Generate a route with arbitrary model

Question

I have a generic table.blade.php that manage the build of a table based on the model collection given.

How can I, for example, build the edit link foreach items with the right controller.

Example:

<!-- stored in /resources/views/includes/table.blade.php -->

all table....

<tbody>
@foreach( $items as $model )
 <tr><td>    
 <a href="{{ route( '****.edit', $model ) }}">Edit</a>
 </td></tr>
@endforeach
</tbody>

where **** should be the correct controller for the model.

Will it give you more flexibility if you handle it in the controller? Thus allowing your logic in the controller to choose which model to interact with and return what you need (data/view/etc) . — jeremykenedy, Oct 01 '18 at 00:28
the table is rendered with the ModelController@index action. — Mindexperiment, Oct 01 '18 at 08:23

Artem Zakharchenko · Answer 1 · 2018-10-01T07:17:27.740

0

In routes/web.php:

Route::get('some-link/{model}', 'EntityController@edit')->name('****.edit');

Or in view can use action() helper:

{{ action('EntityController@edit', $params) }}

edited Oct 01 '18 at 07:17

answered Sep 30 '18 at 20:11

Artem Zakharchenko

11
3

I'm using the built-in Route::resource() – Mindexperiment Oct 01 '18 at 08:24
@Mindexperiment check your route name with command php artisan route:list – Artem Zakharchenko Oct 01 '18 at 11:18

score 0 · Answer 2 · answered Sep 30 '18 at 20:17

0

Why don't you pass the controller name from the controller and it shall be rendered to the desired route.

<a href="{{ route( $controllerName.'.edit', $model ) }}">Edit</a>

You can pass variables from controllers using compact()

answered Sep 30 '18 at 20:17

Abdulkarim Itani

251
2
3

this solution works... but, I would like to know if there's a more elegant way to solve my problem – Mindexperiment Sep 30 '18 at 23:15

Generate a route with arbitrary model

2 Answers2