如何通过条件使用“with”laravel 关系函数来获取数据？

Question

this is my table => 表格图片

//This is my controller
public function listUserStripeActionableRequests(Request $request) {
        return $model = UserRequest::whereNull('deleted_at')
            ->with('oldPackage')
            ->get();
}


userRequest.php

public function oldPackage() {
    return $this->hasOne(Package::class, 'id', 'old_package_id');
}

我需要根据表中的package_type获取数据 如果 package_type == 1 那么我需要从 Package::class 获取数据， 如果 package_type == 2 然后来自 AddOnsPackage::class 然后模型函数如下：- $this->hasOne(AddOnsPackage::class, 'id', 'old_package_id');

如何使用单个查询来相应地获取数据？

Answer 1

你可以试试

class UserRequest extends Model
{


    public function package()
    {
        return $this->hasOne(Package::class, 'id', 'old_package_id');
    }

    public function addon()
    {
        return $this->hasOne(AddonPackage::class, 'type', 'add_on_type');
    }

    public static function sorted()
    {
        return static::whereNull('deleted_at')
            ->get()
            ->map(function ($record) {
                if ($record->package_type === 1) {
                    $record->load('package');
                } elseif ($record->package_type === 2) {
                    $record->load('addon');
                }

                return $record;
            });
    }
}

然后在控制器中你可以做

public function listUserStripeActionableRequests(Request $request)
{
    return $model = UserRequest::sorted();
}