java - 为什么 GET 方法返回对象中的对象

Question

我有 2 个实体:User 和 UserDetails
一个用户只能拥有一个用户详细信息。

    /* class User */
@Entity
@Table(name = "users")
public class User {

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name = "login")
    private String login;

    @Column(name = "password")
    private String password;

    @OneToOne(mappedBy = "user", cascade = CascadeType.ALL,
            fetch = FetchType.LAZY, optional = false)
    private UserDetails userDetails;

    /* getters and setters */
}
    /* class UserDetails */

@Entity
@Table(name = "userdetails")
public class UserDetails {

    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name = "name")
    private String name;

    @Column(name = "surname")
    private String surname;

    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "user_id")
    private User user;

    /* getters and setters */
}

这是类UserDetailsController:

@RestController
@RequestMapping("/userDetails")
public class UserDetailsController {
    @GetMapping("/{id}")
    public UserDetails findById(@PathVariable Long id) {
        return userDetailsRepository.findById(id).orElseThrow(() -> new UserDetailsNotFoundException(id));
    }

    @GetMapping("/")
    public Iterable<UserDetails> findAll() {
        return userDetailsRepository.findAll();
    }

    @PostMapping("/")
    public UserDetails save(@RequestBody UserDetails userDetails) {
        return userDetailsRepository.save(userDetails);
    }

    @PostMapping(value = "/", params = {"name", "surname", "user_id"})
    public UserDetails saveWithParams(@RequestParam("name") String name, @RequestParam("surname") String surname,
                                      @RequestParam("user_id") Long userId) {
        UserDetails userDetails = new UserDetails();
        userDetails.setName(name);
        userDetails.setSurname(surname);
        User user = userRepository.findById(userId).orElseThrow(() -> new UserNotFoundException(userId));
        userDetails.setUser(user);
        return userDetailsRepository.save(userDetails);
    }
}

这就是我尝试在 .http 文件中使用我的 Controller 的方式:

GET http://localhost:8080/userDetails/

###

GET http://localhost:8080/userDetails/4

###

POST http://localhost:8080/userDetails/
Content-Type: application/json

{
  "name": "my name",
  "surname": "my surname",
  "user": {
    "user_id": 12
  }
}

###

POST http://localhost:8080/userDetails/
Content-Type: application/x-www-form-urlencoded

user_id=12&name=testName&surname=testSurname

###

<强>1。为什么第一个 POST 方法不起作用:异常 user_id 不能为空？
2. 为什么userDetails中的GET all方法都返回这样的东西？:

Answer 1

Why first POST method not works: exception user_id can't be null

因为您正在接收数据作为请求参数而不是请求正文。所以所有数据都是空的。所以这就是为什么它给你错误。如果您想作为请求正文发送，则使用这些字段创建一个 DTO 类并使用 @RequestBody

进行映射

public UserDetails saveWithParams(@RequestBody UserDetailsDTO reqestBody) {
    ...
}

Why both GET all method in userDetails returns something like this?

您正在发送以这种方式定义关系的实体，因此它将递归地获取数据。您可以在 UserDetails 类的 user 上使用 @JsonIgnore 来解决它，这样就不会再次获取用户。

@JsonIgnore
@OneToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_id")
private User user;