dpspn60064 2015-12-13 08:41
浏览 46
已采纳

在laravel 4中加载视图时的undefined $ load属性

I need to do the process of loading a view inside my controller which can be done using something like:

$main['menu'] = $this->load->view('myView', NULL, TRUE);

but when executing I receive an error saying Undefined property: MainController::$load

How can this be fixed, or if you can give me another way to do the work

  • 写回答

1条回答 默认 最新

  • dongwu8064 2015-12-13 09:35
    关注

    You can generate view in your controller with:

    // get the view object
    $view = View::make('myView'));
    
    // get view content as string
    $content = $view->render();
    
    // pass the content to another view
    $anotherView = View::make('anotherView', array('content' => $content));
    

    You can read more about how to use views in Laravel 4 here: http://laravel.com/docs/4.2/responses#views. I suggest you have a look here as it seems that what you're trying to do is not the standard way of doing stuff in Laravel.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 乌班图ip地址配置及远程SSH
  • ¥15 怎么让点阵屏显示静态爱心,用keiluVision5写出让点阵屏显示静态爱心的代码,越快越好
  • ¥15 PSPICE制作一个加法器
  • ¥15 javaweb项目无法正常跳转
  • ¥15 VMBox虚拟机无法访问
  • ¥15 skd显示找不到头文件
  • ¥15 机器视觉中图片中长度与真实长度的关系
  • ¥15 fastreport table 怎么只让每页的最下面和最顶部有横线
  • ¥15 java 的protected权限 ,问题在注释里
  • ¥15 这个是哪里有问题啊?