Laravel：（如何）是否可以使用参数链接到命名路由？
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Laravel: (How) is it possible to link to a named route with a parameter?
提问by Pim
I have defined the following route:
我定义了以下路线：
Route::get('/tours/{country_identifier}', ['as' => 'country', 'uses' => 'CountryController@index']);
Now I would like to link to this named route using something like
现在我想使用类似的东西链接到这个命名路线
route('country')
with a parameter for {country_identifier} filled in, e.g. '/tours/Canada'.
填写了 {country_identifier} 的参数，例如“/tours/Canada”。
How can I make this happen?
我怎样才能做到这一点？
采纳答案by Sh1d0w
You can pass the parameters as second argument array:
您可以将参数作为第二个参数数组传递：
route('country', ['country_identifier' => $someValue]);
You can take a look at the documentation, there are a lot of useful examples there :)
你可以看看文档，那里有很多有用的例子:)
回答by Szenis
To link from a view use (assuming you use blade)
从视图使用链接（假设您使用刀片）
<a href="{{ URL::route('country', array('country_identifier' => $var)) }}">$var</a>
回答by Emeka Mbah
Its simple:
这很简单：
For a single link
对于单个链接
<a href="{{ route('country', ['country_identifier' => $country_id]) }}">$country_name</a>
If you are generating some links you can do for example
如果你正在生成一些链接，你可以做例如
@foreach($countries as $country)
<a href="{{ route('country', ['country_identifier' => $country->id]) }}">$country->name</a>
@endforeach
NB: Where $countriesis model object passed to view example $countries = Country::all()
注意：$countries模型对象在哪里传递给视图示例$countries = Country::all()
回答by Ashwini G.
Try this
尝试这个
route('country', array('country_identifier' => 'Canada'))
It will take 'canada'as parameter to route.
它将以 'canada' 作为参数进行路由。
