应用程序时调用 User_model::__construct()。 CodeIgniter 必须先加载所有模型，然后才能使用它们，这意味着您不能或不应该将参数传递给它们。您需要将该代码从 __construct 移动到另一个函数，例如 User_model 类中的 add_user，您可以将数据传递到该函数中。

<?php

class User_model extends MY_Model {

    public $_table = 'pp_user';
    public $primary_key = 'u_id';
    public $firstname;
    public $lastname;
    public $emailAddress;
    public $password;
    public $gender;
    public $deliveryAddress;
    public $address;
    public $city;
    public $zip;
    public $country;
    public $isAdmin;

    public function __construct() {
        parent::__construct();
    }

    public function add($firstname, $lastname, $emailAddress, $password, $gender, $address, $deliveryAddress, $city, $zip, $country, $isAdmin) {
        $this->firstname = $firstname;
        $this->lastname = $lastname;
        $this->emailAddress = $emailAddress;
        $this->password = $password; //TODO!
        if ($gender == 'male') {
            $this->gender = 0;
        } else if ($gender == 'female') {
            $this->gender = 1;
        } else {
            $this->gender = -1;
        }
        $this->deliveryAddress = $deliveryAddress;
        $this->address = $address;
        $this->city = $city;
        $this->zip = $zip;
        $this->country = $country;
        $this->isAdmin = $isAdmin;
    }
}

然后像这样调用模型。

$firstname = $this->input->post('tf_first_name');
$lastname = $this->input->post('tf_last_name');
$emailAddress = $this->input->post('tf_email_address');
$password = $this->input->post('tf_password_base');
$gender = $this->input->post('tf_gender');
$address = $this->input->post('tf_address');
$deliveryAddress = $this->input->post('tf_delivery_addres');
$city = $this->input->post('tf_city');
$zip = $this->input->post('tf_zip');
$country = $this->input->post('tf_country');
$isAdmin = FALSE;

$this->load->model('User_model');
$this->User_model->add($firstname,
                $lastname,
                $emailAddress,
                $password,
                $gender,
                $address,
                $deliveryAddress,
                $city,
                $zip,
                $country,
                $isAdmin);

Answer 2

你必须这样做

public function __construct($firstname="", $lastname="", ...) {

否则你在创建类的对象时传递了参数 因为当你创建类的对象时，构造方法会执行，所以你在创建类的对象期间传递了所有参数值，或者你必须使用上面的方法