php 单个 Route::get() 中的多个路由调用 Laravel 4
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multiple routes in single Route::get() call Laravel 4
提问by AndrewMcLagan
When defining a route in Laravel 4 is it possible to define multiple URI paths within the same route?
在 Laravel 4 中定义路由时，是否可以在同一路由中定义多个 URI 路径？
presently i do the following:
目前我执行以下操作：
Route::get('/', 'DashboardController@index');
Route::get('/dashboard', array('as' => 'dashboard', 'uses' => 'v1\DashboardController@index'));
but this defeats my purpose, i would like to do something like
但这违背了我的目的，我想做类似的事情
Route::get('/, /dashboard', array('as' => 'dashboard', 'uses' => 'DashboardController@index'));
采纳答案by Markus Hofmann
If I understand your question right I'd say:
如果我理解你的问题，我会说：
Use Route Prefixing: http://laravel.com/docs/routing#route-prefixing
使用路由前缀：http: //laravel.com/docs/routing#route-prefixing
Or (Optional) Route Parameters: http://laravel.com/docs/routing#route-parameters
或（可选）路由参数：http: //laravel.com/docs/routing#route-parameters
So for example:
例如：
Route::group(array('prefix' => '/'), function() { Route::get('dashboard', 'DashboardController@index'); });
OR
或者
Route::get('/{dashboard?}', array('as' => 'dashboard', 'uses' => 'DashboardController@index'));
回答by graemec
I believe you need to use an optional parameter with a regular expression:
我相信您需要使用带有正则表达式的可选参数：
Route::get('/{name}', array(
'as' => 'dashboard',
'uses' => 'DashboardController@index')
)->where('name', '(dashboard)?');
*Assuming you want to route to the same controller which is not entirely clear from the question.
*假设您想路由到同一个控制器，但问题中并不完全清楚。
*The current accepted answer matches everything not just /OR /dashboard.
*当前接受的答案匹配所有内容，而不仅仅是/OR /dashboard。
回答by Oluwatobi Samuel Omisakin
I find it interesting for curiosity sake to attempt to solve this question posted by @Alexas a comment under @graemec's answer to post a solution that works:
出于好奇，我发现尝试解决@Alex发布的这个问题作为@graemec发布有效解决方案的答案下的评论很有趣：
Route::get('/{name}', [
'as' => 'dashboard',
'uses' => 'DashboardController@index'
]
)->where('name', 'home|dashboard|'); //add as many as possible separated by |
Because the second argument of where()expects regular expressions so we can assign it to match exactlyany of those separated by |so my initial thought of proposing a whereIn()into Laravel route is resolved by this solution.
因为第二个参数where()期望正则表达式，所以我们可以将它分配为完全匹配任何分隔的那些，|所以我最初提出whereIn()进入 Laravel 路由的想法通过这个解决方案解决。
PS:This example is tested on Laravel 5.4.30
PS：此示例在 Laravel 5.4.30 上测试
Hope someone finds it useful
希望有人觉得它有用
