Laravel 5:如何通过构造函数添加Auth :: user()-> id? [英] Laravel 5: How to add Auth::user()->id through the constructor ?

问题描述

我可以这样获得经过身份验证的用户的ID:

I can get the ID of the authenticated user like this:

Auth::user()->id = $id;

它很好用，但是我有很多需要它的方法，而且我想要一种更干净的方法将它添加到整个类中，因此我可以在每个方法中引用$ id.我本来打算将其放入构造函数中，但是由于Auth :: user是静态的，因此我弄得一团糟，不知道该怎么做.

Great it works, ... but I have a load of methods which need it and I want a cleaner way of adding it to the class as a whole,so I can just reference the $id in each method. I was thinking of putting it into the constructor, but as Auth::user is a static, I am making a mess of things and don't know how to do it.

非常感谢您的帮助！

推荐答案

您可以在整个应用程序中使用Auth :: user().不管你在哪里.但是，针对您的问题，您可以使用controllers文件夹中的'Controller'类.在此处添加构造函数，并引用用户ID.

You can use Auth::user() in the whole application. It doesn't matter where you are. But, in response to your question, you can use the 'Controller' class present in your controllers folder. Add a constructor there and make the reference to the user ID.

<?php namespace App\Http\Controllers;

use Illuminate\Foundation\Bus\DispatchesCommands;
use Illuminate\Routing\Controller as BaseController;
use Illuminate\Foundation\Validation\ValidatesRequests;

/* Don't forget to add the reference to Auth */
use Auth;

abstract class Controller extends BaseController {

    use DispatchesCommands, ValidatesRequests;

    function __construct() {
        $this->userID = Auth::user()?Auth::user()->id:null;
    }
}

然后，在任何控制器的任何方法中，都可以使用$ this-> userID变量.

Then, in any method of any controller you can use the $this->userID variable.

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