当列预计同时为 FK 和 PK 时，嵌入不起作用

Question

我正在尝试学习 JPA，目前遇到了可嵌入关键字的问题。

我的目标是病史，我想将firstName和lastName作为主键和外键，但是我遇到了错误消息。

这是我遇到的错误消息。

Caused by: org.hibernate.AnnotationException: Unable to find column reference in the @MapsId mapping: first_name
    at org.hibernate.cfg.CopyIdentifierComponentSecondPass.createSimpleProperty(CopyIdentifierComponentSecondPass.java:203) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.cfg.CopyIdentifierComponentSecondPass.doSecondPass(CopyIdentifierComponentSecondPass.java:112) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.boot.internal.InFlightMetadataCollectorImpl.processFkSecondPassesInOrder(InFlightMetadataCollectorImpl.java:1691) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.boot.internal.InFlightMetadataCollectorImpl.processSecondPasses(InFlightMetadataCollectorImpl.java:1623) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.boot.model.process.spi.MetadataBuildingProcess.complete(MetadataBuildingProcess.java:295) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.metadata(EntityManagerFactoryBuilderImpl.java:1460) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1494) ~[hibernate-core-5.6.4.Final.jar:5.6.4.Final]
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:58) ~[spring-orm-5.3.15.jar:5.3.15]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365) ~[spring-orm-5.3.15.jar:5.3.15]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:409) ~[spring-orm-5.3.15.jar:5.3.15]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:396) ~[spring-orm-5.3.15.jar:5.3.15]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:341) ~[spring-orm-5.3.15.jar:5.3.15]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1863) ~[spring-beans-5.3.15.jar:5.3.15]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1800) ~[spring-beans-5.3.15.jar:5.3.15]
    ... 16 common frames omitted

这是我试图使其工作的示例代码。

    @Embeddable
    public class PersonId implements Serializable {
        String firstName;
        String lastName;
    }
    
    @Entity
    public class Person implements Serializable {
        @EmbeddedId
        PersonId personId;
    }
    
    @Entity
    public class MedicalHistory implements Serializable {
        @EmbeddedId PersonId id;
    
        @MapsId
        @JoinColumns({
                @JoinColumn(name = "firstName", referencedColumnName = "firstName"),
                @JoinColumn(name = "lastName", referencedColumnName = "lastName")
        })
        @OneToOne
        Person patient;
    }

不确定我做错了什么？

• 注意：我正在使用这个学习材料“JSR 338：JavaTM Persistence API，版本2.2”

Answer 1

@chris 是绝对正确的，Hibernate 正在将列名从“firstName”转换为“first_name”，我猜“lastName”的列名也会相同。

您还可以更改列名称以匹配

  @Column(name = "first_name")
  private String firstName;

@Embeddable 上的命名，并调整 @JoinColumn 上的名称来实现这一目的。

如果您不自动生成数据库架构，则还必须重命名数据库中的列。