[英]Use JSON_EXTRACT to get value from JSON in Spring JPA Custom Query for custom DTO
我想从 JSON 列中获取值并在 spring JPA 中返回自定义 DTO。
表结构
userId (int)
name (string)
street (string)
zipcode (string)
state (string)
country (string)
meta (JSON)
meta
列包含年龄,例如{"age": "45"}
我想获取具有id
、 name
和age
的用户列表。 由于数据可能很大,我创建了一个自定义 DTO UserDataDto
以下是相同的示例:
@Query("SELECT new com.model.UserDataDto(userId, name, FUNCTION('JSON_EXTRACT', meta, '$.age')) " +
"FROM User " +
"WHERE userId IN (:userIds) ")
List<UserDataDto> findUsersByIdIn(@Param("userIds") List<Long> userIds);
User
实体:
@Data
@Table(name = "user")
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long userId;
private String name;
private String street;
private String zipcode;
private String state;
private String country;
@NotNull
@Convert(converter = UserMetaConverter.class)
private UserMeta meta;
}
UserMeta
结构:
@Data
public class UserMeta {
private Integer age;
}
UserDataDto
结构:
public class UserDataDto {
private Long userId;
private String name;
private Integer age;
}
在启动 spring 启动应用程序时,我得到
警告 | 上下文初始化期间遇到异常 - 取消刷新尝试:org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'userRepo': Invocation of init method failed; 嵌套异常是 java.lang.IllegalArgumentException:查询方法 public abstract java.util.List com.db.UserRepo.findUsersByIdIn(List<java.lang.Long>) 的查询验证失败!
我能想出的唯一解决方案是使用本机查询,非常感谢任何其他解决方案。
我能想出的唯一解决方案是使用@NamedNativeQuery
和@SqlResultSetMapping
在User
实体中
@Data
@Table(name = "user")
@Entity
@NamedNativeQuery(
name = "findUsersByIdIn",
query = "SELECT userId, name, meta->>'$.age' as age " +
"FROM user " +
"WHERE user_id = :userIds",
resultSetMapping = "UserDataDto"
)
@SqlResultSetMapping(
name = "UserDataDto",
classes = @ConstructorResult(
targetClass = UserDataDto.class,
columns = {
@ColumnResult(name = "userId", type = Long.class),
@ColumnResult(name = "name", type = String.class),
@ColumnResult(name = "age", type = Integer.class)
}
)
)
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String name;
private String street;
private String zipcode;
private String state;
private String country;
@NotNull
@Convert(converter = UserMetaConverter.class)
private UserMeta meta;
}
在存储库中
public interface UserRepo extends JpaRepository<User, Long> {
@Query("findUsersByIdIn", nativeQuery = true)
List<UserDataDto> findUsersByIdIn(@Param("userIds") List<Long> userIds);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.