我在网上找到了一个用php4编写的登录脚本，并试图修改它，使其符合php5。
我有4个类：用户，数据库，表单，邮件
下面是我的用户类的片段
<?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'];
}
}
我调用数据库并像这样形成
$db = new db($config);
$user = new User($db);
$form = new Form;
但它抛出了一个错误
Catchable fatal error: Argument 2 passed to user::__construct() must be an instance of Form, none given, called in C:'wamp'www'ecornwall3'include'user.php on line 900 and defined in C:'wamp'www'ecornwall3'include'user.php on line 30
但我不知道为什么。如果我从构造函数中删除表单$form，它可以正常工作，但我需要访问表单类
您没有向构造函数提供足够的参数User::__construct()。看看声明：
public function __construct(db $db, Form $form)
它需要（如声明的）2 个参数：db类的实例和Form类的实例。
试试这个：
$db = new db($config);
$form = new Form;
$user = new User($db, $form);
为什么很容易。您的构造函数是这样说的：
public function __construct(db $db, Form $form)
这意味着你必须给它 2 件事：类 db 的东西和类Form的东西。
你称之为：
$user = new User($db);
那没有Form，这正是你的错误所说的。如果从构造函数中删除Form，则不再期望它，因此错误不存在，但您没有正确的功能。
你应该做的是将参数添加到构造函数调用中：
$user = new User($db, $form);