从 Spring MVC 以 JSON 形式发送时动态忽略 Java 对象中的字段

json java hibernate spring-mvc

2022-08-31 08:01:05

我有这样的模型类，用于休眠

@Entity
@Table(name = "user", catalog = "userdb")
@JsonIgnoreProperties(ignoreUnknown = true)
public class User implements java.io.Serializable {

    private Integer userId;
    private String userName;
    private String emailId;
    private String encryptedPwd;
    private String createdBy;
    private String updatedBy;

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "UserId", unique = true, nullable = false)
    public Integer getUserId() {
        return this.userId;
    }

    public void setUserId(Integer userId) {
        this.userId = userId;
    }

    @Column(name = "UserName", length = 100)
    public String getUserName() {
        return this.userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

    @Column(name = "EmailId", nullable = false, length = 45)
    public String getEmailId() {
        return this.emailId;
    }

    public void setEmailId(String emailId) {
        this.emailId = emailId;
    }

    @Column(name = "EncryptedPwd", length = 100)
    public String getEncryptedPwd() {
        return this.encryptedPwd;
    }

    public void setEncryptedPwd(String encryptedPwd) {
        this.encryptedPwd = encryptedPwd;
    }

    public void setCreatedBy(String createdBy) {
        this.createdBy = createdBy;
    }

    @Column(name = "UpdatedBy", length = 100)
    public String getUpdatedBy() {
        return this.updatedBy;
    }

    public void setUpdatedBy(String updatedBy) {
        this.updatedBy = updatedBy;
    }
}

在Spring MVC控制器中，使用DAO，我能够获得对象。并作为 JSON 对象返回。

@Controller
public class UserController {

    @Autowired
    private UserService userService;

    @RequestMapping(value = "/getUser/{userId}", method = RequestMethod.GET)
    @ResponseBody
    public User getUser(@PathVariable Integer userId) throws Exception {

        User user = userService.get(userId);
        user.setCreatedBy(null);
        user.setUpdatedBy(null);
        return user;
    }
}

视图部分是使用AngularJS完成的，因此它将获得如下所示的JSON

{
  "userId" :2,
  "userName" : "john",
  "emailId" : "john@gmail.com",
  "encryptedPwd" : "Co7Fwd1fXYk=",
  "createdBy" : null,
  "updatedBy" : null
}

如果我不想设置加密的密码，我会将该字段也设置为空。

但我不想这样，我不想将所有字段发送到客户端。如果我不希望发送密码，更新，创建字段，我的结果JSON应该像

{
  "userId" :2,
  "userName" : "john",
  "emailId" : "john@gmail.com"
}

我不想发送到来自其他数据库表的客户端的字段列表。因此，它将根据登录的用户而变化。我该怎么做？

我希望你得到了我的问题。

答案 1

将注释添加到 POJO。@JsonIgnoreProperties("fieldname")

或者，您可以在反序列化 JSON 时在要忽略的字段名称之前使用。例：@JsonIgnore

@JsonIgnore
@JsonProperty(value = "user_password")
public String getUserPassword() {
    return userPassword;
}

GitHub 示例

答案 2

我可以动态地做到这一点吗？

创建视图类：

public class View {
    static class Public { }
    static class ExtendedPublic extends Public { }
    static class Internal extends ExtendedPublic { }
}

为模型添加注释

@Document
public class User {

    @Id
    @JsonView(View.Public.class)
    private String id;

    @JsonView(View.Internal.class)
    private String email;

    @JsonView(View.Public.class)
    private String name;

    @JsonView(View.Public.class)
    private Instant createdAt = Instant.now();
    // getters/setters
}

在控制器中指定视图类

@RequestMapping("/user/{email}")
public class UserController {

    private final UserRepository userRepository;

    @Autowired
    UserController(UserRepository userRepository) {
        this.userRepository = userRepository;
    }

    @RequestMapping(method = RequestMethod.GET)
    @JsonView(View.Internal.class)
    public @ResponseBody Optional<User> get(@PathVariable String email) {
        return userRepository.findByEmail(email);
    }

}

数据示例：

{"id":"5aa2496df863482dc4da2067","name":"test","createdAt":"2018-03-10T09:35:31.050353800Z"}

UPD：请记住，使用实体作为响应不是最佳做法。最好为每个案例使用不同的 DTO，并使用modelmapper