Php 拉维Api资源
我正在尝试使用Laravel的资源为我的数据表创建一个API。我有三个关系模型。每次我点击api路由检查结果时,我的sub_专门化中都会得到一个空值。这是已经JSON格式的结果Php 拉维Api资源,php,laravel,api,laravel-5.8,laravel-resource,Php,Laravel,Api,Laravel 5.8,Laravel Resource,我正在尝试使用Laravel的资源为我的数据表创建一个API。我有三个关系模型。每次我点击api路由检查结果时,我的sub_专门化中都会得到一个空值。这是已经JSON格式的结果 { "data":[ { "first_name":"Rusty", "last_name":"Ferry", "specializations":{ "specialization_id":11, "
{
"data":[
{
"first_name":"Rusty",
"last_name":"Ferry",
"specializations":{
"specialization_id":11,
"specialization_name":"Endocrinology"
},
"sub_specializations":null
},
{
"first_name":"Nadia",
"last_name":"Ondricka",
"specializations":{
"specialization_id":22,
"specialization_name":"ENT"
},
"sub_specializations":null
},
{
"first_name":"Erich",
"last_name":"Torphy",
"specializations":{
"specialization_id":2,
"specialization_name":"Cardiologist"
},
"sub_specializations":null
}
]
}
这是我所有的资源。这是博士资源
public function toArray($request)
{
return [
'first_name' => $this->first_name,
'last_name' => $this->last_name,
'specializations' => new SpecializationsResource($this->specializations),
'sub_specializations' => new SubSpecializationsResource($this->sub_specializations),
];
}
专业资源
public function toArray($request)
{
return [
'specialization_id' => $this->specialization_id,
'specialization_name' => $this->specialization_name,
];
}
亚专业化
public function toArray($request)
{
return [
'sub_specialization_id' => $this->sub_specialization_id,
'sub_specialization_name' => $this->sub_specialization_name,
'doctors' => new DoctorsResource($this->doctors),
];
}
最后,这是控制器
protected $user;
public function __construct(Doctors $doctors){
$this->doctors = $doctors;
}
public function index()
{
$doctors = $this->doctors->with('specializations', 'subSpecializations')->get();
return DoctorsResource::collection($doctors);
}
我期望的结果与此类似
{
"data":[
{
"first_name":"Rusty",
"last_name":"Ferry",
"specializations":{
"specialization_id":11,
"specialization_name":"Endocrinology"
},
"sub_specializations": {
"sub_specialization_name":"value"
}
}
]
}
您必须确保有特定医生的子专科数据 如果有数据,则将该数据添加到医生资源中,否则该数据将为空 只需更改医生资源中的行,如:
'sub_specializations' => $this->sub_specializations !== null ? new SubSpecializationsResource($this->sub_specializations) : '',
你也可以对专业做同样的事情。你必须确保有特定医生的子专业数据 如果有数据,则将该数据添加到医生资源中,否则该数据将为空 只需更改医生资源中的行,如:
'sub_specializations' => $this->sub_specializations !== null ? new SubSpecializationsResource($this->sub_specializations) : '',
你也可以对专业化做同样的事情。谢谢兄弟,顺便说一句,很抱歉回复晚了。这帮我解决了我的问题Ethank's bro顺便说一句,很抱歉回复太晚。这帮我解决了我的问题