jpa @Inheritance(strategy=InheritanceType.JOINED) 如何选择不包含子实体的超级实体答案

【问题标题】：jpa @Inheritance(strategy=InheritanceType.JOINED) how to select super entity not contain childjpa @Inheritance(strategy=InheritanceType.JOINED) 如何选择不包含子实体的超级实体
【发布时间】：2017-03-23 08:28:01
【问题描述】：

我有一个问题：假设我有两个类 User 和 person，person 扩展了 User。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 {......}

公司.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);
}

出现以下异常：

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

如何获取不包含 Person,Company 的用户信息
如何获取不包含用户、公司的人员信息

【问题讨论】：

标签： spring-data-jpa

【解决方案1】：

要仅选择用户，您可以使用 JPQL TYPE() 运算符，如下所示：

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

这里是完整的解释：https://stackoverflow.com/a/3766541/534877

【讨论】：

@Query(value="select u from User u where TYPE(u) Person and u.email=?1")
@Query(value="select u from User u where TYPE(u) Person and u.email=?1")
选择 user0_.id 作为 id2_1_，user0_.email 作为 email3_1_，user0_1_.first_name 作为 first_na1_0_，user0_1_.last_name 作为 last_nam2_0_，user0_.user_type 作为 user_typ1_1_ 从用户 user0_ 离开外连接人 user0_1_ 在 user0_.id =user0_1_.id where case when user0_1_.id is not null then Person when user0_.id is not null then 'User' end 'Person' and user0_.email=?
在spring jpa中type(u)不能生效
从查询sql中可以看出还有使用left jon，不只是选择父表，比如select * from user；