参数1传递给Laravel 6

Question

我想使用邀请码在我的网站上注册，所以我在控制器的If语句中添加了用于检查激活码的语句

 protected function create (array $data)
    {
        if ($data['activation_code'] === 'mypassword123') {
            return User::create([
                'name' => $data['name'],
                'lastname' => $data['lastname'],
                'email' => $data['email'],
                'password' => Hash::make($data['password']),
            ]);
        }
            return redirect('/register')->with('message', 'De activatiecode is onjuist');
        }

}

并且当我提交注册时出现此错误

传递给Illuminate \ Auth \ SessionGuard :: login（）的参数1必须是Illuminate \ Contracts \ Auth \ Authenticatable的实例，给定的Illuminate \ Http \ RedirectResponse实例，在C：\ xampp \ htdocs \ Suppliers中调用\ vendor \ laravel \ framework \ src \ Illuminate \ Foundation \ Auth \ RegistersUsers.php，第35行

所以我不知道问题出在哪里，请帮忙

Answer 1

[就像评论所说的protected function create (array $data)不负责重定向到另一个页面，我认为逻辑也不应该存在。

您可以从register覆盖trait RegistersUsers方法，也可以创建中间件。

覆盖register方法，在RegisterController中在此处添加逻辑。

    /**
     * Handle a registration request for the application.
     *
     * @return \Illuminate\Http\Response
     */
    public function register(Request $request)
    {
        if ('mypassword123' !== $request->input('activation_code')) {
            return redirect('/register')->with('message', 'De activatiecode is onjuist');
        }

        $this->validator($request->all())->validate();

        event(new Registered($user = $this->create($request->all())));

        $this->guard()->login($user);

        return $this->registered($request, $user)
                        ?: redirect($this->redirectPath());
    }

中间件

<?php

namespace App\Http\Middleware;

use Closure;

class CheckCode
{
    /**
     * Handle an incoming request.
     *
     * @param \Illuminate\Http\Request $request
     *
     * @return mixed
     */
    public function handle($request, Closure $next)
    {
        if ('mypassword123' !== $request->input('activation_code')) {
            return redirect('/register')->with('message', 'De activatiecode is onjuist');
        }

        return $next($request);
    }
}

然后在需要时添加它。
use App\Http\Middleware\CheckAge;

Route::get('someroutes', function () {
    //some code
})->middleware(CheckCode::class);

希望这会有所帮助。