I've declared this route:
Route::get('category/{id}{query}{sortOrder}',['as'=>'sorting','uses'=>'CategoryController@searchByField'])->where(['id'=>'[0-9]+','query'=>'price|recent','sortOrder'=>'asc|desc']);
I want to get this in url: http://category/1?field=recent&order=desc How to achieve this?
if you have other parameters in url you can use;
request()->fullUrlWithQuery(["sort"=>"desc"])
Query strings shouldn't be defined in your route as the query string isn't part of the URI.
To access the query string you should use the request object. $request->query() will return an array of all query parameters. You may also use it as such to return a single query param $request->query('key')
class MyController extends Controller
{
public function getAction(\Illuminate\Http\Request $request)
{
dd($request->query());
}
}
You route would then be as such
Route::get('/category/{id}');
Edit for comments:
To generate a URL you may still use the URL generator within Laravel, just supply an array of the query params you wish to be generated with the URL.
url('route', ['query' => 'recent', 'order' => 'desc']);
Request::query()<a href={{url('route_url'}}></a> this. I guess now i can't do that.category/{id}','CategoryController@index' and {!! link_to_action('CategoryController@customQuery',null,['id'=>1,'query'=>'price','order'=>'desc]) !!} goes to index method rather than going to customQuery methodRoute::get('category/{id}/{query}/{sortOrder}', [
'as' => 'sorting',
'uses' => 'CategoryController@searchByField'
])->where([
'id' => '[0-9]+',
'query' => 'price|recent',
'sortOrder' => 'asc|desc'
]);
And your url should looks like this: http://category/1/recent/asc. Also you need a proper .htaccess file in public directory. Without .htaccess file, your url should be look like http://category/?q=1/recent/asc. But I'm not sure about $_GET parameter (?q=).
