如何在Laravel 4中为URL友好创建自定义路由？

Question

I am using laravel 4 routing for create friendly url for my project. I want to get url with this format: http://www.mywebsite.com/post/post-slug-content-123 . post-slug-content is my slug, and 123 is my post id. my route:

Route::get('post/{name}-{id}','postController@detail'); 

It not work because laravel detect "post" is my {name} variable, and "slug" is my {id} variable, but i want "post-slug-content" is {name} and 123 is {id}. Any idea ???

Answer 1

using - wont work and which - will it consider breaking in post-slug-content-123

if your url is http://www.mywebsite.com/post/post_slug_content-123

or http://www.mywebsite.com/post/postSlugContent-123 try using

Route::get('post/{name}-{id}','postController@detail');  

because you have multiple - and it will consider it as a separate URL

Or you may try

1: Route::get('post/{nameId}','postController@detail'); where both your name and id is present, further you could break them in your postController .

2: Route::get('post/{name}/{id}','postController@detail'); this will work fine

or use stackoverflow kind id followed by name

3: Route::get('post/{id}/{name}','postController@detail');

you can also use http://www.mywebsite.com/post/post-slug-content_123 as said by @itachi

4: Route::get('post/{name}_{id}','postController@detail');

Further reference for passing parameters Here .