【发布时间】:2023-04-10 12:55:01
【问题描述】:
我有一个玩家类和朋友类,玩家类中有朋友对象。但是朋友类也有与玩家类相同的字段。如何将好友列表添加到播放器。就像在 facebook 上,用户在他的朋友列表中有朋友。如何使 PlayerController 类中的注释方法起作用?
//@CrossOrigin(origins = "http://localhost:4200")
@RestController
@RequestMapping(value = "/players", produces = { MediaType.APPLICATION_JSON_VALUE })
public class PlayerController {
@Autowired
private PlayerReopsitory playerReopsitory;
@GetMapping(produces = "application/json")
public List<Player> getAllPlayers(){
return playerReopsitory.findAll();
}
@PostMapping
// @ResponseStatus(HttpStatus.OK)
public void create(@RequestBody Player player) {
playerReopsitory.save(player);
}
@GetMapping("/{id}")
public Player getOne(@PathVariable("id") Long id) {
return playerReopsitory.getOne(id);
}
@DeleteMapping("/{id}")
void deletePlayer(@PathVariable Long id) {
playerReopsitory.deleteById(id);
}
@PutMapping("/{id}")
Player updatePlayer(@RequestBody Player newPlayer, @PathVariable Long id) {
return playerReopsitory.findById(id).map(player -> {
player.setName(newPlayer.getName());
player.setEmail(newPlayer.getEmail());
player.setPhone(newPlayer.getPhone());
return playerReopsitory.save(player);
}).orElseGet(() -> {
newPlayer.setId(id);
return playerReopsitory.save(newPlayer);
});
}
// how to make this function work?
// @GetMapping("/{id}/friends")
// public List<Friend> getAllFriendFromList(@PathVariable Long id) {
// return playerReopsitory.findById(id).get();
// }
//
}
@Entity
@Table(name= "user", schema = "rabbit")
@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
public class Friend {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name="userName")
private String userName;
@Column(name="playername")
private String name;
@Column(name="email")
private String email;
@Column(name="phone")
private String phone;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPhone() {
return phone;
}
public void setPhone(String phone) {
this.phone = phone;
}
@Override
public String toString() {
return "Friend [id=" + id + ", userName=" + userName + ", name=" + name + ", email=" + email + ", phone="
+ phone + "]";
}
}
@Entity
@Table(name= "user", schema = "rabbit")
@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
public class Player {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name="userName")
private String userName;
@Column(name="playername")
private String name;
@Column(name="email")
private String email;
@Column(name="phone")
private String phone;
private Friend friend;
public Friend getFriend() {
return friend;
}
public void setFriend(Friend friend) {
this.friend = friend;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPhone() {
return phone;
}
public void setPhone(String phone) {
this.phone = phone;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
@Override
public String toString() {
return "Player [id=" + id + ", userName=" + userName + ", name=" + name + ", email=" + email + ", phone="
+ phone + ", friend=" + friend + "]";
}
public interface FriendReopsitory extends JpaRepository<Player, Long> {
}
public interface PlayerReopsitory extends JpaRepository<Player, Long> {
}
@Component
public class FriendService {
private static Map<Friend, Long> getFriends = new HashMap<>();{
}
}
【问题讨论】:
-
如果一个玩家只有一个好友字段,你只能从一个玩家id中获取一个好友。如果您希望多个朋友与您的 Player 相关联,则 Player 类必须有一个 List
以及关联的 getter、setter 和 Spring-JPA 映射注释。您在寻找代码改进建议吗?如果朋友也可以成为玩家,那么只有一个类会更有意义,如果不是,他们应该扩展一个父类,因为它们非常相似。 -
注意,如果你有通常的Spring Boot配置,可以直接说
@PathVariable("id") Player player。 -
朋友也是玩家
标签: java spring-boot spring-data-jpa crud