Laravel 中的动态网址？
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Dynamic urls in laravel?
提问by Hailwood
I am looking at switching to laravelfor my next project.
我正在考虑为我的下一个项目切换到 Laravel。
My next project is probably going to be a small site with a few static pages, a blog and a projects manager and will be using controllers not routes.
我的下一个项目可能是一个包含几个静态页面、一个博客和一个项目经理的小站点，并且将使用控制器而不是路由。
What I am curious about is how I can manage dynamic routes in Laravel.
我很好奇的是如何在 Laravel 中管理动态路由。
Basically, I want to build in an admin section so I can easily create the static pages on the fly, and the static pages will have SEO focussed urls, e.g. http://domain.com/when-it-startedI do not want to have to create a new controller or route manually for each page.
基本上，我想建立一个管理部分，这样我就可以轻松地动态创建静态页面，并且静态页面将具有以 SEO 为重点的 url，例如http://domain.com/when-it-started我不想要必须为每个页面手动创建一个新的控制器或路由。
So I am wondering what the cleanest way is to handle this.
所以我想知道处理这个问题的最干净的方法是什么。
essentially all static pages are going to share the same view, just a few variables to change.
基本上所有静态页面都将共享相同的视图，只需更改几个变量。
The dynamic routing should work withthe controllers not instead of.
动态路由应该与控制器一起工作而不是。
E.g. if we have a controller aboutwith a function staffthen this should be loaded via http://domain.com/about/staff
例如，如果我们有一个about带有功能的控制器，staff那么它应该通过http://domain.com/about/staff加载
but we dont have the function players, so a call to http://domain.com/about/playersshould check the database to see if a dynamic route exists and matches. If it does display that, otherwise show the 404 page. Likewise for a non-existant controller. (e.g. there would not be a when-it-startedcontroller!)
但是我们没有这个功能players，所以调用http://domain.com/about/players应该检查数据库以查看是否存在动态路由并匹配。如果确实显示，否则显示 404 页面。对于不存在的控制器也是如此。（例如，不会有when-it-started控制器！）
The chosen answer doesn't seem to work in Laravel 4. Any help with that?
所选答案似乎在 Laravel 4 中不起作用。对此有帮助吗？
采纳答案by Deji S
For Laravel 4 do this
对于 Laravel 4，请执行此操作
Route::get('{slug}', function($slug) {
$page = Page::where('slug', '=', $slug)->first();
if ( is_null($page) )
// use either one of the two lines below. I prefer the second now
// return Event::first('404');
App::abort(404);
return View::make('pages.show', array('page' => $page));
});
// for controllers and views
Route::get('{page}', array('as' => 'pages.show', 'uses' => 'PageController@show'));
回答by TLGreg
You could use the route wildcards for the job, you can start with an (:any)and if you need multiple url segments add an optional (:all?), then identify the page from the slug.
您可以为作业使用路由通配符，您可以从 an 开始，(:any)如果您需要多个 url 段，请添加一个 optional (:all?)，然后从 slug 中识别页面。
For example:
例如：
Route::get('(:any)', function($slug) {
$page = Page::where_slug($slug)->first();
if ( is_null($page) )
return Event::first('404');
return View::make('page')->with($page);
});
回答by Adam Marshall
Very similar to Charles' answer, but in the controller:
与查尔斯的回答非常相似，但在控制器中：
public function showBySlug($slug) {
$post = Post::where('slug','=',$slug)->first();
// would use app/posts/show.blade.php
return View::make('posts.show')->with(array(
'post' => $post,
));
}
Then you can route it like this:
然后你可以像这样路由它：
Route::get('post/{slug}', 'PostsController@showBySlug')
->where('slug', '[\-_A-Za-z]+');`
...which has the added bonus of allowing you an easy way to link straight to the slug routes on an index page, for example:
...它有一个额外的好处，让您可以轻松地直接链接到索引页面上的 slug 路线，例如：
@foreach ($posts as $post)
<h2>{{ HTML::link(
action('PostsController@showBySlug', array($post->slug)),
$post->title
)}}</h2>
@endforeach
