Question

我正在努力理解laravel是如何工作的，而且我很难用它来完成

模型 - User.php用户模型

  <?php

 use Illuminate\Auth\UserInterface;
 use Illuminate\Auth\Reminders\RemindableInterface;

 class User extends Eloquent implements UserInterface, RemindableInterface {

protected $fillable = array('email' , 'username' , 'password', 'code');
/**
 * The database table used by the model.
 *
 * @var string
 */
protected $table = 'users';

/**
 * The attributes excluded from the model's JSON form.
 *
 * @var array
 */
protected $hidden = array('password');

public function Characters()
{
      return $this->hasMany('Character');
}

/**
 * Get the unique identifier for the user.
 *
 * @return mixed
 */
public function getAuthIdentifier()
{
    return $this->getKey();
}

/**
 * Get the password for the user.
 *
 * @return string
 */
public function getAuthPassword()
{
    return $this->password;
}

/**
 * Get the e-mail address where password reminders are sent.
 *
 * @return string
 */
public function getReminderEmail()
{
    return $this->email;
}

}

模型 - Character.php角色模型

  <?php

   class Character extends Eloquent {

protected $table = 'characters';

protected $fillable = array('lord_id','char_name', 'char_dynasty', 'picture');

public function user()
    {
        return $this->belongsTo('User');
    }

public function Titles()
    {
        return $this->hasMany('Title');
    }
 }

  ?>

routes.php文件

Route::group(array('prefix' => 'user'), function()
{

    Route::get("/{user}", array(
        'as' => 'user-profile',
        'uses' => 'ProfileController@user'));

});

ProfileController.php

  <?php
  class ProfileController extends BaseController{



public function user($user) {
    $user = User::where('username', '=', Session::get('theuser') );

    $char = DB::table('characters')
            ->join('users', function($join)
            {
                $join->on('users.id', '=', 'characters.user_id')
                     ->where('characters.id', '=', 'characters.lord_id');
            })
            ->get();

    if($user->count()) {
        $user = $user->first();
        return View::make('layout.profile')
        ->with('user', $user)
        ->with('char', $char);
    }

    return App::abort(404);
}


 }

在我的代码中，我将使用以下内容重定向到此路线：

  return Redirect::route('user-profile', Session::get('theuser'));

在视图中我只想做：欢迎回来，{{$ user-＆gt; username}}，您的主角是{{$ char-＆gt; char_name}}

我的问题是我会收到此错误：尝试在我的视图中获取非对象的属性。我确信它指的是$ char-＆gt; char_name。出了什么问题？我很难理解Laravel。我不知道为什么。提前谢谢！

Answer 1

您应该使用Auth类来获取登录用户的会话信息。

$user = Auth::user();

$welcome_message = "Welcome back, $user->username, your main character is $user->Character->char_name";

您也不需要将任何内容传递给该路线。只需检查用户是否已登录，然后检索数据。您可以从应用程序的任何位置访问此数据。

if (Auth::check())
{
    //the user is logged in
    $user = Auth::user();

要在评论中回答您的问题，请阅读documentation解决所有这些问题，但是：

public function user()
{
    if (Auth::check())
    {
        $user = Auth::user();
        return View::make('rtfm', compact('user'));
    }
    else
    {
         return "The documentation explains all of this very clearly.";
    }
}

Laravel试图获得非对象的属性

1 个答案: