英文:
Hibernate IS-A relation mapping
问题
我有一个实体模型(a)，还有一些其他的实体(x)，比如手机、平板电脑、汽车等。
这些实体(x)都有一个主键，该主键引用了模型(a)的主键，因此实体(x)只能取模型(a)实体的值。
我在谈论一个IS-A关系。
我还需要从两端都能访问。
我需要在Hibernate中进行映射方面的帮助。
我现在所做的但不起作用的是：
- 模型实体
@Data
@Entity(name = "Model")
@Table(name = "model", schema = "mysch")
public class Model {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "model_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@Column(name = "description", length = 255, nullable = false, unique = true)
private String description;
@OneToOne(mappedBy = "model")
private Mobile mobile;
- 手机实体
@Data
@Entity(name = "Mobile")
@Table(name = "mobile", schema = "mysch")
public class Mobile {
@Id
@Column(name = "mobile_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@OneToOne
@JoinColumn(name = "mobile_id", referencedColumnName = "model_id", nullable = false, foreignKey = @ForeignKey(name = "FK_Mobile_Model"))
private Model model;
我想要的是在手机表中创建一个引用模型表主键的主键。
英文:
I have an entity model(a) and some other entities(x) like mobile, tablet, car etc.
The entities(x) have a primary key that references to the primary of the model(a), so the entities(x) can take only the values of the model(a) entity.
I'm talking about an IS-A relationship.
I also need to have access from both ends.
I need help with the mapping in hibernate.
What i do right now and does not work:
- model entity
@Data
@Entity(name = "Model")
@Table(name = "model", schema = "mysch")
public class Model {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "model_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@Column(name = "description", length = 255, nullable = false, unique = true)
private String description;
@OneToOne(mappedBy = "model")
private Mobile mobile;
- mobile entity
@Data
@Entity(name = "Mobile")
@Table(name = "mobile", schema = "mysch")
public class Mobile {
@Id
@Column(name = "mobile_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@OneToOne
@JoinColumn(name = "mobile_id", referencedColumnName = "model_id", nullable = false, foreignKey = @ForeignKey(name = "FK_Mobile_Model"))
private Model model;
What i want is to create a PK in mobile table that references to the PK of the model table.
答案1
得分: 0
我可能已经找到了解决方案。
目前似乎可以正常工作。稍后我将进行jpa查询以进行测试。尽管数据库中的模式似乎正是我想要的。
以下是我所做的事情：
- 模型实体
@Data
@Entity(name = "Model")
@Table(name = "model", schema = "sch")
public class Model {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "model_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@Column(name = "description", length = 255, nullable = false, unique = true)
private String description;
@OneToOne(mappedBy = "model")
private Mobile mobile;
}
- 移动设备
@Data
@Entity(name = "Mobile")
@Table(name = "mobile", schema = "sch")
public class Mobile {
@Id
@Column(name = "mobile_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@OneToOne
@MapsId
private Model model;
}
不过，我不喜欢移动设备生成的主键列名...它是这样的：
"model_model_id"。
尽管它似乎工作得很好...我遇到了一个问题。我得到了：
Exception in thread "JavaFX Application Thread" java.lang.StackOverflowError
at java.base/java.lang.String.equals(String.java:1009)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile.equals(Mobile.java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile.equals(Mobile.java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile.equals(Mobile.java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
英文:
I may have found a solution.
Seems to work for now. I will make jpa queries later to test it. Though the schema in the DB, seems to be exactly what i wanted.
Here is what i did:
- model entity
@Data
@Entity(name = "Model")
@Table(name = "model", schema = "sch")
public class Model {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "model_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@Column(name = "description", length = 255, nullable = false, unique = true)
private String description;
@OneToOne(mappedBy = "model")
private Mobile mobile;
}
- mobile
@Data
@Entity(name = "Mobile")
@Table(name = "mobile", schema = "sch")
public class Mobile {
@Id
@Column(name = "mobile_id", columnDefinition = "BIGINT UNSIGNED")
private long id;
@OneToOne
@MapsId
private Model model;
}
I don't like the generated PK column name of the mobile though... It goes like:
"model_model_id".
Though it seemed to be working well... I fell into a problem. I get:
Exception in thread "JavaFX Application Thread" java.lang.StackOverflowError
at java.base/java.lang.String.equals(String.java:1009)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile .equals(Mobile .java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile .equals(Mobile .java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
at com.dc.sch.entity.Mobile .equals(Mobile .java:6)
at com.dc.sch.entity.Model.equals(Model.java:6)
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