Spring Data JPA：按联接表中的属性排序后的重复记录

Question

我在我的项目中遇到了一个问题，因此我将提供一个使用演示项目的示例。假设有两个实体：教师和学生。

@Entity
@Table(name = "STUDENTS")
public class Student {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    private String surname;

    @ManyToOne
    private Teacher teacher;

    // getters and setters
}

@Entity
@Table(name = "TEACHERS")
public class Teacher  {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    private String surname;

    @OneToMany(mappedBy = "teacher")
    List<Student> students = new ArrayList<>();

    // getters and setters

}

有一个存储库和一个服务类：

public interface TeacherRepository extends PagingAndSortingRepository<Teacher, Long>, JpaSpecificationExecutor<Teacher> {
}

@Service
public class TeacherService {

    @Autowired
    private TeacherRepository teacherRepository;

    @Autowired
    private TeacherToDTOMapper teacherToDTOMapper;

    public Page<TeacherDTO> findAll(Pageable pageable) {
        Specification<Teacher> specification = (root, criteriaQuery, criteriaBuilder) -> {

            return criteriaBuilder.lessThanOrEqualTo(root.get("id"), 2);
        };
        return teacherRepository.findAll(specification, pageable).map(teacherToDTOMapper::mapEntityToDTO);
    }
}

最后是一个控制器类：

@RestController
public class TeacherController {

    @Autowired
    private TeacherService teacherService;

    @GetMapping("/teachers")
    @ResponseStatus(HttpStatus.OK)
    public PagedModel<EntityModel<TeacherDTO>> findAll(PagedResourcesAssembler<TeacherDTO> resourcesAssembler, Pageable pageable) {
        Page<TeacherDTO> page = teacherService.findAll(pageable);
        return resourcesAssembler.toModel(page);
    }
}

在 UI 上有一个table of teachers。

用户过滤掉 id > 2 的教师，并按学生姓名对他们进行排序。预期的结果应该完全按照该顺序包含 Aliyah Wiggins 和 Macy Reyes（因为学生 Amina Parks 应该是排序后的第一个）。 http://localhost:8080/teachers?page=0&size=3&sort=students.name,asc

但是，实际结果是：

{
  "_embedded": {
    "teacherDTOList": [
      {
        "id": 1,
        "name": "Aliyah",
        "surname": "Wiggins",
        "students": [
          {
            "id": 1,
            "name": "Amina",
            "surname": "Parks"
          },
          {
            "id": 2,
            "name": "Caleb",
            "surname": "Baker"
          },
          {
            "id": 3,
            "name": "Valentino",
            "surname": "Kent"
          },
          {
            "id": 4,
            "name": "Jaron",
            "surname": "Luna"
          },
          {
            "id": 5,
            "name": "Cayden",
            "surname": "Hoover"
          }
        ]
      },
      {
        "id": 1,
        "name": "Aliyah",
        "surname": "Wiggins",
        "students": [
          {
            "id": 1,
            "name": "Amina",
            "surname": "Parks"
          },
          {
            "id": 2,
            "name": "Caleb",
            "surname": "Baker"
          },
          {
            "id": 3,
            "name": "Valentino",
            "surname": "Kent"
          },
          {
            "id": 4,
            "name": "Jaron",
            "surname": "Luna"
          },
          {
            "id": 5,
            "name": "Cayden",
            "surname": "Hoover"
          }
        ]
      },
      {
        "id": 2,
        "name": "Macy",
        "surname": "Reyes",
        "students": [
          {
            "id": 6,
            "name": "Jasmin",
            "surname": "Friedman"
          },
          {
            "id": 7,
            "name": "Randall",
            "surname": "Archer"
          },
          {
            "id": 8,
            "name": "Caroline",
            "surname": "Davila"
          },
          {
            "id": 9,
            "name": "Sarah",
            "surname": "Pace"
          }
        ]
      }
    ]
  },
  "_links": {
    "self": {
      "href": "http://localhost:8080/teachers?page=0&size=3&sort=students.name,asc"
    }
  },
  "page": {
    "size": 3,
    "totalElements": 3,
    "totalPages": 1,
    "number": 0
  }
}

Aliyah Wiggins 有重复记录。我还没有找到一种适当的方法来获得预期的结果，所以现在我认为搜索与一个实体相关的对象并在另一个实体中按属性排序是不正确的。 演示项目可用here。

Answer 1

我开始认为尝试按学生（即“许多”）的名字对教师（即一对多关系中的“一个”）进行排序是错误的。尝试按学生姓名排序后，我们将得到以下结果：

ROW    TEACHER_NAME     TEACHER_SURNAME     STUDENT_FULL_NAME
1      Evan             Fuentes             Allyson Noble
2      Aliyah           Wiggins             Amina Parks
3      Evan             Fuentes             Ashlyn Berg
4      Aliyah           Wiggins             Caleb Baker
...

一位老师可以同时是第一名、第三名和最后一名。所以，那个排序之王是没有意义的。 此外，Many-To-Many 更适合该示例，但仍不能改变事实。

Answer 2

您可以将查询设置为返回规范中的不同值：

criteriaQuery.distinct(true);

这通常由 Spring Data 在 findAll 方法上设置为 true，但在您使用自己的规范时并不暗示。您现在可以看到数据库连接查询的结果（教师将为每个学生重复一次）。

我建议在模型中添加一个标志属性，而不是按 id 跳过。引入hidden / deleted / private 之类的东西会简化代码并使其更具可读性。然后您可以使用一个空的存储库接口方法：findByActiveTrue();

Answer 3

您可以使用命名实体图来避免重复记录。

@NamedEntityGraph(name = "Teacher.students" ,attributeNodes = @NamedAttributeNode("students"))
@Entity
@Table(name = "TEACHERS")
public class Teacher  {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    private String surname;

    @OneToMany(mappedBy = "teacher")
    List<Student> students = new ArrayList<>();

    // getters and setters

}

并将 findAll 方法覆盖为

@EntityGraph(value = "Teacher.students" , type = EntityGraphType.LOAD)
Page<Teacher> findAll(Specification<Teacher> spec, Pageable pageable);