Laravel 路由:如何将参数传递给控制器(不是通过 URL)?
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Laravel routing: how to pass parameter to controller (not via URL)?
提问by Andrew Vershinin
I have two routes: A/{param}/C
and B/{param}/C
. Also I have a controller with method:
我有两条路线:A/{param}/C
和B/{param}/C
。我还有一个带有方法的控制器:
public function index($param, $param1 = false)
{
//...
}
In case of A/{param}/C
I pass only one parameter - one specified in URL, and the function uses the default for second one. In case of the second route, I want to pass true
as second parameter. Since it isn't specified in URL, how to pass it to the function?
如果A/{param}/C
我只传递一个参数 - 在 URL 中指定的一个,并且该函数使用第二个的默认值。在第二条路线的情况下,我想true
作为第二个参数传递。由于没有在 URL 中指定,如何将其传递给函数?
回答by c-griffin
If i understand you correctly, without using $_POST, you won't be able to pass a parameter without be being somewhere in the URI (outside of setting a session variable, which i wouldn't advise)
如果我理解正确,如果不使用 $_POST,您将无法传递参数而不位于 URI 中的某处(设置会话变量之外,我不建议这样做)
Another option may be to pass it as a query string parameter. It will still be in the url, but won't necessarily be caught in the route pattern
另一种选择可能是将其作为查询字符串参数传递。它仍然会在 url 中,但不一定会被捕获在路由模式中
URLs:
网址:
A/{param}/C
B/{param}/C?param1=true
--
Controller:
--
控制器:
public function index($param)
{
$param1 = Request::query('param1'); // if not present false, if present, {value}
}
--
——
Alternatively, if you'd like a friendly URL, you can place a question mark after your second parameter, indicating that it may or may not exist. This will match both of the below URLs
或者,如果您想要一个友好的 URL,您可以在第二个参数后放置一个问号,表明它可能存在也可能不存在。这将匹配以下两个 URL
URL:
网址:
A/{param}/C
B/{param}/C/true
Route:
路线:
Route::get('B/{param}/C/{param1?}', 'YourController@index');
Controller:
控制器:
public function index($param, $param1 = false)
{
//
}