spring-data-jpa-guide
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zhangzhenhuajack
如何用JPQL 操作 MySQL json 类型的字段？
这个有点高端，贴点例子，感觉就是正常的属性操作呀？和普通的应该没什么区别
MySQL 有个字段属性叫 json ，和varchar一样是基本属性。这个属性意味着能存复杂格式，让MySQL 有点像Mongodb 这种NOSQL。目前已经找到方法了
呵呵 shaopro [email protected] 於 2020年12月8日 週二 21:10 寫道：
这个有点高端，贴点例子，感觉就是正常的属性操作呀？和普通的应该没什么区别
MySQL 有个字段属性叫 json ，和varchar一样是基本属性。这个属性意味着能存复杂格式，让MySQL 有点像Mongodb 这种NOSQL。目前已经找到方法了
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@shaopro 怎么处理的？贴个例子学习下
替换json 中某个字段。，
需要的依赖
<dependency> <groupId>com.vladmihalcea</groupId> <artifactId>hibernate-types-52</artifactId> <version>2.10.0</version> </dependency>
实体类
`
import com.vladmihalcea.hibernate.type.json.JsonStringType;
@Entity @Table(name = "UserInfoDO") @DynamicInsert @DynamicUpdate @EntityListeners({AuditingEntityListener.class}) @TypeDef(name = "JSON", typeClass = JsonStringType.class) public class UserInfoDO extends BaseEntity {
private static final long serialVersionUID = 460626250371942731L;
@Column(name = "name", nullable = false, columnDefinition = "varchar(20) default '默认名字' comment '我是username注释'")
private String username;
@Column(name = "age", nullable = false, columnDefinition = "INT(4) default 18 comment '我是age注释'")
private Integer age;
@Type(type = "JSON")
@Column(name = "roles", columnDefinition = "JSON")
private Set<RoleInfoBO> roles;
@Type(type = "JSON")
@Column(columnDefinition = "JSON")
private AddressBO address;
}
值类
@Data
@Slf4j
@Builder
@ToString
@NoArgsConstructor
@AllArgsConstructor
public class AddressBO implements Serializable {
private static final long serialVersionUID = 8330351916838251803L;
private String name;
private String path;
}
数据仓储层
public interface IUserInfoRepository extends JpaRepository<UserInfoDO, Long> {
@Modifying
@Transactional
@Query("UPDATE UserInfoDO AS user SET user.address = FUNCTION('JSON_REPLACE', user.address,'$.name', :name2) WHERE user.username = :name1")
int replace(@Param("name1") String name1, @Param("name2") String name2);
}
`