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我如何在 Laravel 授权中将数字传递给中间件“can”

[英]How do i pass number to middleware “can” in Laravel Authorization

based on documentation https://laravel.com/docs/5.5/authorization#via-middleware基于文档https://laravel.com/docs/5.5/authorization#via-middleware

use App\Post;

Route::put('/post/{post}', function (Post $post) {
    // The current user may update the post...
})->middleware('can:update,post');

the 'post' in 'can:update,post' is variable passed from '{post}'. 'can:update,post' 中的 'post' 是从 '{post}' 传递的变量。

im trying to use middleware('can:update,1') .我正在尝试使用中间件('can:update,1') its not working.它不工作。 maybe its search for '$1' variable, how to pass number '1' to 'can' middleware?也许它搜索 '$1' 变量,如何将数字 '1' 传递给 'can' 中间件?

update this is the gate:更新这是门:

    Gate::define('update', function ($user, $role){
        $id = $user->id;
        $roles = $user::find($id)->roles()->get();
        $roleid = $roles[0]->pivot->role_id;
        //above code just for get role id, and $role_id is 1
       return $roleid === $role;
    });

Probably you don't have Policy created for Post.可能您没有为 Post 创建策略。

You can create by command:您可以通过命令创建:

php artisan make:policy PostPolicy --model=Post

and then implement method update in that policy.然后在该策略中实现方法更新。

I to had this same problem so I did some digging into the 'can' middleware (Which maps to Illuminate\\Auth\\Middleware\\Authorize )我遇到了同样的问题,所以我对“can”中间件进行了一些挖掘(它映射到Illuminate\\Auth\\Middleware\\Authorize

Once in the class we see the following code进入类后,我们会看到以下代码

 /**
     * Get the model to authorize.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  string  $model
     * @return \Illuminate\Database\Eloquent\Model|string
     */
    protected function getModel($request, $model)
    {
        if ($this->isClassName($model)) {
            return trim($model);
        } else {
            return $request->route($model, null) ?:
                ((preg_match("/^['\"](.*)['\"]$/", trim($model), $matches)) ? $matches[1] : null);
        }
    }

What this means is...这意味着……

  • If our string passed in is a class name then return that class name如果传入的字符串是类名,则返回该类名
  • If it is not a class name then...如果它不是类名,那么...
    • 1 Try to get it from the route, then return the route param 1 尝试从路由中获取,然后返回路由参数
    • 2 Try to get the model from the string via the regex "/^['\\"](.*)['\\"]$/" 2 尝试通过正则表达式"/^['\\"](.*)['\\"]$/"从字符串中获取模型

So now lets say we have the middleware call of $this->middleware(sprintf("can:create,%s,%s", User::class, Role::SUPPORT));所以现在假设我们有$this->middleware(sprintf("can:create,%s,%s", User::class, Role::SUPPORT));的中间件调用$this->middleware(sprintf("can:create,%s,%s", User::class, Role::SUPPORT));
This will not work because the Role::SUPPORT does not match the regex这将不起作用,因为Role::SUPPORT与正则表达式不匹配
To match it we simply need to place the Role::SUPPORT into quotes.为了匹配它,我们只需要将Role::SUPPORT放入引号中。
TAKE NOTE OF THE "'" around the second %s $this->middleware(sprintf("can:create,%s,'%s'", User::class, Role::SUPPORT));注意第二个 %s 周围的 "'" $this->middleware(sprintf("can:create,%s,'%s'", User::class, Role::SUPPORT));

To answer your question specifically, quote your '1' value要具体回答您的问题,请引用您的“1”值

middleware("can:update,'1'").

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