如何使用本机查询在 JPA 中执行嵌套联接

Question

我有现有的学生，临时演员

Table student
------------------------------------------
id  |   name    
------------------------------------------
1   | John          
------------------------------------------

Table extras
-----------------------------------------------------------------
id  |   student_id  |  extras_key               | extras_value
-----------------------------------------------------------------
1   | 1             | class                     | X3
2   | 1             | address_id                | addr-2
-----------------------------------------------------------------

Table address
--------------------------------------------
addr_id     |   name            | city
--------------------------------------------
addr-2      | Office            | San Jose
--------------------------------------------

如何在JPA Hibernate上加入这些表？我当前的代码是

学生实体类

@Entity
@Table(name = "student")
@Data
public class Student implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "name", nullable = false, columnDefinition = "TEXT")
    private String name;

//    @OneToOne
//    @JoinColumn(name = "id", referencedColumnName = "student_id")
//    private Extras extras;

    @OneToMany(mappedBy = "studentId")
    private Set<Extras> extras;
}

附加实体类:

@Data
@Entity
@Table(name="extras")
public class Extras implements Serializable {
    @Id
    @Column(name="id")
    private int id;

    @Column(name="student_id", nullable = false)
    private int studentId;

    @Column(name="extras_key", nullable = false)
    private String extrasKey;

    @Column(name="extras_value", nullable = false)
    private String extrasValue;

    @OneToOne(optional = false)
    @NotFound(action = NotFoundAction.IGNORE)
    @JoinColumn(name = "extras_value", referencedColumnName = "addr_id", insertable = false, updatable = false)
    private Address address;
}

和地址实体类

@Entity
@Table(name="address")
@Data
public class Address implements Serializable {
    @Id
    @Column(name="addr_id")
    private String addrId;

    @Column(name="name")
    private String name;

    @Column(name="city")
    private String city;

}

这是我的存储库类

@Repository
public interface StudentRepository extends CrudRepository<Student, Integer> {
    @Query(value = "select * from student " +
            "JOIN extras ON student.id=extras.student_id " +
            "JOIN address ON address.addr_id = extras.extras_value", nativeQuery = true)
    List<Student> findAllData();
}

但是当我在学生实体中使用OneToOne时，我得到了异常“org.hibernate.LazyInitializationException：未能懒惰地初始化角色集合：com.jpa.belajarjpa.enitities.Student.extras，无法初始化代理-没有会话”。但是当我在学生实体中使用OneToOne时，我没有遇到异常，但得到了错误的结果，学生（id=1，名称=john，附加项=null），当我执行此查询时，它显示正确的结果SELECT*From学生加入附加项ONstudent.id=extras.student_idJOIN地址ONaddress.addr_id=extras.extras_value

Answer 1

通过在属性中添加spring . JPA . properties . hibernate . enable _ lazy _ load _ no _ trans = true解决

Answer 2

您需要将联接提取与 HQL 查询结合使用，如下所示：

@Repository
public interface StudentRepository extends CrudRepository<Student, Integer> {
    @Query("FROM Student s " +
            "JOIN FETCH s.extras e " +
            "JOIN FETCH e.address")
    List<Student> findAllData();
}

< code > spring . JPA . properties . hibernate . enable _ lazy _ load _ no _ trans = true 是一个反模式。如果我是你，我就不会用它。