Question

我有疑问：我们假设我有两个类用户和人，人扩展用户。在用户中我有@Inheritance(strategy=InheritanceType.JOINED)。有时我只想选择不包含子表信息的用户信息，

User.java（超类）

@Entity
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name = "user_type")
@Table(name = "users")
public class User {.........}

Person.java

@Entity
public class Person extends User {......}

Company.java

@Entity
public class Company extends User {.............}

UserRepository.java

@Transactional
public interface UserRepository extends UserBaseRepository<User> {

//@Query(value="select u from User u where u.email=?1")
@Query(value="select u.id,u.email from User u where u.email=?1")
User getUser(String email);

@Query("select u from User u where u.email=?1")
User getUser2(String email);

@Query(value="select * from Users where email=:email",nativeQuery=true)
User getUser3(String email);
}

发生以下异常：

Hibernate: select user0_.id as col_0_0_, user0_.email as col_1_0_ from users user0_ where user0_.email=?
org.springframework.core.convert.ConversionFailedException: Failed to convert from type java.lang.Object[] to type netgloo.models.User for value '{1, 635@qq.com}';
nested exception is org.springframework.core.convert.ConverterNotFoundException: No converter found capable of converting from type java.lang.Long to type netgloo.models.User

如何获取用户信息不包含Person，Company
如何获取人员信息不包含用户，公司

Answer 1

要仅选择您可以使用JPQL TYPE()运算符的用户，请执行以下操作：

SELECT u FROM User u WHERE TYPE(u) := User

以下是完整的解释：https://stackoverflow.com/a/3766541/534877

jpa @Inheritance（strategy = InheritanceType.JOINED）如何选择不包含子的超实体

1 个答案: