bounty将在3天后过期。回答此问题可获得+100声望奖励。abr希望吸引更多人关注此问题。

假设我有一个名为Research的模型，每个研究都属于许多Location模型，每个Location模型也属于许多Contact模型，但是，每个Contact也与Research相关。

class Research extends Model {
   protected $table = 'researches';

   public function locations()
   {
      return BelongsToMany( Location::class, 'research_locations_list', 'research_id', 'location_id' );
   }
}

class Location extends Model {
   protected $table = 'locations';

   public function researches()
   {
      return BelongsToMany( Research::class, 'research_locations_list', 'research_id', 'location_id' );
   }

   public function contacts()
   {
      return BelongsToMany( Contact::class, 'location_contacts_list', 'location_id', 'contact_id' );
   }
}

class Contact extends Model {
   protected $table = 'contacts';

   public function locations()
   {
      return BelongsToMany( Location::class, 'location_contacts_list', 'location_id', 'contact_id' );
   }
}

researches表格：

+----+------------+
| id |  research  |
+----+------------+
|  1 | Research 1 |
|  2 | Research 2 |
+----+------------+

locations表格：

+----+---------------+
| id |   location    |
+----+---------------+
|  1 | United States |
|  2 | Great Britain |
|  3 | Germany       |
+----+---------------+

contacts表格：

+----+---------+
| id | contact |
+----+---------+
|  1 | Jack    |
|  2 | John    |
|  3 | Hanz    |
+----+---------+

research_locations_list表格：

+----+-------------+-------------+
| id | research_id | location_id |
+----+-------------+-------------+
|  1 |           1 |           1 |
|  2 |           1 |           2 |
|  3 |           2 |           2 |
|  4 |           2 |           3 |
+----+-------------+-------------+

因此，研究1在美国和英国进行，研究2在英国和德国进行
location_contacts_list表格：

+----+-------------+------------+-------------+
| id | location_id | contact_id | research_id |
+----+-------------+------------+-------------+
|  1 |           1 |          1 |           1 |
|  2 |           1 |          2 |           1 |
|  3 |           2 |          1 |           2 |
|  4 |           3 |          3 |           2 |
+----+-------------+------------+-------------+

研究1应将Jack和John作为在美国的联系人，而在其他地方没有联系人;
研究2应该让John作为英国的联系人，让Hanz作为德国的联系人;
现在，使用延迟加载可以实现：

$researches = Research::all();

foreach( $researches as $research )
{
    foreach( $research->locations as $location )
    {
        $contacts = $location->contacts()->wherePivot( 'research_id', $research->id )->get();
        // Will return John and Jack in United States for Research 1 and John in Great Britain and Hanz in Germany for Research 2
    }
}

现在，问题是：我如何通过快速加载实现这一点？

$researches = Research::with( 'locations.contacts' )->all();

foreach( $researches as $research )
{
    foreach( $research->locations as $location )
    {
        $contacts = $location->contacts;
        // Will return John and Jack in United States, John in Great Britain ( which is not supposed to happen ) for Research 1 and John in Great Britain and Hanz in Germany for Research 2
    }
}

也许我可以指示以某种方式为联系人尊重祖先id？像：

$research = Research::with( 'locations.contacts' )->where( 'researches.id = location_contacts_list.research_id' )->all();

更新

最接近解决这个问题的方法是修改Location模型，如下所示：

class Location extends Model {
   protected $table = 'locations';

   public function researches()
   {
      return BelongsToMany( Research::class, 'research_locations_list', 'research_id', 'location_id' );
   }

   public function contacts()
   {
      return BelongsToMany( Contact::class, 'location_contacts_list', 'location_id', 'contact_id' );
   }

   // Modify contacts attribute getter
   public function getContactsAttribute()
   {
      $contacts = $this->contacts();
      
      if( !empty( $this->pivot->research_id ) )
      {
         $contacts = $contacts->wherePivot( 'research_id', $this->pivot->research_id );
      }
      
      return $contacts->get();
   }
}

但看起来有点脏...