如何在JPA中映射复合主键，其中主键的一部分是外键

小编典典

如何在JPA中映射复合主键，其中主键的一部分是外键

hibernate

我试图弄清楚如何构建JPA实体bean，以使数据适用于我的设备。该数据库是旧的，一成不变的，所以我不能更改架构。设备模型具有复合主键，其中的一列是设备类型的FK。

我尝试了几种不同的方法。首先是设备具有DeviceModel和DeviceType，但是这给了我一个错误，那就是太多的东西在引用dev_type。因此，然后我尝试让DeviceModel引用DeviceType，但遇到了相同的错误。

如果有帮助/问题，我正在使用Spring Data 4.2.x和Hibernate 4.3.8.Final来备份所有内容。

我在网上找到的其他答案对我没有帮助，因为它们仅映射到基本数据类型。实际上，上面的答案是在下面的代码中实现的…但是我需要再上一级。

架构：

create table devices
(
  device_nbr serial(1),
  device_id nchar(20) not null unique,
  dev_type integer not null,
  model_nbr integer default 1,
  unit_addr nchar(32),
  primary key (device_nbr),
  foreign key (dev_type) references devtypes (dev_type),
  foreign key (dev_type, model_nbr) references devmodels (dev_type, model_nbr)
);

create table devmodels
(
  dev_type      integer  not null,
  model_nbr     integer  not null,
  model_desc    nchar(20),
  primary key (dev_type, model_nbr),
  foreign key (dev_type) references devtypes (dev_type)
);

create table devtypes
(
  dev_type integer not null,
  dev_desc nchar(16) not null unique,
  primary key (dev_type)
);

到目前为止，我的Bean（它们不会将DeviceType绑定到Device或DeviceModel，这是我需要的帮助）：

@Entity
@Table(name = "devices")
public class Device
{
    @Id
    @GeneratedValue
    @Column(name = "device_nbr")
    private Long                number;

    @Column(name = "device_id", length = 30)
    private String          id;

    @Column(name = "unit_addr", length = 30)
    private String          unitAddress;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumns({
        @JoinColumn(name = "dev_type"),
        @JoinColumn(name = "model_nbr")
    })
    private DeviceModel deviceModel;

...Getters and setters
}

public class DeviceModelPK implements Serializable
{
    private static final long   serialVersionUID    = -8173857210615808268L;
    protected Integer                   deviceTypeNumber;
    protected Integer                   modelNumber;

...Getters and setters
}

@Entity
@Table(name = "devmodels")
@IdClass(DeviceModelPK.class)
public class DeviceModel
{
    @Id
    @Column(name = "dev_type")
    private Integer         deviceTypeNumber;

    @Id
    @Column(name = "model_nbr")
    private Integer         modelNumber;

    @Column(name = "model_desc")
    private String          description;

...Getters and setters
}

@Entity
@Table(name = "devtypes")
public class DeviceType
{
    @Id
    @GeneratedValue
    @Column(name = "dev_type")
    private Integer number;

    @Column(name = "dev_desc", length = 30)
    private String  description;

...Getters and setters
}

阅读 664

2020-06-20

共1个答案

小编典典

嗯，您遇到的基本问题是根据列而不是实体进行思考，尽管由于问题有点棘手，所以这可能是不公平的陈述。基本问题是如何将实体作为复合键的一部分包含在内，我在这里找到了答案：如何在JPA中创建包含@ManyToOne属性作为@EmbeddedId的复合主键？。设备：

@Entity
@Table(name = "devices")
public class Device
{
    @Id
    @Column(name = "device_nbr")
    private Long number;

    @Column(name = "device_id", length = 20)
    private String deviceId;

    @ManyToOne(fetch=FetchType.EAGER, cascade=CascadeType.ALL)
    @JoinColumns({@JoinColumn(name="dev_type", referencedColumnName="dev_type"), @JoinColumn(name="model_nbr", referencedColumnName="model_nbr")})
    private DeviceModel deviceModel;

    // This creates a foreign key constraint, but otherwise doesn't function
    // deviceType must be accessed through deviceModel
    // note, it can be used for explicit selects, e.g., "select d.deviceType from Device d"
    @OneToOne(fetch=FetchType.EAGER, cascade=CascadeType.ALL)
    @JoinColumn(name="dev_type", referencedColumnName="dev_type", insertable=false, updatable=false)
    private DeviceType deviceType;

    @Column(name = "unit_addr", length = 32)
    private String unitAddress;

设备型号：

@Entity
@Table(name = "devmodels")
public class DeviceModel
{
    @EmbeddedId
    private DeviceModelId id;

    @ManyToOne(fetch=FetchType.EAGER, cascade=CascadeType.ALL)
    @JoinColumn(name="dev_type")
    @MapsId("deviceType")
    private DeviceType deviceType;

    @Column(name = "model_desc", length=20)
    private String  description;

DeviceModelId：

@Embeddable
public class DeviceModelId implements Serializable
{
    private static final long   serialVersionUID    = -8173857210615808268L;
    private Integer deviceType;
    @Column(name="model_nbr")
    private Integer modelNumber;

请注意，我使用@Embeddable和@EmbeddedId。它只是更新的，我已经阅读了JPA提供程序的评论，它比首选@IdClass。我认为这也使列命名更容易一些，但我不记得了。

设备类型：

@Entity
@Table(name = "devtypes")
public class DeviceType
{
    @Id
    @GeneratedValue
    @Column(name = "dev_type")
    private Integer deviceType;

    @Column(name = "dev_desc", length = 16)
    private String  description;

诀窍是@MapsId在DeviceModel。这样就可以在CompositeKey中使用实体。该字段上的@JoinColumn启用了该字段的命名。使用它的唯一技巧是手动创建DeviceTypeId：

DeviceModel model = new DeviceModel();
DeviceModelId modelId = new DeviceModelId();
modelId.setModelNumber(654321);
// have to have a DeviceType to create a DeviceModel
model.setDeviceType(type);
model.setId(modelId);

这将创建以下模式，该模式似乎与您的模式匹配。

create table devices (device_nbr bigint not null, device_id varchar(20), unit_addr varchar(32), dev_type integer, model_nbr integer, primary key (device_nbr))
create table devmodels (dev_type integer not null, model_nbr integer not null, model_desc varchar(20), primary key (dev_type, model_nbr))
create table devtypes (dev_type integer not null, dev_desc varchar(16), primary key (dev_type))
alter table devices add constraint FK8q0a886v04gg0qv261x1b2qrf foreign key (dev_type, model_nbr) references devmodels
alter table devices add constraint FKb72a7hq5phwjtbhaglobdkgji foreign key (dev_type) references devtypes
alter table devmodels add constraint FK4xlwyd2gwpbs4g4hdckyb11oj foreign key (dev_type) references devtypes

2020-06-20