我在laravel中很难建立起非常嵌套的关系。

所需的行为如下，

我通过ID选择一个事件，我想查看哪些人已订阅该事件。 现在的问题是事件和人员之间有一些表格。

这是有效的查询！

SELECT persons.id, 
       persons.firstname, 
       persons.lastname, 
       event_scores.score 
FROM   events 
       JOIN cities 
         ON cities.id = events.city_id 
       JOIN companies 
         ON cities.id = companies.city_id 
       JOIN persons 
         ON companies.id = persons.company_id 
       JOIN event_scores 
         ON event_scores.person_id = persons.id 
WHERE  event_scores.event_id = 1 
GROUP  BY persons.id

这些是我的关系

事件模型

class Event extends Eloquent
{
    protected $table = 'events';

    public function city()
    {
        return $this->belongsTo('City');
    }
}

城市模型

class City extends Eloquent
{
    protected $table = 'cities';

    public function companies()
    {
        return $this->hasMany('Company');
    }

    public function event()
    {
        return $this->hasMany('Event');
    }
}

公司模式

class Company extends Eloquent {

    protected $table = 'companies';

    public function persons()
    {
        return $this->hasMany('Person');
    }

    public function city()
    {
        return $this->belongsTo('City');
    }
}

人物模型

class Person extends Eloquent
{
    protected $table = 'persons';

    public function company()
    {
        return $this->belongsTo('Company');
    }

    public function eventscore()
    {
        return $this->belongsToMany('Event', 'event_scores', 'person_id', 'event_id')
            ->withPivot('score')
            ->withTimestamps();
    }
}

我尝试过的

return Event::with('city')->with('company')->get();

和

return Event::with('city')
    ->whereHas('companies', function($query) use ($company_id){
        $query->where('company_id', $company_id);
    })->get();

还有很多其他可能性，我真的很坚持。在laravel中实现这种嵌套关系链接如此困难吗？

谢谢！