java - 在 Hibernate 表中保存 POJO 时发生 hibernate.MappingException

Question

UserDO.java

@Entity
@Table(name = "UserDO")
public class UserDO {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long userId;

    private boolean successfullyLinked;

    private UserInformation userInformation;
}

UserInformation.java

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonPropertyOrder({ "address", "country_code", "currency_code", "email_address", "name", "phone" })
public class UserInformation {

    @JsonProperty("address")
    @Valid
    private Address address;

    @JsonProperty("country_code")
    @NotNull
    private String countryCode;

    @JsonProperty("currency_code")
    @Size(min = 3, max = 3)
    private String currencyCode;

    @JsonProperty("email_address")
    @NotNull
    private String emailAddress;

    @JsonProperty("name")
    @Valid
    @NotNull
    private Name name;

    @JsonProperty("phone")
    @Valid
    private Phone phone;

}

我正在尝试将 UserInformation POJO 保存为 Hibernate 中 UserDO 的一部分。然而，在将其作为 Spring Boot 应用程序的一部分运行时，我收到错误。以下是堆栈跟踪。

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaAutoConfiguration.class]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory

Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory

Caused by: org.hibernate.MappingException: Could not determine type for: com.paypal.marketplaces.vaas.api.models.UserInformation, at table: Tracking, for columns: [org.hibernate.mapping.Column(userInformation)]

注意:UserInformation POJO 非常复杂，其中包含其他对象以及这些对象中的对象(等等)。任何不需要将 UserInformation POJO 显式映射到 UserDO 表的列的解决方案都更好。

任何帮助将不胜感激!

Answer 1

持久性提供者不知道该类，也不知道如何处理它。我建议将其设为可嵌入并可选择指定列名称:

import javax.persistence.Embeddalbe;
import javax.persistence.Column;

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonPropertyOrder({ "address", "country_code", "currency_code", "email_address", "name", "phone" })
@Embeddable
public class UserInformation {

    @JsonProperty("country_code")
    @NotNull
    @Column(name = "COUNTRY_CODE")
    private String countryCode;

您必须对每个嵌套类重复该过程。

最后用以下方式注释userInformation:

@Embedded
private UserInformation userInformation;