为一个班级提供2个其他班级的访问权限？

Question

我有一个在网上找到的登录脚本，该脚本是用php4编写的，并尝试对其进行修改，使其符合php5。

我有4个类别：user，db，form，mailer

以下是我的用户类别的摘要

<?php
include("include/database.php");
include("include/mailer.php");
include("include/form.php");

include("constants.php");

class user
{
var $username;     //Username given on sign-up
var $firstname;
var $lastname;
var $userid;       //Random value generated on current login
var $userlevel;    //The level to which the user pertains
var $time;         //Time user was last active (page loaded)
var $logged_in;    //True if user is logged in, false otherwise
var $userinfo = array();  //The array holding all user info
var $url;          //The page url current being viewed
var $referrer;     //Last recorded site page viewed
var $num_active_users;   //Number of active users viewing site
var $num_active_guests;  //Number of active guests viewing site
var $num_members;        //Number of signed-up users

/**
    * Note: referrer should really only be considered the actual
    * page referrer in process.php, any other time it may be
    * inaccurate.
    */

public function __construct(db $db, Form $form)
{
    $this->database = $db;
    $this->form = $form;
    $this->time = time();
    $this->startSession();

    $this->num_members = -1;

    if(TRACK_VISITORS)
    {
        /* Calculate number of users at site */
        $this->calcNumActiveUsers();

        /* Calculate number of guests at site */
        $this->calcNumActiveGuests();
    }


}   
   /**
   * startSession - Performs all the actions necessary to 
   * initialize this session object. Tries to determine if the
   * the user has logged in already, and sets the variables 
   * accordingly. Also takes advantage of this page load to
   * update the active visitors tables.
   */
  function startSession()
  {
    session_start();   //Tell PHP to start the session

    /* Determine if user is logged in */
    $this->logged_in = $this->checkLogin();

    /**
    * Set guest value to users not logged in, and update
    * active guests table accordingly.
    */
    if(!$this->logged_in)
    {
        $this->username = $_SESSION['username'] = GUEST_NAME;
        $this->userlevel = GUEST_LEVEL;
        $this->addActiveGuest($_SERVER['REMOTE_ADDR'], $this->time);
    }
    /* Update users last active timestamp */
    else
    {
        $this->addActiveUser($this->username, $this->time);
    }

    /* Remove inactive visitors from database */
    $this->removeInactiveUsers();
    $this->removeInactiveGuests();

    /* Set referrer page */
    if(isset($_SESSION['url']))
    {
         $this->referrer = $_SESSION['url'];
    }
    else
    {
        $this->referrer = "/";
    }

    /* Set current url */
    $this->url = $_SESSION['url'] = $_SERVER['PHP_SELF'];
  }
}

我像这样调用数据库和表格

$db = new db($config);
$user = new User($db);
$form = new Form;

但是它抛出一个错误

Catchable fatal error: Argument 2 passed to user::__construct() must be an instance of Form, none given, called in C:\wamp\www\ecornwall3\include\user.php on line 900 and defined in C:\wamp\www\ecornwall3\include\user.php on line 30

但是我不确定为什么。 如果我从构造函数中删除表单$ form，它可以正常工作，但是我需要访问表单类

Answer 1

为什么很容易。 您的构造函数说：

public function __construct(db $db, Form $form)

这意味着您必须给它两件事： db类和Form类。

您称之为：

$user = new User($db);

那没有Form ，而这正是您错误所言。 如果从构造函数中删除Form ，您将不再期望它，因此不会出现错误，但是您没有正确的功能。

您应该做的是将参数添加到构造函数调用中：

    $user = new User($db, $form);

Answer 2

您没有为User::__construct()构造函数提供足够的参数。 看一下声明：

public function __construct(db $db, Form $form)

它需要（声明）两个参数： db类的实例和Form类的实例。

尝试这个：

$db = new db($config);
$form = new Form;
$user = new User($db, $form);