Php 类别';App\Http\Controllers\admin\Auth';在拉维5中未发现
登录时在laravel 5中找不到类“App\Http\Controllers\admin\Auth”之类的错误。我是新来的,所以请帮助我或给我一些教程链接,为完整的laravel应用程序开发与管理方面 Routes.phpPhp 类别';App\Http\Controllers\admin\Auth';在拉维5中未发现,php,Php,登录时在laravel 5中找不到类“App\Http\Controllers\admin\Auth”之类的错误。我是新来的,所以请帮助我或给我一些教程链接,为完整的laravel应用程序开发与管理方面 Routes.php Route::group(array('prefix'=>'admin'),function(){ Route::get('login', 'admin\AdminHomeController@showLogin'); Route::post('che
Route::group(array('prefix'=>'admin'),function(){
Route::get('login', 'admin\AdminHomeController@showLogin');
Route::post('check','admin\AdminHomeController@checkLogin');
});
<?php namespace App\Http\Controllers\admin;
use App\Http\Requests;
use App\Http\Controllers\Controller;
use Illuminate\Http\Request;
class AdminHomeController extends Controller {
//
public function showLogin()
{
return view('admin.login');
}
public function checkLogin(Request $request)
{
$data=array(
'username'=>$request->get('username'),
'password'=>$request->get('password')
);
if(Auth::attempt($data))
{
return redirect::intended('admin/dashboard');
}
else
{
return redirect('admin/login');
}
}
public function logout()
{
Auth::logout();
return redirect('admin/login');
}
public function showDashboard()
{
return view('admin.dashboard');
}
}
<html>
<body>
{!! Form::open(array('url' => 'admin/check', 'id' => 'login')) !!}
<input type="text" name="username" id="username" placeholder="Enter any username" />
<input type="password" name="password" id="password" placeholder="Enter any password" />
<button name="submit">Sign In</button>
{!! Form::close() !!}
</body>
</html>
AdminHomeController.php
Route::group(array('prefix'=>'admin'),function(){
Route::get('login', 'admin\AdminHomeController@showLogin');
Route::post('check','admin\AdminHomeController@checkLogin');
});
<?php namespace App\Http\Controllers\admin;
use App\Http\Requests;
use App\Http\Controllers\Controller;
use Illuminate\Http\Request;
class AdminHomeController extends Controller {
//
public function showLogin()
{
return view('admin.login');
}
public function checkLogin(Request $request)
{
$data=array(
'username'=>$request->get('username'),
'password'=>$request->get('password')
);
if(Auth::attempt($data))
{
return redirect::intended('admin/dashboard');
}
else
{
return redirect('admin/login');
}
}
public function logout()
{
Auth::logout();
return redirect('admin/login');
}
public function showDashboard()
{
return view('admin.dashboard');
}
}
<html>
<body>
{!! Form::open(array('url' => 'admin/check', 'id' => 'login')) !!}
<input type="text" name="username" id="username" placeholder="Enter any username" />
<input type="password" name="password" id="password" placeholder="Enter any password" />
<button name="submit">Sign In</button>
{!! Form::close() !!}
</body>
</html>
因为除非您专门导入Auth
名称空间,否则您的控制器是有名称空间的,因此PHP将假定它位于类的名称空间下,并给出此错误
要解决此问题,请添加使用Auth
位于AdminHomeController
文件的顶部,以及其他use语句,或者在Auth
的所有实例前面加上反斜杠:\Auth
,让PHP知道如何从全局命名空间加载它 但是我不知道为什么它会转到else条件公共函数checkLogin(Request$Request){$data=array('username'=>$Request->get('username'),'password'=>$Request->get('password');如果(Auth::true($data)){echo“我不过来”;die;return redirect::designed('admin/dashboard');}否则{echo“始终在这里”;die;return redirect('admin/login');}}在auth.php文件'table'=>'sysadmin'和sysadmin中只有两个字段“username”主键和“password”