CODEWITHSUNDEEP
laravelmodel-view-controllerroutes
Lovelock
Lovelock

Reputation: 8075

Laravel - multiple urls for one route

cant find anything on this but im sure its simple.

I have a route that I need to duplicate 4 times as I use the url to change the query.

Currently:

Route::get('/', function()
{
  $builds = Blog::findBuilds();
  return View::make('pages/home', compact('builds'));  
});

What I want to do is e.g.:

Route::get(array('/', '/trending', 'staff-picks'), function()
{
  $builds = Blog::findBuilds();
  return View::make('pages/home', compact('builds'));  
});

But that of course does not work. Whats the trick?

Upvotes: 3

Views: 5366

Answers (2)

lukasgeiter
lukasgeiter

Reputation: 152860

As Laravelian's answer demonstrates you can and should use a dynamic route parameter. However if you just do 

Route::get('{slug?}', function($slug = null){}

you will catch every request (except the ones that matches a route that was defined before this one)

To have more control use a regular expression:

Route::get('{slug?}', function($slug = 'index')
{
    $builds = Blog::findBuilds();
    return View::make('pages/home', compact('builds'));  
})->where('slug', '(trending|staff-picks)');

Now slug has to be either: nothing (because its an optional parameter {...?}) trending or staff-picks

Upvotes: 6

Laravelian
Laravelian

Reputation: 598

Route::get('/{slug}', function($slug = null) {
    $builds = Blog::findBuilds($slug);
    return View::make('pages/home', compact('builds')); 
});

You're better off though, putting the function in a controller, and passing it this way:

Route::get('/{slug}', 'BlogController@getSlug');
Route::get('/', 'BlogController@getSlug');

Upvotes: 0

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