如何将 ObjectNode 转换为 Java POJO 和 Java POJO 转换为 JsonNode？

Question

我有一个 controller 接受ObjectNode作为@RequestBody 。

该ObjectNode代表json以及一些用户数据

{
    "given_name":"ana",
    "family_name": "fabry",
    "email": "fabry@gmail.com",
    "password": "mypass",
    "gender": "FEMALE"
}

Controller.java

@PostMapping(produces = MediaType.APPLICATION_JSON_VALUE)
    public JsonNode createUser(@RequestBody ObjectNode user){
        return userService.addUser(user);
 }

我想让用户作为ObjectNode将其转换为 Java POJO 将其保存到数据库并再次将其作为JsonNode返回。

UserServiceImpl.java

    private final UserRepository userRepository;
    private final UserMapper userMapper;

    @Override
    public JsonNode addUser(@RequestBody ObjectNode user) {
        try {
            return userMapper.fromJson(user)
                    .map(r -> {
                        final User created = userRepository.save(r);
                        return created;
                    })
                    .map(userMapper::asJson)
                    .orElseThrow(() -> new ResourceNotFoundException("Unable to find user"));
        } catch (RuntimeException re) {
            throw re;
        }
    }

如何将 ObjetcNode 转换为 POJO？

我尝试过这样的事情：

public Optional<User> fromJson(ObjectNode jsonUser) {
  User user = objectMapper.treeToValue(jsonUser, User.class);
}

但是.treeToValue下划线是 read 说：

Add exception to method signature.

这甚至是正确的做法吗？

用户.java

@Builder
@AllArgsConstructor
public class User {

    @Indexed(unique = true)
    public final String email;
    public final String password;
    public final String firstName;
    public final String lastName;
    public final Gender gender;
    public final Instant registrationTime;

    public static User createUser(
            String email,
            String password,
            String firstName,
            String lastName,
            Gender gender,
            Instant registrationTime){
        return new User(email, password, firstName, lastName, gender, registrationTime);
    }
}

Answer 1

.treeToValue是正确的方法。 您得到的编译器错误是因为该方法抛出了一个已检查的异常，您需要重新抛出或处理该异常。 有关已检查异常的更多信息，请参见此处。