我正在尝试显示与该成员对应的Designation角色，
我想要实现的目标，
Chairman |
|
President - Heading
- member name
Vice President - Heading
- member name
我有一个下属的主席，根据指定ID划分，
为实现这一目标，我创建了3个表，
国家主席
id | first_name | from | to
1 | Chairman | 2019 | 2020
statemembers
id | first_name | chairman_id | designation_id
1 | Member 1 | 1 | P (inherted from designation_code of statedesignation)
1 | Member 2 | 1 | VP (inherted from designation_code of statedesignation)
statedesignation
id | designation_code | designation_name | designation_desc
1 | P | President | President Description.
1 | VP | Vice President | President Description.
数据库设计可以更好吗？ 。任何其他建议
我想显示成员名单，然后是他们的名字：
我试过订购桌子但是，我无法引入相应会员的名称，
我的控制人员带来了主席下的国家成员
$statemembers = StateChairman::find($id)->statemembers()->orderBy('designation_id')->get();
dd($statemembers);
我得到的结果 - 所有州成员的名单 -
But when I try this
$statedesignation = Statemembers::find($id)->statedesignations()->get();
dd($statedesignation);
对于这个我只能在Statedesignation获得Id 1的结果，即总统，副总统被忽略，
我只获得了Designation总裁，因为find（$ id）只带来1或相应的id
模型：
国家主席：
class Statechairman extends Model
{
public function statemembers(){
return $this->hasMany('App\Statemembers','chairman_id','id');
}
}
StateMembers：
class Statemembers extends Model
{
public function statechairman(){
return $this->belongsTo('App\Statechairman','chairman_id','id');
}
州指定：
public function statemembers(){
return $this->hasMany('App\Statemembers');
我对如何达到上述要求毫无头绪，
我还想从statemembers表中删除designation_id并添加designation_name并直接在页面上显示它。更新表格这样的事情，
id | first_name | chairman_id | designation_id
1 | Member 1 | 1 | President
2 | Member 2 | 1 | Vice President
但这不是一个好习惯。
请提供一些有关上述问题的见解。