结果

Question

我将Laravel与Resources结合使用来创建REST Api。我想获得一个包含有关关系元素的信息的json，例如示例

{
"id": 30,
"calf": 23,
"chest": 27
"user_id": {
    "id": 30,
    "email": "test@example.com"
}}

但是当我在docs中创建代码时，出现错误“在int上调用成员函数first（）”。有人可以告诉我我做错了什么吗？我做了完全一样的事情，就像在文档中一样，我也尝试使用whenLoaded，但是出现了同样的错误。这是我的代码：

 class CircuitMeasurementResource extends JsonResource
{
    /**
     * Transform the resource into an array.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return array
     */
    public function toArray($request)
    {
//        return parent::toArray($request);
        return [
            'id' => $this->id,
            'calf' => $this->calf,
            'chest' => $this->chest,
            'user_id' => UserResource::collection($this->user_id)
        ];
    }
}

我的方法在控制器中显示：

public function show(CircuitMeasurement $circuitMeasurement): CircuitMeasurementResource {
        return new CircuitMeasurementResource($circuitMeasurement);
    }

我的模特：

    class CircuitMeasurement extends Model
{
    protected $fillable = [
        'user_id', 'calf', 'thigh', 'hips', 'waist', 'chest', 'neck', 'biceps'
    ];
    public function users(){
        return $this->belongsTo(User::class);
    }
}

class User extends Authenticatable
{
    use Notifiable, HasApiTokens;

    /**
     * The attributes that are mass assignable.
     *
     * @var array
     */

    protected $fillable = [
        'name', 'email', 'password',
    ];

    /**
     * The attributes that should be hidden for arrays.
     *
     * @var array
     */
    protected $hidden = [
        'password', 'remember_token',
    ];

    /**
     * The attributes that should be cast to native types.
     *
     * @var array
     */
    protected $casts = [
        'email_verified_at' => 'datetime',
    ];

    public function weightMeasurement(){
        return $this->hasMany(WeightMeasurement::class);
    }
    public function circuitMeasurement(){
        return $this->hasMany(CircuitMeasurement::class);
    }

}

Answer 1

您的代码中有些错误：

在您的CircuitMeasurement中，更改为：

// since it gets just one
    public function user(){
        return $this->belongsTo(User::class);
    }

您将ResourceName::collection()用于多个项目，将new ResourceName()用于单个项目。将您的资源更改为：

public function toArray($request)
    {
//        return parent::toArray($request);
        return [
            'id' => $this->id,
            'calf' => $this->calf,
            'chest' => $this->chest,
            'user_id' => new UserResource($this->user)
           //I think you should name user instead of user_id
        ];
    }

Answer 2

您根本不需要使用UserResource，只需通过该ID查找用户

public function toArray($request)
{
    return [
        'id' => $this->id,
        'calf' => $this->calf,
        'chest' => $this->chest,
        'user_id' => User::find($this->user_id),
    ];
}

结果

{
    "data": {
        "id": 30,
        "calf": 23,
        "chest": 27,
        "user_id": {
            "id": 30,
            "name": "Charity Douglas",
            "email": "demmerich@example.org",
            "email_verified_at": "2019-11-06 11:25:07",
            "created_at": "2019-11-06 11:25:07",
            "updated_at": "2019-11-06 11:25:07"
        }
    }
}

希望这会有所帮助

无法在json Laravel中返回关系元素

2 个答案:

结果