带有闭包和名称的 Laravel 路由
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Laravel routing with Closures and Names
提问by Lisa
I'm curious, because I can't find much in the way of documentation on this - How would I name a route if I also want to call a closure?
我很好奇，因为我找不到太多关于此的文档 - 如果我还想调用闭包，我将如何命名路由？
I've found how I can call a controller function, but not how to name the route.
我找到了如何调用控制器函数，但没有找到如何命名路由。
Named Route:
命名路线：
Route::get( '{foo}', ['as' => 'foo.home', 'uses' => 'FooController@home'] );
Route::get( '{foo}', ['as' => 'foo.home', 'uses' => 'FooController@home'] );
Closure Route w/ controller call:
带有控制器调用的关闭路由：
Route::get( '{foo}', function() {
$fooController = $app->make('FooController');
return $fooController->callAction('home', $parameters = array());
});
But I can't find how to incorporate the name of the route into the second example.
但是我找不到如何将路线名称合并到第二个示例中。
回答by lukasgeiter
You can use an array with nameand usesas well:
您也可以将数组与name和一起使用uses：
Route::get('{foo}', array('name' => 'foo.home', 'uses' => function(){
$fooController = $app->make('FooController');
return $fooController->callAction('home', $parameters = array());
}));
It also works without uses(Laravel recognizes the type Closure)
它也可以在没有uses（Laravel 识别类型Closure）的情况下工作
Route::get('{foo}', array('name' => 'foo.home', function(){
$fooController = $app->make('FooController');
return $fooController->callAction('home', $parameters = array());
}));
回答by Mohsen
Easiest way is call it as class
最简单的方法是将其称为类
Route::get( '{foo}', function() {
return (new $fooController)->homeMethod($parameters);
})->name('foohome');;
