Laravel API 资源

Question

我正在尝试使用 Laravel 的资源为我的数据表创建一个 API。我有三个有关系的模型。每次我点击我的 api 路由来检查结果时，我的 sub_specializations 中都会得到一个空值。这是已经格式化为 JSON 的结果。

    { 

"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。这是医生资源

    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,

    ];

}

子专业

    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);

}

Answer 1

您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 Doctor Resource，否则将为空白。只需要更改医生资源中的行，例如：'sub_specializations' => $this->sub_specializations !== null ? new SubSpecializationsResource($this->sub_specializations) : '',你也可以用专业做同样的事情。