Laravel 5-雄辩的关系

Question

我有以下三个表格：

 ------------------------------
| users  | projects  | clients |
|--------|-----------|-------- |
| id     | id        | id      |
| name   | client_id | name    |
|        | user_id   |         |
|        | name      |         |
 ------------------------------

所以基本上这种关系就像：

用户有许多项目
项目属性到客户端

我的问题是：
如何使用eloquent获得用户的所有客户端？

似乎我不能用户hasManyPass方法，因为在客户端表中没有project_id。

因此，我希望达到的目标是：

foreach($users as $user)
     foreach($user->clients as $client) //HOW TO SET UP THIS RELATION?
         foreach($client->projects($user)->get() as $project)

Answer 1

    $users = User::with('projects')->get();
    
    // organize cleints' project and add for each user  
    foreach ($users as $user) {
        // order projects by client_id
        $projects = Collect($user->projects->toArray());
        $clients = $projects->groupBy('client_id');

        // keep projects data in 'projects' index 
        $temp = array();
        foreach($clients as $client => $projects){
            $temp[$client] = array('projects' => $projects);
        }

        // add clinets data for each user
        $user['clients'] = $temp;
    }

    // output
    foreach($users as $user)
        foreach($user->clients as $client)
            foreach($client['projects'] as $project)
                echo 'user-'.$project['user_id'].', clinet-'.$project['client_id'].', project-'.$project['name'].'<br>';

上面的代码为我生成了这个输出：