Question

我有/app/resources/views/category/index.blade.php

在index.blade.php我想知道它位于“ category ”文件夹中。是否有laravel函数将返回当前视图的路径？

类似的东西：

 $folder = view_path(); // "/category"

Answer 1

在班级顶部添加名称空间：use Illuminate\Support\Facades\File; 运行：

$files = File::allFiles(resource_path('views'));

  foreach ($files as $file) {
     echo (string) $file, "\n";
  }

或您的情况：

$files = File::files(resource_path('views/category'));

foreach ($files as $path => $value) {
     echo $value->getPath() . "\n";
}

Answer 2

您可以使用Laravel View Composers

视图作曲家是a时调用的回调或类方法视图呈现。如果您有要绑定到视图的数据每次渲染视图时，视图编辑器都可以帮助您进行组织把这个逻辑放到一个地方。

在文件app/Providers/AppServiceProvider.php中，修改public function boot()方法

/**
 * Bootstrap any application services.
 *
 * @return void
 */
public function boot(Application $app)
{

  view()->composer('*', function($view) {
       // get view paths from config - this is absolute path
       $configPaths      = config('view.paths');

       // get view's full path
       $absoluteViewPath = $view->getPath();

       foreach($configPaths as $configPath) {
         // see if current view's path matches path from config
         // both are absolute paths
         if ( strpos($absoluteViewPath, $configPath) !== false ) {
           $path = str_replace($configPath, '', $absoluteViewPath);

           // remove view name - from last position of `/`
           $dir  = substr($path, 0, strrpos( $path, '/'));

           // attach to views
           view()->share('view_folder', $dir);

           // view matched, stop looking
           break;
         }

       }

       // if nothing was found
       view()->share('view_folder', '');
   });
}

在此之后，在任何刀片模板中，您现在都可以访问变量

{{ $view_folder }}

获取laravel视图

2 个答案: