如何创建一个包含原始和复合主键作为表复合键（Spring JPA）的 JoinTable？答案

【问题标题】：How do I create a JoinTable which includes both a primitive and a composite primary key as the tables composite key (Spring JPA)?如何创建一个包含原始和复合主键作为表复合键（Spring JPA）的 JoinTable？
【发布时间】：2020-04-28 23:24:12
【问题描述】：

我有一个用户、玩家、CoopGames 和 CoopGamesPlayers 表。从概念上讲，我本质上是在尝试重新创建 Order/OrderProducts 类型关系。在我的例子中，每个 CoopGamesPlayers 条目都有 CoopGame id 和 Player id 作为其键。但是，我被卡住了，因为在我的情况下，我的 Player 类本身有一个复合主键：生成的玩家 ID，以及用户 ID 外键。我有点迷路了。我已经按照 Order.OrderProducts 的教程进行了设置，但我不是 JPA 专家，也从未真正在 JoinTable 中使用过复合键。代码：

public class PlayerPK implements Serializable {

    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumn(name = "id")
    private Player player;

    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumn(name = "userid")
    private User user;
}

@Entity
@Table(name = "players")
public class Player implements Serializable {

    @EmbeddedId
    private PlayerPK pk;

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

@Entity
@Table(name = "coop_games")
public class CoopGame {

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

    @OneToMany(mappedBy = "pk", fetch = FetchType.EAGER)
    private List<CoopGamesPlayers> players;
}

public class CoopGamesPlayersPK implements Serializable {

    @JsonBackReference
    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumn(name = "coop_game_id")
    private CoopGame game;

    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumns({
        @JoinColumn(name="playerid", referencedColumnName="id"),
        @JoinColumn(name="userid", referencedColumnName="userid")
    })
    @MapsId("pk")
    private Player player;
}

@Entity
@Table(name = "coop_games_players")
public class CoopGamesPlayers {

    @EmbeddedId
    private CoopGamesPlayersPK pk;

    public CoopGamesPlayers(CoopGame game, Player player) {
        pk = new CoopGamesPlayersPK();
        pk.setGame(game);
        pk.setPlayer(player);
    }
}

@Entity
@Table(name = "coop_games")

public class CoopGame {

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

    @OneToMany(mappedBy = "pk", fetch = FetchType.EAGER)
    private List<CoopGamesPlayers> players;
}

【问题讨论】：

标签： java hibernate spring-boot jpa spring-data-jpa

【解决方案1】：

我认为您最好重新考虑所有关系的数据模型（应该解决哪个问题？..）。

至于您的 CoopGamesPlayers 表：如果您只想将 CoopGamesPlayers 用作可连接件，则不需要 CoopGamesPlayersPK。

CoopGame 和 CoopGamesPlayers 应该如下所示：

@Entity
@Table(name = "coop_games")
public class CoopGame {

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

    @OneToMany(mappedBy = "game", fetch = FetchType.EAGER)
    private List<CoopGamesPlayers> players;
}

@Entity
@Table(name = "coop_games_players")
public class CoopGamesPlayers {

    @EmbeddedId
    private PlayerPK pk;

    @JsonBackReference
    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumn(name = "coop_game_id")
    private CoopGame game;

    @ManyToOne(optional = false, fetch = FetchType.EAGER)
    @JoinColumns({
            @JoinColumn(name = "playerid", referencedColumnName = "id"),
            @JoinColumn(name = "userid", referencedColumnName = "userid")
    })
    @MapsId("id") 
    private Player player;
}

@MapsId("id") - id 因为你希望玩家 id 为 fk 或？

【讨论】：