如何使用休眠投影检索复杂类及其成员

Question

我有一个如下类，需要使用Hibernate从数据库中检索。问题是我的类有多个成员，其中大多数是类，如何检索它们？

@Entity
public class Student {
  @Id
  long id;
  String name;
  String fname;
  @OneToMany
  List<Course> courses;
  @ManyToOne
  Dealer dealer;
  ...
}
@Entity
public class Dealer {
   @Id
   long id;
   String name; 
   @OneToMany(fetch = FetchType.LAZY, mappedBy = "cr.dealer", cascade = CascadeType.ALL)
   Set<Car> cars = new HashSet<Cars>(0);
   ..
}

我需要检索学生证 1 及其所有课程、经销商和经销商汽车列表。

我的投影如下，但它没有返回任何内容。

  ...
    .setProjection(Projections.projectionList()
    .add(Projections.property("friends.cars").as("cars")
    ...

Answer 1

    // Projection is not needed, Hibernate will load child values as shown below
    Student student = session.get(Student.class);
    List<Course> courses = student.getCourses();
    Dealer dealer = student.getDealer();
    //  If u want records only where child records are present, u can use LEFT_OUTER_JOIN
    Criteria criteria = getHibernateSession().createCriteria(Student.class);
    criteria.createAlias("Course", "Course", JoinType.LEFT_OUTER_JOIN);
    // If u want to use Projections for performance, u have to add each and every column in projection
    Criteria criteria = getHibernateSession().createCriteria(A.class);
    criteria.createAlias("b", "b", JoinType.INNER_JOIN);
    criteria.createAlias("b.r", "b.r", JoinType.INNER_JOIN);
    criteria.createAlias("b.c", "b.c", JoinType.LEFT_OUTER_JOIN);
    ProjectionList projectionList = Projections.projectionList();
    projectionList.add(Projections.groupProperty("column1"));
    projectionList.add(Projections.property("column2"));
    projectionList.add(Projections.property("column3"));
    criteria.setProjection(projectionList);
    criteria.setResultTransformer(Transformers.aliasToBean(Table.class));

Answer 2

因为您有一个课程列表和一组汽车，所以您可以在单个查询中简单地获取整个图表：

select s
from Student s
left join fetch s.courses
left join fetch s.dealer d
left join fetch d.cars
where s.id = :id

由于您正在提取两个集合，因此此查询将生成笛卡尔积，因此您需要确保所选子集合没有太多条目。

如果您不想遇到笛卡尔乘积，只需运行此查询：

select s
from Student s
left join fetch s.courses
left join fetch s.dealer d
where s.id = :id

然后，您访问 dealer.cars 以使用单独的查询获取该集合：

Student s = ...;
s.getDealer().getCars().size();

Answer 3

如果高性能不是问题，那么你应该让Hibernate做他的工作。只需使用实体的获取者即可。例如：

Student student1 = session.get(Student.class, 1L);
List<Course> courses = student1.getCourses();
Dealer dealer = student1.getDealer();
Set<Car> cars = dealer.getCars();

Answer 4

我不确定你是否可以使用 QueryOver，但对于这类任务来说会非常容易。

Student student = null;
Dealer dealer = null;
Course course = null;
Car car = null;
var myStudent = Session.QueryOver<Student>(() => student)
.Left.JoinQueryOver(() => student.courses, () => courses)
.Left.JoinQueryOver(() => student.dealer, () => dealer)
.Left.JoinQueryOver(() => dealer.cars, () => car)
.SelectList(list => list
  .Select(() => student.Name)
  .Select(() => student.Age)
  .Select(() => courses.Description)
  .Select(() => dealer.locaiton)
  .Select(() => car.Model))
  .TransformUsing(Transformers.AliasToBean<StudentModel>())
  .List<StudentModel>().AsQueryable();

创建学生模型 DTO 以获得结果。这只是一个提示，您可以根据需要对其进行修改。我希望这会奏效。:)