使用laravel和blade正确地为img src分配{{route（function）}}和参数

Question

I got a question using laravel, I am trying to call a function and passing some parameters along with it using the {{ route }} method something along these lines here:

<img src="{{route('cacheImage', ['general', 'mini-logo.png']) }}" /> 

My route in my web.php file looks like this:

Route::get('/cache/images', ['uses'=>'HomeController@cache_image','as'=>'cacheImage']);

As a result, I am getting something like this here when the page loads in my img src:

http://localhost:8000/cache/images?general&mini-logo.png

But this is not what I want, what I would like is to have something along these lines here in my img src:

http://localhost:8000/cache/images/general/mini-logo.png

At the same time I want to call the cacheImage function in my controller and pass it both parameters, which are "general", and "mini-logo.png"

How do I achieve this? Can someone give me an example?

Answer 1

Change your route to this.

Route::get('/cache/images/{type}/{name}', [
    'uses' => 'HomeController@cache_image',
    'as' => 'cacheImage'
]);

Controller method to this to be able to receive the url segments as parameters.

public function cache_image($type, $name)
{
    dd($type, $name);
}

Then you can generate links like so.

$link = route('cacheImage', ['general', 'mini-logo.png']);
// generates http://example.dev/cache/images/general/mini-logo.png