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java - JPA 从父实体填充复合键 id 部分

问题描述

假设我有这两个实体，个人和保险。一个人可以拥有多份保险，保险唯一性由（保险类型、保单编号和人员 ID）的组合键组合维护。下面的代码代表场景......

父类

@Entity
@Setter
@Getter
@RequiredArgsConstructor
public class Person implements Serializable {

  @Id
  @GeneratedValue(strategy = "GenerationType.IDENTITY")
  @Column(name "person_id")
  private Long personId;

  @Column(name = "fst_nm")
  private String fstName;

  @Column(name = "lst_nm")
  private String lstNm;
  
  // ..Other columns & relationships
  
  @OneToMany(mappedBy = "person")
  private List<Insurance> insurances;

  public void addInsurance(Insurance toAdd) {
    getInsurances().add(toAdd);
    toAdd.setPerson(this);
  }
}

儿童班

@Entity
@Setter
@Getter
@RequiredArgsConstructor
public class Insurance implements Serializable {

  @EmbeddedId
  private insurancePK id;

  //other data

  @ManyToOne
  @MapsId("personId")
  private Person person;
}

复合PK类

@Setter
@Getter
@Embeddable
public class InsurancePK implements Serializable {

  @Column(name = "person_id", insertable = false, updatable = false)
  private Long personId;

  @Column(name = "insurance_type")
  private String insuranceType;

  @Column(name = "pol_num")
  private String polNum;
}

现在，我的数据映射器看起来像这样......

  Person newPerson = new Person();
  newPerson.setInsurances(new ArrayList<>());

  // fill out Person Model data

  // incoming insurance data
  while (incomingData.hasNext()) {
    Insurance insuranceData = new Insurance();
    InsurancePK pk = new InsurancePK();

    // set other insurance data

    pk.setInsuranceType("Dental");
    pk.setPolNum("123Abc00");

    insuranceData.setId(pk);
    person.addInsurance(insuranceData);
  }

问题是我在复合键中的 person_id 总是得到一个空值，不知道为什么（@MapsId 不应该处理那个值）？我需要动态获取该值，大多数 JPA 复合键解决方案只是手动设置所有值，但这不是我的场景。

从 saveAndflush() 返回对象

{
  person: {
    person_id: 55,
    fst_nm: blah,
    lst_nm: blah,
    insurances: [
      {
        insurance_pk: {
          person_id: null,
          insurance_type: "Dental",
          pol_num: "123Abc00"
        }
       //other insurance data
      }
    ]
  }
}

关于我缺少什么的任何建议？先感谢您！

标签： javaspring-boothibernatejpaspring-data-jpa

解决方案

  1. 从. @Column(name = "person_id", insertable = false, updatable = false)_InsurancePK.personId

  2. 添加以下注释：

@JoinColumn(name = "name = "person_id"")

到Insurance.person.

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