我仍然按照登录/注册教程我已经浏览了几次视频,看看我是否在某个地方犯了打字错误,但我似乎无法找到任何错误我&#39 ;得到的是"注意:试图获得非对象的财产"
我确实环顾四周,建议尝试以下解决方案,但它没有工作:
echo $user[0]->data()->username;
我知道它与我的数据类有关,我认为这是我的数据类错误。
的index.php
<?php
require_once 'core/init.php';
if(Session::exists('home')){
echo '<p>' . Session::flash('home', 'You have been registered and can now log in!') . '</p>';
}
$user = new User();
echo $user->data()->username;
?>
<a href="login.php">Login</a>
user.php的
<?php
class User{
private $_db,
$_data,
$_sessionName;
public function __construct($user = null){
$this ->_db = DB::getInstance();
$this->_sessionName = Config::get('session/session_name');
if(!$user){
if(Session::exists($this->_sessionName)){
$user = Session::get($this->_sessionName);
if($this->find($user)){
$this->_isLoggedIn = true;
}else{
//logout
}
}
} else{
$this->find($user);
}
}
public function create($fields = array()){
if($this->_db->insert('users', $fields)){
throw new Exception('There was a problem creating account');
}
}
public function find($user = null){
if($user){
$field = (is_numeric($user)) ? 'id' : 'username';
$data = $this->_db->get('users', array($field, '=', $user));
if($data->count()) {
$this->_data = $data->first();
return true;
}
}
return false;
}
public function login($username = null, $password = null){
$user = $this->find($username);
if($user){
if($this->data()->password ===Hash::make($password, $this->data()->salt)){
Session::put($this->_sessionName, $this->data()->id);
return true;
}
}
return false;
}
public function data(){
return $this->_data;
}
}
的login.php
<?php
require_once 'core/init.php';
if(Input::exists()){
if(Token::check(Input::get('token'))){
$validate = new Validate();
$validation = $validate->check($_POST, array(
'username' => array('required' => true),
'password' => array('required' => true)
));
if ($validation->passed()){
//log user in
$user = new User();
$login = $user->login(Input::get('username'), Input::get('password'));
if($login){
echo 'Success';
}else{
echo'<p>Sorry invalid details</p>';
}
} else{
foreach($validation->errors() as $error)
echo $error, '<br />';
}
}
}
?>
<form action="" method="POST">
<div class="field">
<label for="username" id="username"> Username </label>
<input type="text" name="username" id="username" autocomplete="off">
</div>
<div class="field">
<label for="password" id="password"> Password </label>
<input type="password" name="password" id="password" autocomplete="off">
</div>
<input type="hidden" name="token" value="<?php echo Token::generate();?>">
<input type="submit" value="Login">
</form>
<a href ="index.php">Home</a>
session.php文件
<?php
class Session {
public static function put($name, $value){
return @$_SESSION[$name] = $value;
}
public static function get($name)
{
return self::exists($name) ? @$_SESSION[$name] : null;
}
public static function exists($name)
{
return @$_SESSION[$name] !== null;
}
public static function delete($name){
if(self::exists($name)){
unset($_SESSION[$name]);
}
}
public static function flash($name, $string =''){
if(self::exists($name)){
$session = self::get($name);
self::delete($name);
return $session;
} else {
self::put($name, $string);
}
}
}
答案 0 :(得分:1)
我假设这是违规代码:
$user = new User();
echo $user->data()->username;
这段代码实例化了一个User对象,我们可以看到构造函数需要给出用户名(或者可能是id?)。如果没有给出用户名,则看起来find函数只返回false,因此数据数组将为空。这就是您收到错误消息的原因。
尝试使用当前用户名调用构造函数,例如
$user = new User('test');
如果可以在没有用户名的情况下加载索引页面,那么在尝试使用它之前,您只需要在数据对象上添加额外的检查,例如
if (@$user->data())
echo $user->data()->username;
else
echo "Not logged in";
答案 1 :(得分:0)
我的init.php文件中的基本输入错误,我真的很讨厌自己......