Laravel 在關系方法中使用模型屬性

Question

我有一個模型的關系方法，它對模型本身的屬性（created_at）進行了比較。 我想將“約會”模型的 created_at 日期與“治療”的 created_at 日期進行比較。 如果“created_at”日期晚於“約會”日期，我只想要“治療”

約會控制器.php

public function index(): JsonResource
{
    $users = User::with([
        'appointment' => function ($query) {
            $query->with(['appointment_events']);
        },
    ])
        ->active()
        ->get();
    
    return UserResource::collection($users);
}

約會.php

public function appointment_events(): HasMany
{
    return $this->hasMany(AppointmentEvent::class)
        ->with([
            'customer' => function ($query) {
                $query->with([
                    'section' => function ($subquery) {
                        $subquery
                            ->with([
                                'advice',
                                'treatment' => function ($subquery) {
                                  $subquery->where('created_at', '>', blank); //How can I get the model's created_at date.
                                },
                            ]);
                    },
                ]);
            },
        ]);
}

有沒有辦法做到這一點？

Answer 1

我認為您應該將查詢移至Appointment模型的范圍並保持關系定義簡單。

注意：下面假設表名和列名符合標准 Laravel 約定。 如果您有不同的約定，請相應地修改表名和列名。

class Appointment extends Model
{
    public function scopeWithFutureTreatments($query)
    {
        return $query->with([
            'appointment_events' => function($query) {                
                $query->with([
                    'customers' => function($query) {
                        $query->with([
                            'section' => function($query) {
                                $query->with([
                                    'advice',
                                    'treatment' => function($query){
                                        $query->join('sections', 'section.id', '=', 'treatments.section_id')
                                            ->join('customers', 'customers.id', '=', 'sections.customer_id')
                                            ->join('appointment_events', 'appointment_events.id', '=', 'customer.appointment_events_id')
                                            ->join('appointments', 'appointments.id', '=', 'appointment_events.appointment_id')
                                            ->whereColumn('treatments.created_at', '>', 'appointments.created_at')
                                            ->select('treatments.*')
                                    }
                                ]);
                            }
                        ]);
                    }
               ]);
            }
        ]);
    }

    public function appointment_events(): HasMany
    {
        return $this->hasMany(AppointmentEvent::class);
    }
}

然后在AppointmentController中，以下內容應該可以為您提供所需的結果

public function index(): JsonResource
{
    $users = User::with([
        'appointment' => function ($query) {
            $query->withFutureTreatments();
        },
    ])
    ->active()
    ->get();
    
    return UserResource::collection($users);
}

Answer 2

如果我理解你的關系是

用戶 ->HasManyAppointments ->HasManyAppointmentsEvents。

您可以直接訪問來自用戶的約會事件數據

namespace App\Models;
 
use Illuminate\Database\Eloquent\Model;
 
class User extends Model
{
    /**
     * Get all of the appointment_events for the user.
     */
    public function appointment_events()
    {
        return $this->hasManyThrough(Appointment::class, AppointmentEvent::class);
    }
}

來自 Laravel Doc：有很多通過

所以可以查詢

$collection = User::with(['appointments','appointment_events'])->where()->get();

它為您提供特定用戶的所有約會和相關的約會事件。 您可以點擊該集合或過濾器或您想要的任何業務邏輯。