Question

我有桌面表演和时间。一个表演可以在某些时间播放多次。一对多的关系。

@Entity
@Table(name = "performance_type")
@Data
public class PerformanceType {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(nullable = false, unique = true)
    private Integer performanceTypeId;
    @Column(length=127)
    private String performOptions;
}

和

@Entity
@Table(name = "performance")
@Data
public class Performance {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(nullable = false, unique = true)
    private Integer performanceId;

    @OneToMany(mappedBy = "performance")
    private List<Hours> performanceHours = new ArrayList<>();
}

在数据库hibernate中，创建具有performanceId的表时间只有一个值。我如何插入列表值，同时播放id为1,4,7的表演。我需要存储hourId和perfomanceId的附加表hours_performance？

Answer 1

由于您希望实现的关系是（根据我的理解）多对多关系，因此您确实需要第三个表格将小时数映射到关系。 Here is a very good example.您需要设置第三个表，其中包含两个要连接的表的外键。

@Entity
@Table(name = "performance")
@Data
public class Performance {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(nullable = false, unique = true)
    private Integer performanceId;

    @ManyToMany
    @JoinTable(
        name = "hours_performance", 
        joinColumns = { @JoinColumn(name = "performance_id") }, 
        inverseJoinColumns = { @JoinColumn(name = "hour_id") }
    )
    private List<Hours> performanceHours = new ArrayList<>();
}

@Entity
@Table(name = "hours")
@Data
public class Hours {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(nullable = false, unique = true)
    private Integer hourId;
    ZonedDateTime time;

    @ManyToMany(mappedBy = "performanceHours")
    private List<Performance> performances;
}

将列表存储在表列中

1 个答案: