无法使用 POSTMAN 删除（映射）我的数据库的对象答案

【问题标题】：Cannot Delete(Mapping) an object of my DB with POSTMAN无法使用 POSTMAN 删除（映射）我的数据库的对象
【发布时间】：2020-04-12 11:11:02
【问题描述】：

我有一个名为 Player 的实体和另一个名为 Game 的实体，每个玩家都有 X 场比赛，所以 Games 有一个 idPlayer 的外键。我试图用他的所有 GamesTries 从逻辑上删除一个玩家，并且 spring 说“java.sql.SQLIntegrityConstraintViolationException：无法删除或更新父行：外键约束失败（dices.game，CONSTRAINT @987654323 @FOREIGN KEY (player_id_player) REFERENCES player (id_player)) ”，而 POSTMAN 通过 JSON 说：“消息”：“无法执行语句；SQL [n/a]；约束 [null]；嵌套异常是 org.hibernate.exception.ConstraintViolationException: 无法执行语句”。

这是两个实体的代码：

@Entity
@Table(name="player")
public class Player {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @JoinColumn(name = "id_player")
    private Integer idPlayer;
    private String name;
    private LocalDateTime registerDate;
    private float successRate;

    public Player() {
    }

    public Player(Integer idPlayer, String name, LocalDateTime registerDate, float successRate) {
        this.idPlayer = idPlayer;
        this.name = name;        
        this.registerDate = registerDate;
        this.successRate = successRate;
    }

//getters n setters
}

@Entity
@Table(name = "game")
public class Game {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @JoinColumn(name = "id_game")
    private Integer idGame;
    @JoinColumn(name = "id_player")
    private Integer idPlayer;
    private Integer valueDice1;
    private Integer valueDice2;
    private boolean won;

    public Game() {
    }

    public Game(Integer idGame, Integer idPlayer, Integer valueDice1, Integer valueDice2, boolean won) {
        super();
        this.idGame = idGame;
        this.idPlayer = idPlayer;
        this.valueDice1 = valueDice1;
        this.valueDice2 = valueDice2;
        this.won = won;

    }
//getters n setters
}

然后是播放器控制器和播放器服务

//CONTROLLER
    @DeleteMapping("dices/players/id")
    public void deletePlayerById(@RequestBody Player player){
        playerService.deletePlayerById(player.getId());
    }

//SERVICE
    @Transactional
    public void deletePlayerById(Integer id) {
        gameRepository.deleteByIdPlayer(id);
        playerRepository.deleteById(id);
    }

非常感谢你们！！

【问题讨论】：

标签： java spring hibernate jpa

【解决方案1】：

如果你想在游戏中拥有一个玩家外键，只需添加一个@ManyToOne RelationShip：

@Entity
@Table(name="player")
public class Player {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer idPlayer;
    private String name;
    private LocalDateTime registerDate;
    private float successRate;

    public Player() {
    }

    public Player(Integer idPlayer, String name, LocalDateTime registerDate, float successRate) {
        this.idPlayer = idPlayer;
        this.name = name;        
        this.registerDate = registerDate;
        this.successRate = successRate;
    }

//getters n setters
}

对于游戏，你可以拥有这个：

@Entity
@Table(name = "game")
public class Game {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Integer idGame;

    @ManyToOne
    @JoinColumn(name="player_id", nullable=false)
    @OnDelete(action = OnDeleteAction.CASCADE)
    private Player player;

    private Integer valueDice1;
    private Integer valueDice2;
    private boolean won;

    public Game() {
    }

    public Game(Integer idGame, Integer idPlayer, Integer valueDice1, Integer valueDice2, boolean won) {
        super();
        this.idGame = idGame;
        this.idPlayer = idPlayer;
        this.valueDice1 = valueDice1;
        this.valueDice2 = valueDice2;
        this.won = won;

    }
//getters n setters
}

所以现在如果您删除播放器，OnDelete 注释将连同您的播放器一起删除孤儿游戏。我不知道你为什么要这样尝试 JoinColumn。值得注意的是，您也可以将其建模为双向并将孤立删除功能设置为 true。你可以阅读更多here

【讨论】：