dsgw3315 2015-11-12 17:34
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Laravel身份验证:如何获取用户属性?

I would like to check account status in Middleware. The status is represented by tinyint. if it is 1 then it is activated, otherwise 0.

In App\Http\Middleware\Authenticate, I have:

public function handle($request, Closure $next)
{
    if ($request->input('status') == 0)
    {
        Session::flash('message', 'Your account hasn\'t been activated, please try again later.');
        return redirect('/');
    }

    if ($this->auth->guest()) {
        if ($request->ajax()) {
            return response('Unauthorized.', 401);
        } else {
            return redirect()->guest('auth/login');
        }
    }

    return $next($request);
}

$request->input('status') == 0 is probably wrong. How should I do this?

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1条回答 默认 最新

  • douxianglu4370 2015-11-12 17:45
    关注

    First, I would recommend checking if the user is a guest before you check the status. Otherwise, you'd be trying to grab the status of guests, which is not possible.

    Second, you have access to the user with the user method so you can just grab the status off the user like this:

    public function handle($request, Closure $next)
    {
        if ($this->auth->guest()) {
            if ($request->ajax()) {
                return response('Unauthorized.', 401);
            } else {
                return redirect()->guest('auth/login');
            }
        }
    
        if ($this->auth->user()->status === 0) {
            Session::flash('message', 'Your account hasn\'t been activated, please try again later.');
            return redirect('/');
        }
    
        return $next($request);
    }
    

    Extra tidbit: Laravel has a $casts property you can add to your models. So, if you add this to your User model:

    protected $casts = [
        'status' => 'boolean'
    ];
    

    It will automatically convert "cast" status into a boolean value so you can just do this:

    if (!$this->auth->user()->status) {
        // User's status is false
    }
    
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