我希望能够通过短链接(如articles/ID)到达文章,但我也希望通过完整链接articles/ID/category/slug到达那里.但是我无法执行以下操作:
I want to be able to get to an article by a short-link, like this articles/ID, but I also want to get there with the full link articles/ID/category/slug. But I can not get the following to work:
// Route file: Route::pattern('id', '[0-9]+'); Route::pattern('cat', '^(?!create).*'); Route::pattern('slug', '^(?!edit).*'); Route::get('articles/{id}/{cat?}/{slug?}', ['as' => 'articles.show', 'uses' => 'ArticlesController@show']); // Controller file: public function show($id, $cat = null, $slug = null) { dd('1: ' . $cat . ' | 2:' . $slug); }以下链接articles/28/ullam/vel-repellendus-aut-est-est-esse-fugiat给出结果:
string(53) "1: ullam/vel-repellendus-aut-est-est-esse-fugiat | 2:"我不明白为什么不进行拆分,如果我在路由定义中删除了?,它会起作用.
I don't understand why it's not split, if I remove the ? in my route definition it works.
我已经尝试过此解决方案 stackoverflow/a/21865488/3903565 可以,但是并非直接针对控制器.为什么?
I have tried this solution stackoverflow/a/21865488/3903565 and that works, but not when directed at a controller. Why?
更新;我最终重新排列了路线文件:
Route::pattern('id', '[0-9]+'); // Articles Route::get('articles/create', ['as' => 'articles.create', 'uses' => 'ArticlesController@create']); Route::get('articles/edit/{id}', ['as' => 'articles.edit', 'uses' => 'ArticlesController@edit']); Route::get('articles/{id}/{category?}/{slug?}', ['as' => 'articles.show', 'uses' => 'ArticlesController@show']); Route::get('articles/{category?}', ['as' => 'articles.index', 'uses' => 'ArticlesController@index']); Route::resource('articles', 'ArticlesController', ['only' => ['store', 'update', 'destroy']]);推荐答案
问题仅在于超前模式.
您需要$和不包含/的类才能使其正常工作.
You need $ and class excluding / in order to make it work.
所以它们在这里:
Route::pattern('id', '[0-9]+'); Route::pattern('cat', '^(?!create$)[^/]*'); Route::pattern('slug', '^(?!edit$)[^/]*'); Route::get('articles/{id}/{cat?}/{slug?}', function($id, $cat, $slug) { dd($id.': ' . $cat . ' | 2:' . $slug); });更多推荐
Laravel中默认有多个路由到控制器
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