[英]Use JSON_EXTRACT to get value from JSON in Spring JPA Custom Query for custom DTO
我想從 JSON 列中獲取值並在 spring JPA 中返回自定義 DTO。
表結構
userId (int)
name (string)
street (string)
zipcode (string)
state (string)
country (string)
meta (JSON)
meta
列包含年齡,例如{"age": "45"}
我想獲取具有id
、 name
和age
的用戶列表。 由於數據可能很大,我創建了一個自定義 DTO UserDataDto
以下是相同的示例:
@Query("SELECT new com.model.UserDataDto(userId, name, FUNCTION('JSON_EXTRACT', meta, '$.age')) " +
"FROM User " +
"WHERE userId IN (:userIds) ")
List<UserDataDto> findUsersByIdIn(@Param("userIds") List<Long> userIds);
User
實體:
@Data
@Table(name = "user")
@Entity
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long userId;
private String name;
private String street;
private String zipcode;
private String state;
private String country;
@NotNull
@Convert(converter = UserMetaConverter.class)
private UserMeta meta;
}
UserMeta
結構:
@Data
public class UserMeta {
private Integer age;
}
UserDataDto
結構:
public class UserDataDto {
private Long userId;
private String name;
private Integer age;
}
在啟動 spring 啟動應用程序時,我得到
警告 | 上下文初始化期間遇到異常 - 取消刷新嘗試:org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'userRepo': Invocation of init method failed; 嵌套異常是 java.lang.IllegalArgumentException:查詢方法 public abstract java.util.List com.db.UserRepo.findUsersByIdIn(List<java.lang.Long>) 的查詢驗證失敗!
我能想出的唯一解決方案是使用本機查詢,非常感謝任何其他解決方案。
我能想出的唯一解決方案是使用@NamedNativeQuery
和@SqlResultSetMapping
在User
實體中
@Data
@Table(name = "user")
@Entity
@NamedNativeQuery(
name = "findUsersByIdIn",
query = "SELECT userId, name, meta->>'$.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;
}
在存儲庫中
public interface UserRepo extends JpaRepository<User, Long> {
@Query("findUsersByIdIn", nativeQuery = true)
List<UserDataDto> findUsersByIdIn(@Param("userIds") List<Long> userIds);
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.