Laravel 5.4:在__construct()中登录用户ID [英] Laravel 5.4 : Get logged in user id inside __construct()

问题描述

我正在尝试访问构造函数内的Auth::user()->id;，但它总是返回错误试图获取非对象的属性.我在laravel文档中研究到，在构造函数内部无法访问Session，并且还在SO上进行搜索.我需要在构造函数中登录用户ID，因为我必须从数据库中获取数据并使其可用于其所有方法.我当前的代码是:

I am trying to access Auth::user()->id; inside constructor but it always return the error Trying to get property of non-object. I study in the laravel documentation that Session is not accessible inside constructor and also search on SO for this. I need logged in user id inside constructor because I have to fetch data from database and make it available for all its method. My current code is :

public function __construct(){
    $this->middleware('auth');
    $induction_status = TrainingStatusRecord::where('user_id',Auth::user()->id)->where('training','=','induction')->get();
    View::share('ind_status',$induction_status);
}  

是否有任何方法(简便方法)在构造函数中获取用户ID.

Is there any way (easy way) to get logged in user id inside constructor.

我将不胜感激.

谢谢

推荐答案

在视图 AppServiceProvider 中共享变量是一种很好的方法

To share a variable in view AppServiceProvider is a good approach

转到 App \ Providers \ AppServiceProvider.php

在页面顶部包含外观

use Illuminate\Support\Facades\Auth;

use App\TrainingStatusRecord;

并在启动方法中粘贴以下代码

and paste below code in boot method

view()->composer('*', function($view){
        if(Auth::user()){
            $induction_status = TrainingStatusRecord::where('user_id',Auth::user()->id)->where('training','=','induction')->get();
            View::share('induction_status',$induction_status);
        }
    });

现在，您将可以在应用程序中获取变量$induction_status.

Now you will be able to get your variable $induction_status in your app.

参考 https://laravel.com /docs/5.4/providers#the-boot-method

希望它会对您有所帮助.

Hope it will help you.

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