如何从laravel 5中的两个相关表中获取数据？

我有两个模型，员工和部门，我想从员工那里获取所有数据，包括it部门，如下所示：

SELECT e.Firstname, e.Surname, e.Age, d.Name as Department FROM Employees as e INNER JOIN Departments as d ON e.Department = d.ID;

$data = Employees::all();

其结果是：

------------------------------------------
| Firstname | Surname | Age | Department |
------------------------------------------
| John      | Doe     | 25  | Finance    |

我怎样才能在拉威尔5中得到同样的结果

我知道我可以得到这样一个模型的所有数据：

SELECT e.Firstname, e.Surname, e.Age, d.Name as Department FROM Employees as e INNER JOIN Departments as d ON e.Department = d.ID;

$data = Employees::all();

但是我需要Department列作为父表的值

使用查询生成器，它如下所示

 DB::table('Employees')
            ->join('Departments', 'Employees.Department', '=', 'Departments.ID')
            ->select('Employees.Firstname', 'Employees.Surname', 'Employees.Age', 'Departments.Name')
            ->get();

用雄辩的语言看起来是这样的

 DB::table('Employees')
            ->join('Departments', 'Employees.Department', '=', 'Departments.ID')
            ->select('Employees.Firstname', 'Employees.Surname', 'Employees.Age', 'Departments.Name')
            ->get();

模型

控制器

$data->employees = Employees::with('department')->get();

---

使用查询生成器，我可以通过

@include

传递数据，并按如下方式访问数据：

SELECT e.Firstname, e.Surname, e.Age, d.Name as Department FROM Employees as e INNER JOIN Departments as d ON e.Department = d.ID;

$data = Employees::all();

app.blade.php

@include('Club.index.employees',['employees'=>$data->employees])

@foreach($employees as $employee)
    @include('Club.index.member',['employee'=>$employee])
@endforeach

<h3>{{ $employee->Firstname }}</h3>
<p class="member-surname">{{ $employee->Surname }}</p>
...

employees.blade.php

@include('Club.index.employees',['employees'=>$data->employees])

@foreach($employees as $employee)
    @include('Club.index.member',['employee'=>$employee])
@endforeach

<h3>{{ $employee->Firstname }}</h3>
<p class="member-surname">{{ $employee->Surname }}</p>
...

member.blade.php

@include('Club.index.employees',['employees'=>$data->employees])

@foreach($employees as $employee)
    @include('Club.index.member',['employee'=>$employee])
@endforeach

<h3>{{ $employee->Firstname }}</h3>
<p class="member-surname">{{ $employee->Surname }}</p>
...

{{$employee->Firstname}
{{$employee->姓氏}
...

它是有效的，但是当我使用Elounce时，它不会显示来自父级的字段，比如

Departments.Name

在这种情况下，我不知道在视图中调用它时是否遗漏了一些内容。

---

我还想知道如何为表列使用别名。

在模型中，员工创建职能部门

public function department()
  {
      return $this->hasMany('App\Department');
  }

然后在你的控制器中，你可以这样做

$results = Employee::with('department')->get();

---

这是获取员工所有部门的方式，但请确保这两个表都在外键基础上相关，以及如何设置别名？这不起作用，查询生成错误，它生成此查询：

select*from departments where departments.employees\u id in（1，2）

。我想要的是来自员工而不是部门的数据