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Php Laravel include原因错误:方法Illumb\View\View::\uu toString()不能引发异常_Php_Laravel_Laravel 4 - Fatal编程技术网

Php Laravel include原因错误:方法Illumb\View\View::\uu toString()不能引发异常

Php Laravel include原因错误:方法Illumb\View\View::\uu toString()不能引发异常,php,laravel,laravel-4,Php,Laravel,Laravel 4,我在laravel中包含了一个文件,它向我抛出了以下错误: Method Illuminate\View\View::__toString() must not throw an exception 我已将该文件包括在内,如下所示: @include('users.opentasks') 我在同一页上使用了两个include,但如果我使用一个,就没有什么区别了 我真的不确定这意味着什么,以及如何解决这一点的新手在这里。希望有人能帮忙 我的代码如下: UserController.php pu

我在laravel中包含了一个文件,它向我抛出了以下错误:

Method Illuminate\View\View::__toString() must not throw an exception
我已将该文件包括在内,如下所示:

@include('users.opentasks')
我在同一页上使用了两个include,但如果我使用一个,就没有什么区别了

我真的不确定这意味着什么,以及如何解决这一点的新手在这里。希望有人能帮忙

我的代码如下:

UserController.php

public function profile() {
     $status = "closed";
    $data["projects"] = $projects = Auth::user()->projects()->where('status', '!=', $status) 
                                ->paginate(4);

    //$data["tasks"] = $tasks = Auth::user()->tasks->paginate(4);

      //  $data["tasks_pages"] = $tasks->links();

        //Comments pagination
        $data["projects_pages"] = $projects->links();


         if(Request::ajax())
            {

            $html = View::make('users.openprojects', $data)->render();
            return Response::json(array('html' => $html));
        //    $html = View::make('users.opentasks', $data)->render();
          //  return Response::json(array('html' => $html));
        }

        echo View::make('users.profile')->with('projects', $projects);

}

public function opentasks() { 

            $user = User::with(array('tasks', 'tasks.status'))->find(Auth::user()->id); 
            return View::make('users.opentasks')->with('user', $user); 

}
profile.blade.php

@extends("layout")
@section("content")
@include('users.openprojects')
@include('users.opentasks')
@stop
@foreach($user->tasks as $task) 
    {{ $task->task_name }} 
    {{ $task->task_brief}} 
            @if(!is_null($task->status)) 
              {{ $task->status->status_name }}<br/><br/><br/><br/> 
            @endif 
@endforeach
opentasks.blade.php

@extends("layout")
@section("content")
@include('users.openprojects')
@include('users.opentasks')
@stop
@foreach($user->tasks as $task) 
    {{ $task->task_name }} 
    {{ $task->task_brief}} 
            @if(!is_null($task->status)) 
              {{ $task->status->status_name }}<br/><br/><br/><br/> 
            @endif 
@endforeach
@foreach($user->tasks as$task)
{{$task->task_name}
{{$task->task_-brief}
@如果(!为空($task->status))
{{$task->status->status_name}}




@恩迪夫 @endforeach
您可能希望这样做:

if(Request::ajax())
{
  // --- this part of the code is odd
  $html = View::make('users.openprojects', $data)->render();
  return Response::json(array('html' => $html));
  // ---
} else {
  $user = User::with(array('tasks', 'tasks.status'))->find(Auth::user()->id); 

  $data = array(
     'user'     => $user,
     'projects' => $projects
  );

  return View::make('users.profile', $data);
}

echo视图::make('users.profile')->带有('projects',$projects)
echo
更改为
return
,这会向我抛出一个
未定义变量:user
error,这是因为您@include
opentasks
openprojects
也需要数据。您还必须将这些数据与返回的视图一起传递。这仍然给了我一个未定义的变量:user errror。将在4分钟内接受。还有,代码的这一部分有什么奇怪之处?我在ajax的在线教程中看到了它call@g_9020我只是不熟悉它。。。但我现在有点明白了。(也许)。它试图返回呈现视图的json响应:D