将数据从中间件传递到视图或以其他方式在每个页面中显示特定数据

passing data from middleware to view or alternative way to show specific data in every page

在我的网站上,我有一个相当复杂的类别,我必须在每个视图(在客户端)中显示它,所以我想我将用于创建类别的代码放在中间件中并将结果传递给视图

所以我创建了我的中间件,但我无法弄清楚如何将它的数据传递到我的视图而无需在控制器中执行某些操作 我已经在我的中间件中尝试了这些方法

<?php

namespace App\Http\Middleware;

use Closure;

class CtegoryMiddleware
{
    /**
     * Handle an incoming request.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Closure  $next
     * @return mixed
     */
    public function handle($request, Closure $next)
    {
        $request->merge(array("all_categories" => "abc"));
        $request['all_categories']= 'abc';
        return $next($request);
    }
}

路线:

Route::group(['middleware' => ['category' ]], function () {
     Route::get('/', 'HomeController@index');
});

但在我看来,当我回显 all_categories 时,我得到

Undefined variable: all_categories 

顺便说一句,我已经通过回显进行了检查,中间件在请求时被触发

我认为在您的用例中,使用全局可用的视图变量就足够了。

<?php

namespace App\Http\Middleware;

use Closure;

class CtegoryMiddleware
{
    /**
     * Handle an incoming request.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Closure  $next
     * @return mixed
     */
    public function handle($request, Closure $next)
    {
        $request->merge(array("all_categories" => "abc"));
        $request['all_categories']= 'abc';

        /**
         * This variable is available globally on all your views, and sub-views
         */
        view()->share('global_all_categories', 'abc');

        return $next($request);
    }
}

变量被加载一次(如果你做数据库查询,查询只会执行一次),然后变量存储在视图工厂中。