java - Hibernate 复合键连接

Question

我正在尝试使用 Spring Data 执行连接查询，但我的一个表有一个复合键，我不确定如何映射实体。

这是数据模型的类比:

table: device
pk=model_id
pk=serial_id
...

table: device_settings
pk=device_settings_id
fk=model_id
fk=serial_id
...

这是代码的类比，由于不存在“mappedby”属性，该代码无法编译。

@Entity
@Table(name = "device_settings")
public class DeviceSettings {

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

    // Pretty sure this is the problem
    @OneToMany(targetEntity = Device.class, mappedBy = "deviceKey", cascade = {CascadeType.MERGE}, fetch = FetchType.EAGER)
    @JoinColumns({
        @JoinColumn(name = "model_id", referencedColumnName = "model_id"),
        @JoinColumn(name = "serial_id", referencedColumnName = "serial_id")})
    private List<Device> devices;
}

@Entity
@Table(name = "device")
public class Device {
    @Id
        private DeviceKey deviceKey;
    }
    ...
}


@Embeddable
public class DeviceKey implements Serializable {
    private static final long serialVersionUID = -1943684511893963184L;

    @Column(name = "model_id")
    private Long modelId;

    @Column(name = "serial_id")
    private Short serialId;
}

Answer 1

Associations marked as mappedBy must not define database mappings like @JoinTable or @JoinColumn

要实现您的场景，您必须定义@ManyToOne:

@ManyToOne(cascade = {CascadeType.MERGE}, fetch = FetchType.EAGER)
    @JoinColumns({
        @JoinColumn(name = "model_id", referencedColumnName = "model_id"),
        @JoinColumn(name = "serial_id", referencedColumnName = "serial_id")})
     private Device device;

这最终将是model_id、serial_id、device_settings_id

或

在设备实体中定义@JoinColumn实体:

设备设置:

    @Entity
    @Table(name = "device_settings")
    public class DeviceSettings {

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


        @OneToMany( mappedBy = "deviceSettings", cascade = {CascadeType.MERGE}, fetch = FetchType.EAGER)
         private List<Device> devices;
}

设备实体:

@Entity
@Table(name = "device")
public class Device {

        @EmbeddedId
        private DeviceKey deviceKey;

        @ManyToOne
        @JoinColumn(name="device_settings_id")
        private DeviceSettings deviceSettings;
       //getters and setters
}

注意:您可以决定哪个是关系的所有者，并相应地放置您的映射，或者一个设备有多个设备设置，或者其他方式。