被难住了,完全难住了。。。
假设有两个实体,父实体和子实体,其中多个子实体对应一个父实体。父级的主键的类型为 java.util.UUID
,子级的主键是父级的uuid和序列号的组合。
问题的关键是当我试图用 childRepository.save(child)
,我得到以下异常:
原因:java.lang.illegalargumentexception:无法将类型[com.package.entities.parententity$$\u jvst149\u 0]的值转换为属性“parent”所需的类型[java.util.uuid]:propertyeditor[org.springframework.beans.propertyeditors.uuideditor]返回了类型[com.package.entitities.parententity$$\u jvst149\u 0]的不正确值
请看下面我的课。据我所知,我正在跟踪 JPA
规范正确,所以我想知道这是否是 Spring Data JPA
,可能特定于uuid类型id(类似的事情以前也发生过,请参阅datajpa-269)
我正在使用的注解 spring-boot-starter-data-jpa 1.4.1.RELEASE
父.java:
@Entity
@Table(name = "parent")
public class Parent implements Serializable {
@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid2")
private UUID id;
//...other fields, getters + setters...
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Parent that = (Parent) o;
return Objects.equals(id, that.id);
}
@Override
public int hashCode() {
return Objects.hash(id);
}
}
子.java
@Entity
@Table(name = "child")
@IdClass(ChildKey.class)
public class Child implements Serializable {
@Id
@ManyToOne
@JoinColumn(name = "parent_id", referencedColumnName = "id", insertable = false, updatable = false)
private Parent parent;
@Id
private Integer seqNum;
//...other fields, getters + setters...
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Child that = (Child) o;
return Objects.equals(parent, that.parent) &&
Objects.equals(seqNum, that.seqNum);
}
@Override
public int hashCode() {
return Objects.hash(parent, seqNum);
}
}
childkey.class类
public class ChildKey implements Serializable {
private UUID parent;
private Integer seqNum;
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
ChildKey that = (ChildKey) o;
return Objects.equals(parent, that.parent) &&
Objects.equals(seqNum, that.seqNum);
}
@Override
public int hashCode() {
return Objects.hash(parent, seqNum);
}
}
父存储库.java
@Repository
public interface ParentRepository extends JpaRepository<Parent, UUID> {
}
childrepository.java文件
@Repository
public interface ChildRepository extends CrudRepository<Child, ChildKey> {
}
最后,我执行的代码:
@Transactional
public void createChild(Parent parent) {
// needed to do this to get over "detached entity passed to persist"
parent = parentRepository.getOne(parent.getId());
child = new Child();
child.setParent(parent);
child.setSeqNum(1);
childRepository.save(child);
}
3条答案
按热度按时间xnifntxz1#
在我发布这个问题的几个月里,我一直没有找到合适的答案。不幸的是,我不得不通过不使用
@ManyToOne
而是通过uuid引用父级:我让jpa不知道外键,如果我违反了引用完整性,就让数据库抛出一个错误。
xmjla07d2#
您需要更改childkey类:
upd:我读了jpa规范,明白它是不正确的。但对我来说是有效的。
vngu2lb83#
在多对一关系中,子实体有自己的id,父实体的id是fk而不是pk的一部分。例子