Spring Boot 2. Jackson json 序列化或 Spring Data JPA FetchType.Lazy
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【中文标题】Spring Boot 2. Jackson json 序列化或 Spring Data JPA FetchType.Lazy【英文标题】:Spring Boot 2. Jackson json serialization or Spring Data JPA FetchType.Lazy 【发布时间】:2018-11-07 17:12:55 【问题描述】:我的情况非常奇怪,坦率地说,我不完全确定它是 Jackson 序列化问题还是与 Spring Data JPA 如何获取 Lazy 对象有关
春季 2.0.2
spring-boot-starter-data-jpa spring-boot-starter-web spring-boot-devtools mysql-connector-java当您搜索大多数人对 Jackson 的问题时,它是序列化会话之外的延迟提取,导致以下异常。 但是,我根本没有遇到这个异常......
org.hibernate.LazyInitializationException:无法初始化代理 - 没有会话
问题
-
它实际上在我的控制器中获取数据并返回 json 对象(下面的休眠查询)。急切地获取它是一个问题,因为它会获取未使用的不必要的大数据。
结构(省略getter和setter以使线程更短)
User
| <---[ OneToMany ]
Employee hierarchies entity
|
+-----+-----+
| |
PartTime FullTime
用户
@Entity
public class User
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String name;
@OneToMany(mappedBy="user")
@JsonIgnoreProperties("user")
private List<Employee> employee = new ArrayList<>();
员工
@Entity
@Inheritance(strategy = InheritanceType.JOINED)
public abstract class Employee
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@ManyToOne
@JsonIgnoreProperties("employee")
private User user;
FullTime / PartTime(在 PartTime 时,现场工资更改为每小时工资)
@Entity
public class FullTime extends Employee
private BigDecimal salary;
UserRepository 扩展 CrudRepository
用户服务
public User getUserById(Long id)
Optional<User> optionalUser = userRepo.findById(id);
User user = optionalUser.get();
logger.info("\n User -> ", user);
return user;
控制器
@RestController
@RequestMapping(value="/api/user")
public class UserController
@GetMapping(value="/userId" )
public User getUser(@PathVariable("userId") Long id)
return userService.getUserById(id);
在我的应用程序上扩展了 CommandLineRunner,它允许我在应用程序启动时运行命令
@Override
public void run(String... args) throws Exception
logger.info("users 1 -> " , userService.getUserById(1L));
Hibernate:
select
user0_.id as id1_3_0_,
user0_.name as name2_3_0_
from
user user0_
where
user0_.id=?
INFO 14434 --- [ restartedMain] ication$$EnhancerBySpringCGLIB$$6cf0457c :
users 1 ->
User [id=1, name=Jack]
但是当我通过我的控制器http://localhost:8080/api/user/1
时,我得到了 2 个单独的休眠调用,这两个调用似乎都在我的服务层内。请记住,我的服务层没有事务,所以它真的很奇怪......
Hibernate: << 1st
select
user0_.id as id1_3_0_,
user0_.name as name2_3_0_
from
user user0_
where
user0_.id=?
com.example.app.service.UserService : << -- In service
User ->
User [id=1, name=Jack]
Hibernate:
select
employee0_.user_id as user_id2_0_0_,
employee0_.id as id1_0_0_,
employee0_.id as id1_0_1_,
employee0_.user_id as user_id2_0_1_,
employee0_1_.salary as salary1_1_1_,
employee0_2_.hourly_wage as hourly_w1_2_1_,
case
when employee0_1_.id is not null then 1
when employee0_2_.id is not null then 2
when employee0_.id is not null then 0
end as clazz_1_
from
employee employee0_
left outer join
full_time employee0_1_
on employee0_.id=employee0_1_.id
left outer join
part_time employee0_2_
on employee0_.id=employee0_2_.id
where
employee0_.user_id=?
现在,如果它是 Jackson 序列化问题,我访问过 Avoid Jackson serialization on non fetched lazy objects 、Configure Jackson to omit lazy-loading attributes in Spring Boot 等等,但所有这些都是通过扩展 WebMvcConfigurerAdapter 来完成的,我认为它在 spring5 中已被弃用?
如果它与 Spring Data JPA 有关...那么请说明一下,因为我一直认为您需要 @Transaction 注释来进行惰性 Fetch 以将关系与休眠会话关联...
对不起,长线程...
【问题讨论】:
你应该使用 DTO。请检查***.com/questions/1612334/… 和baeldung.com/… 您正在返回休眠对象,并且在序列化期间,没有会话,因为它已经关闭。 感谢收看。但我实际上没有得到 LazyInitializationException....我不喜欢 DTO,因为它为应用程序添加了额外的层.... 【参考方案1】:我认为您启用了属性spring.jpa.open-in-view
(默认为true
)。检查您是否在日志中看到以下消息:
spring.jpa.open-in-view is enabled by default. Therefore, database queries may be performed during view rendering. Explicitly configure spring.jpa.open-in-view to disable this warning
如果禁用它,您应该会看到所需的异常。
【讨论】:
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