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Php 拉维子域解析_Php_.htaccess_Laravel_Routing_Subdomain - Fatal编程技术网

Php 拉维子域解析

Php 拉维子域解析,php,.htaccess,laravel,routing,subdomain,Php,.htaccess,Laravel,Routing,Subdomain,我想从laravel中的url获取子域。现在就用这个 App::before(function($request) { if(Request::path() == '/' && !Request::ajax()){ $urlParts = explode('.', $_SERVER['HTTP_HOST']); if($urlParts[0] == 'fasfin' || $urlParts[1] == 'fasfin' &

我想从laravel中的url获取子域。现在就用这个

App::before(function($request)
{
        if(Request::path() == '/' && !Request::ajax()){
        $urlParts = explode('.', $_SERVER['HTTP_HOST']);
        if($urlParts[0] == 'fasfin' || $urlParts[1] == 'fasfin' && $urlParts[0] == 'www') {Route::get('/', 'HomeController@index');} //check if url is main site
        elseif($urlParts[0] == 'www') { $subdomain = $urlParts[1]; return App::make('SubdomainController')->getIndex($subdomain);} //fix for www.subdomain.mydomain.com
        else {$subdomain = $urlParts[0]; return App::make('SubdomainController')->getIndex($subdomain);}
        } //get subdomain
});
但我知道这是一种非常糟糕的代码。来自官方文档的{subdomain}.mydomain.com的技巧不起作用。我使用启用了apache中的所有子域

Server Alias *.mydomain.com
使现代化 把我的代码改成这个

if(!Request::ajax()){
        $isSubdomain = false;
        $urlParts = explode('.', $_SERVER['HTTP_HOST']);
        if($urlParts[0] == 'fasfin' || $urlParts[1] == 'fasfin' && $urlParts[0] == 'www') {Route::get('/', 'HomeController@index');}
        elseif($urlParts[0] == 'www') { $subdomain = $urlParts[1]; $isSubdomain = true;}
        else {$subdomain = $urlParts[0]; $isSubdomain = true;}
        if($isSubdomain)
        {
            $user = App::make('SubdomainController')->checkIndex($subdomain);
            if($user instanceof Exception) return View::make('subdomain.notExist');
            $shop = App::make('SubdomainController')->getIndex($subdomain);
            if($shop instanceof Exception) return 'Shop was already registered, but doesn't created yet';
        }
    }
现在新的问题是 如何每次都将变量传递给视图? 最后,我有一个变量$shop,每次调用它时都希望将它传递到view home.blade.php中

View::share('shop', $shop);
你试过了吗