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您好，我从 2018 年看到了一些类似的问题，但没有太多信息

我有以下课程

  1. 员工
@Entity
@Getter
@Setter
@NoArgsConstructor
@NamedEntityGraph(name = "employee.complete", attributeNodes = {
        @NamedAttributeNode(
                value = "addresses",
                subgraph = "address_city"
        )},
        subgraphs = {
                @NamedSubgraph(
                        name = "address_city",
                        attributeNodes = {
                                @NamedAttributeNode("city")
                        })
        },
        subclassSubgraphs = {
                @NamedSubgraph(
                        name = "noUse",
                        type = Engineer.class,
                        attributeNodes = {
                                @NamedAttributeNode("laptop")
                        })
        }
)
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "profession")
public class Employee {

    @Id
    String id;

    String name;

    String email;

    @OneToMany
    @JoinColumn(name = "employee_id", updatable = false)
    List<Address> addresses;

}
  1. 地址
@Entity
@Getter
@Setter
@ToString
@NoArgsConstructor
public class Address {

    @Id
    private String addressId;
    private String streetName;
    private String streetNumber;

    @OneToOne
    @JoinColumn(name = "city_id")
    private City city;

}
  1. 城市
@Entity
@Setter
@Getter
public class City {

    @Id
    String id;
    String name;
}
  1. 工程师

@Entity
@Getter
@Setter
@DiscriminatorValue("engineer")
public class Engineer extends Employee {

    private String seniority;

    @OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "id")
    Laptop laptop;
}
  1. 笔记本电脑
@Entity
@Getter
@Setter
public class Laptop {

    @Id
    @Column(name = "employee_id")
    String employeeId;

    String brand;
}

使用 NamedEntityGraph，我想在单个查询中实现检索员工的所有信息。城市的属性节点和子图工作正常，但子类子图不适用于此示例。

我的期望是我将在一个带有 join 子句的查询中检索笔记本电脑信息，但实际上有两个 select 语句

select employee0_.id as id2_3_0_, addresses1_.address_id as address_2_0_1_, city2_.id as id1_1_2_, employee0_.email as email3_3_0_, employee0_.name as name4_3_0_, employee0_.seniority as seniorit5_3_0_, employee0_.profession as professi1_3_0_, addresses1_.city_id as city_id5_0_1_, addresses1_.street_name as street_n3_0_1_, addresses1_.street_number as street_n4_0_1_, addresses1_.address_type as address_1_0_1_, addresses1_.employee_id as employee6_0_0__, addresses1_.address_id as address_2_0_0__, city2_.name as name2_1_2_ from employee employee0_ left outer join address addresses1_ on employee0_.id=addresses1_.employee_id left outer join city city2_ on addresses1_.city_id=city2_.id where employee0_.id=? and (employee0_.id is not null)

select laptop0_.employee_id as employee1_4_0_, laptop0_.brand as brand2_4_0_ from laptop laptop0_ where laptop0_.employee_id=?

我找不到任何报告有关此错误的页面，我做错了什么吗？

代码仓库-> https://github.com/dkasiaras/pocs/tree/main/jpa-graph-entity

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0 回答 0