检查页面中的用户角色

Question

I have parent User class whose instance is returned when the user is authenticated. 我有父User类，当用户通过身份验证时返回其实例。 That User instance has role_id which indicates the user's role. 该User实例具有role_id ，表示用户的角色。 There are three roles in the system; 系统中有三个角色; standarduser, trainer and gym (each has different table in database) and each has different functions and privileges. 标准用户，培训师和健身房（每个都有不同的数据库表），每个都有不同的功能和特权。

There are some possibilities I can do; 我可以做一些事情;

1) I can write role checks in User 's model class; 1）我可以在User的模型类中编写角色检查;

public function getName()
{
    if($user->role == 0)
    {
        $user->standarduser()->name;
    }
    else if($user->role == 1)
    {
        $user->gym()->name;
    }
}

public function getGymUsers()
{
    if($user->role == 2)
    {
        $user->gym()->users;
    }
    else
    {
        return null;
    }
}

2) I can write functions to each role class but in view (or controller), I have to do that control; 2）我可以为每个角色类编写函数，但在视图（或控制器）中，我必须执行该控制;

if($user->isStandarduser())
{
    echo $user->standarduser()->name;
}
. . .

if($user->isGym())
{
    echo $user->gym()->users;
}
else
{
    echo "";
}

3) Making specific views for each model so there will be no need to check roles. 3）为每个模型制作特定视图，因此不需要检查角色。 For example, if I'm in gym's page, I don't check if the object is a gym object because I know it's the gym's page. 例如，如果我在健身房的页面，我不会检查对象是否是健身房对象，因为我知道这是健身房的页面。

I can't decide which way to go. 我无法决定走哪条路。 If there are some better designs I can also try them. 如果有更好的设计，我也可以试试。

Answer 1

If the gym and the user's view are extremely similar and you are using the blade engine provided by laravel then you can use blade's template system 如果健身房和用户的视图非常相似，并且您正在使用laravel提供的刀片引擎，那么您可以使用刀片的模板系统

http://laravel.com/docs/5.1/blade http://laravel.com/docs/5.1/blade

/resources/layouts/default.blade.php /resources/layouts/default.blade.php

<body>
    <h1>@yield('name')</h1>
    <div>@yield('content')</div>
    <p>stuff that is the same for all users</p>
</body>

/resources/gym.blade.php /resources/gym.blade.php

@extends('layouts.default')

@section('name',$user->gym()->name)

@section('content')
<p>This is content unique to the gym</p>
@endsection

/resources/standarduser.blade.php /resources/standarduser.blade.php

@extends('layouts.default')

@section('name',$user->standarduser()->name)

@section('content')
<p>This is content unique to a standard user</p>
@endsection

Controller 调节器

public function show($id){
    $user = User::where($id)->first();
    if($user->isGym())
        return view('gym', ['user' => $user]);
    else if($user->isStandardUser())
        return view('standarduser', ['user' => $user]);
    else
        abort(500,'Not implemented yet');
}

Because each view is extending the default layout you will be able to have shared components while providing unique components for each type of user 因为每个视图都扩展了默认布局，所以您可以拥有共享组件，同时为每种类型的用户提供独特的组件

检查页面中的用户角色

问题描述

1 个解决方案

解决方案1
1 已采纳 2015-10-23 22:56:00

检查页面中的用户角色

问题描述

1 个解决方案

解决方案1 1 已采纳 2015-10-23 22:56:00

解决方案1
1 已采纳 2015-10-23 22:56:00