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使用正则表达式匹配Laravel路由中的子字符串

[英]Using a regex to match a substring in a Laravel route

My URL is: www.foo.com/some-bar-slug-here/page 我的网址是:www.foo.com/some-bar-slug-here/page

I can't get a route to catch if the string "bar" is found in the slug shown above: 如果在上面显示的slug中找到字符串“bar”,我无法获取捕获的路径:

Route::any('{myslug}/page/', array('as'=>'bar-page', 'uses'=>'Controllers\MyBar@index'))
     ->where('myslug','/bar/');

If I use the regex expression [0-9A-Za-z\\-]+ it works, but it doesn't work for /bar/ . 如果我使用正则表达式[0-9A-Za-z\\-]+它可以工作,但它不适用于/bar/ Any ideas? 有任何想法吗?

I got it working with ^([0-9A-Za-z\\-]+)?bar([0-9A-Za-z\\-]+)? 我得到它与^([0-9A-Za-z\\-]+)?bar([0-9A-Za-z\\-]+)?

So the updated route code looks like this: 所以更新的路由代码如下所示:

Route::any('{myslug}/page/', array('as'=>'bar-page', 'uses'=>'Controllers\MyBar@index'))
 ->where('myslug','^([0-9A-Za-z\-]+)?bar([0-9A-Za-z\-]+)?');

Bonus: To make it case insensitive, I do this: ^([0-9A-Za-z\\-]+)?(?i)bar([0-9A-Za-z\\-]+)? 额外:为了使它不区分大小写,我这样做: ^([0-9A-Za-z\\-]+)?(?i)bar([0-9A-Za-z\\-]+)?

Note: If you are using a copy of this route further down in the your routes file, but searching for a different sub-string then you will need to name your {myslug} to something different like {myslug2} , otherwise Laravel will not run all of the routes. 注:如果您使用这条路线在你的路由文件进一步下跌的副本,而是寻找一个不同的子串,那么你将需要命名{myslug}喜欢的东西不同{myslug2}否则Laravel将无法运行所有的路线。

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