Spring和Hibernate:为什么获取1个惰性字段会触发加载所有其他惰性字段

问题描述 投票:0回答:1

在我的应用程序中,我有一个Staff实体,其中包含许多像这样的延迟加载字段。

@Entity(name="CommonStaff")
@Table(name="staff")
@Getter @Setter
public class Staff implements Serializable {
    ...

    @ManyToOne(fetch=FetchType.LAZY)
    @LazyToOne(LazyToOneOption.NO_PROXY)
    @JoinColumn(name="nationality", referencedColumnName="code", insertable=false, updatable=false)
    private Nationality nationality;

    @ManyToOne(fetch=FetchType.LAZY)
    @LazyToOne(LazyToOneOption.NO_PROXY)
    @JoinColumn(name="marital_status", referencedColumnName="code", insertable=false, updatable=false)
    private MaritalStatus maritalStatus;

    ...
}

当我加载Staff记录时,所有这些字段都没有按期望的那样被快速加载。但是,当我触发例如getNationality()时,我看到框架也正在执行SQL来加载MaritalStatus。我一直在尝试找到解决此问题的方法,但是找不到任何有用的资源。如果您能指出我的方向,我将不胜感激。

一些示例代码。

@Autowired
@Qualifier("dataModuleStaffRepo")
private StaffRepo staffRepo;

@CustomerTransactional
@GetMapping("/profile")
public void testProfile(@RequestParam String userId) {
    Optional<Staff> staff = staffRepo.findByUserId(userId);
    if (staff.isPresent()) {
        System.out.println(staff.get().getName());
        System.out.println(staff.get().getNationality().getName());
    }
}

下面是我在控制台中看到的内容。在打印出职员姓名之后,getNationality()也触发了加载MaritalStatus

Edgar Rey Tann
2020-04-16 14:29:40,283 DEBUG [http-nio-9000-exec-2] org.hibernate.SQL   : 
    /* sequential select
        com.ft.common.db.customer.domain.Staff */ select
            staff_.marital_status as marital19_23_,
            staff_.nationality as nationa22_23_
        from
            staff staff_ 
        where
            staff_.id=?
2020-04-16 14:29:40,283 TRACE [http-nio-9000-exec-2] org.hibernate.type.descriptor.sql.BasicBinder   : binding parameter [1] as [BIGINT] - [660]
2020-04-16 14:29:40,291 DEBUG [http-nio-9000-exec-2] org.hibernate.SQL   : 
    /* load com.ft.common.db.customer.domain.MaritalStatus */ select
        maritalsta0_.id as id1_9_0_,
        maritalsta0_.code as code2_9_0_,
        maritalsta0_.description as descript3_9_0_,
        maritalsta0_.name as name4_9_0_,
        maritalsta0_.order_id as order_id5_9_0_,
        maritalsta0_.short_name as short_na6_9_0_ 
    from
        marital_status maritalsta0_ 
    where
        maritalsta0_.code=?
2020-04-16 14:29:40,291 TRACE [http-nio-9000-exec-2] org.hibernate.type.descriptor.sql.BasicBinder   : binding parameter [1] as [VARCHAR] - [MAR_2]
2020-04-16 14:29:40,298 DEBUG [http-nio-9000-exec-2] org.hibernate.SQL   : 
    /* load com.ft.common.db.customer.domain.Nationality */ select
        nationalit0_.id as id1_16_0_,
        nationalit0_.code as code2_16_0_,
        nationalit0_.description as descript3_16_0_,
        nationalit0_.name as name4_16_0_,
        nationalit0_.order_id as order_id5_16_0_,
        nationalit0_.short_name as short_na6_16_0_ 
    from
        nationality nationalit0_ 
    where
        nationalit0_.code=?
2020-04-16 14:29:40,298 TRACE [http-nio-9000-exec-2] org.hibernate.type.descriptor.sql.BasicBinder   : binding parameter [1] as [VARCHAR] - [NAT_I]
Indonesian
java spring hibernate spring-data-jpa lazy-loading
1个回答
0
投票

默认情况下,实体类的所有惰性属性都属于一个名为DEFAULT的组。并且获取DEFAULT组的任何属性也将获取其他属性。为了解决此问题,我们需要使用@LazyGroup注释定义希望单独获取的组。

因此,我们通过@LazyGroup注释同时注释国籍和maritalStatus。

@ManyToOne(fetch=FetchType.LAZY)
@LazyToOne(LazyToOneOption.NO_PROXY)
@LazyGroup("nationality")
@JoinColumn(name="nationality", referencedColumnName="code", insertable=false, updatable=false)
private Nationality nationality;

@ManyToOne(fetch=FetchType.LAZY)
@LazyToOne(LazyToOneOption.NO_PROXY)
@LazyGroup("maritalStatus")
@JoinColumn(name="marital_status", referencedColumnName="code", insertable=false, updatable=false)
private MaritalStatus maritalStatus;

并且希望您正在使用字节码增强功能进行No-proxy懒惰提取

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