我在我的数据库中将一个Java对象保存为json。对于这个实现,当我保存它时,它工作得很好,并且我可以在我的数据库中找到新行。但是,每当我尝试获取存储的同一对象时,我都会收到以下错误:
java.lang.NullPointerException: null
at java.lang.Class.isAssignableFrom(Native Method)
at com.vladmihalcea.hibernate.type.json.internal.JsonTypeDescriptor.fromString(JsonTypeDescriptor.java:104)
at com.vladmihalcea.hibernate.type.json.internal.JsonTypeDescriptor.wrap(JsonTypeDescriptor.java:165)
at com.vladmihalcea.hibernate.type.json.internal.AbstractJsonSqlTypeDescriptor$1.doExtract(AbstractJsonSqlTypeDescriptor.java:34)
at org.hibernate.type.descriptor.sql.BasicExtractor.extract(BasicExtractor.java:47)...
我的休眠版本是:
Hibernate Core {5.3.7.Final}
Hibernate Commons Annotations {5.0.4.Final}
我正在使用这个库将我的Java类序列化为json。
com.vladmihalcea:hibernate-types-52:2.17.3
奇怪的是,我可以将实体保存到数据库(可以反序列化),但当我尝试获取对象时,它无法重新构建对象。
@Getter
@Setter
@TypeDefs({
@TypeDef(name = "json", typeClass = JsonType.class)
})
public class NotificationMessage implements Serializable {
private Long id;
private String name = "";
private String type;
}
@Entity
@Table(name = "notification")
@Getter
@Setter
@EntityListeners(AuditingEntityListener.class)
public class Notification {
@Id
@Column(name = "notification_id")
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_notification_sequence")
@SequenceGenerator(name = "id_notification_sequence", sequenceName = "id_notification_sequence", allocationSize = 1)
private Long id;
@Column(name = "type", nullable = false)
@Enumerated(EnumType.STRING)
@NotNull
private NotificationType type;
@NotNull
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_uuid", referencedColumnName = "user_uuid")
private User user;
@Type(type = "json")
@Column(name = "message", columnDefinition = "json")
@NotNull
private NotificationMessage message;
@Column(name = "acknowledged")
private boolean acknowledged = false;
@Column(name = "created_date", nullable = false, updatable = false)
@CreatedDate
private LocalDateTime createdDate;
@Column(name = "modified_date")
@LastModifiedDate
private LocalDateTime modifiedDate;
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Notification that = (Notification) o;
return id.equals(that.id);
}
@Override
public int hashCode() {
return Objects.hash(id);
}
}
以下是我用来获取通知的查询:
@Repository
public interface NotificationRepository extends JpaRepository<Notification, Long>, JpaSpecificationExecutor<Notification> {
@Query("SELECT n FROM Notification n WHERE n.type = :type AND n.acknowledged = false")
List<Notification> findAllByTypeAndAcknowledgedIsFalse(@Param("type") NotificationType type);
}
1条答案
按热度按时间qyzbxkaa1#
您还必须使用@TypeDef注解将Hibernate类型声明为Entity类,如下所示: