Api平台- Symfony -在持久化函数中返回json

use Symfony\Component\HttpFoundation\JsonResponse;
use App\Entity\User;

class UserController extends AbstractController
{
    public function getUserById($id)
    {
        // 从数据库中获取用户
        $user = $this->getDoctrine()->getRepository(User::class)->find($id);

        if (!$user) {
            return new JsonResponse(['error' => 'User not found'], 404);
        }

        // 将用户对象转换为数组
        $userData = [
            'id' => $user->getId(),
            'name' => $user->getName(),
            'email' => $user->getEmail(),
        ];

        // 返回JSON响应
        return new JsonResponse($userData, 200);
    }
}

$jsonData = json_encode($userData);
return new Response($jsonData, 200, ['Content-Type' => 'application/json']);

$user = $this->getDoctrine()->getRepository(User::class)->find($id);

if (!$user) {
    return new JsonResponse(['error' => 'User not found'], 404);
}

// 启用缓存
$cache = $this->get('cache.adapter.array');

$key = 'user_' . $id;
if ($cache->has($key)) {
    return new JsonResponse($cache->get($key), 200);
}

$user = $this->getDoctrine()->getRepository(User::class)->find($id);
if (!$user) {
    return new JsonResponse(['error' => 'User not found'], 404);
}

$userData = [
    'id' => $user->getId(),
    'name' => $user->getName(),
    'email' => $user->getEmail(),
];

$cache->set($key, $userData, 3600); // 缓存1小时

return new JsonResponse($userData, 200);