메신저 꽤 phop 클래스 등등에 새로워주세요. 내게 맨손으로 제발 :)다른 2 개의 클래스에 대한 액세스 권한이있는 클래스를 제공합니까?
나는 php4로 작성되었으며 php5 complient가되도록 수정하려고 시도한 온라인 로그인 스크립트가 있습니다.
전 4 개 수업을 : 사용자, DB, 양식, 우편물을 아래
내 사용자 클래스<?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'];
}
}
의 조각이고 나는 데이터베이스를 호출하고 그래서
$db = new db($config);
$user = new User($db);
$form = new Form;
처럼 형성
하지만 오류가 발생 함
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
하지만 확실하지 않은 이유는 무엇입니까? 내가 그것을 잘 작동하는 구조 함수에서 양식 $ 양식을 제거 할 수 있지만 내가 폼 클래스
에 액세스해야하는 경우 도움 주시면 감사하겠습니다 :)
도움을 주셔서 대단히 감사드립니다! – user2886669