在创建时使用laravel返回模型

Question

我需要将保存为json的新模型发送到前面,但我无法看到列组织响应

这是我的模特

class Organization extends Model
{
    protected $table = "core.organizations";
    protected $fillable = ['description'];
    public $primaryKey = "organizationid";
    public $incrementing = false;
    public $timestamps = false;
}

这是我的功能

public function saveOrganization(Request $request)
    {
        try {
            $description = $request->input('description');
            $organization = new Organization();
            $organization->description = $description;
            $organization->save();
            if (!$organization) {
                throw new \Exception("No se guardo la organizacion");
            }           
            return response()->json([
            'organization' => $organization,
            ], 200);
        } catch (\Exception $ex) {
            return response()->json([
                'error' => 'Ha ocurrido un error al intentar guardar la organización',
            ], 200);
        }
    }

这是回应

{"organization":{"description":"Restobar"}}

我能怎么做？

谢谢!!

Answer 1

由于您已创建了一个新对象,而未从数据库中检索到该对象,因此它将了解的唯一属性是您设置的属性.

如果您想要获取表中的其余字段,则需要在保存后重新检索该对象.

// create the new record.
// this instance will only know about the fields you set.
$organization = Organization::create([
    'description' => $description,
]);

// re-retrieve the instance to get all of the fields in the table.
$organization = $organization->fresh();