Question

我有以下情况：有公司和员工。每家公司都有一组员工。每位员工都可以为多家公司工作。所以我实现了以下关系：

Company.class：

@JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "company_id") , inverseJoinColumns = @JoinColumn(name = "employee_id") )
@ManyToMany(fetch = FetchType.LAZY)
private List<Employee> employees;

Employee.class：

@JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "employee_id") , inverseJoinColumns = @JoinColumn(name = "company_id") )
@ManyToMany(fetch = FetchType.LAZY)
private List<Company> companies;

显然，要为多家公司工作，每位员工应该为他或她工作的每家公司分配几个不重叠的时间表。此外，每个组合公司 - 员工应该有一个计划列表，因为有时旧计划到期，新计划生效。所以我也有Schedule.class，它应该与child to parent @ManyToOne和Company建立Employee关系，并且应该遵循以下方式：每个Schedule，因此，List<Schedule>应该只对应Company和Employee个实例的一个组合。如何实现这种关系？

更新1

我只考虑为每个@OneToMany Schedule和Company添加Employee关系，但我需要将Schedule的实例都添加到Company和Employee每次Schedule，这种方式看起来不对，现在对我来说也不是很明显如何取回它。所以任何帮助都将不胜感激。

此帖子已更新，以显示我所拥有的真实场景，而不仅仅是通用的Entity1，Entity2，类的Entity3名称。

更新2

我接受了答案，但如果Schedule包含列表，我就无法使用它。根据我的计划，List<Vacation>应该包含Vacations，以了解一年内List和Days Days的集合，每个Schedule显示特别是工作日，休息和结束这一天。那些schedule_id对于每个Schedule实例也是唯一的。

它应该是类似下面的内容，但显然现在我没有@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, orphanRemoval = true) @JoinColumn(name = "schedule_id") private List<Vacation> vacations; @JoinTable(name = "schedule_week", joinColumns = @JoinColumn(name = "schedule_id") , inverseJoinColumns = @JoinColumn(name = "day_id") ) @OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, orphanRemoval = true) private List<Day> week;，那么如何将这些列表连接到en-GB？

eng

如何正确包含这些列表？

Answer 1

我想建议以下解决方案。

embeddable类，其中包含特定时间表的Company和Employee。

@Embeddable
public class ScheduleOwner implements Serializable{

    @MapsId("id")
    @ManyToOne(cascade = CascadeType.ALL)
    Company c;

    @MapsId("id")
    @ManyToOne(cascade = CascadeType.ALL)
    Employee e;
}

Schedule类正在嵌入ScheduleOwner个实例。

@Entity
public class Schedule {

    @EmbeddedId
    ScheduleOwner owner;

    String description;
}

Company和Employee类（未对其进行任何更改）

@Entity
public class Company {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "company_id") , inverseJoinColumns = @JoinColumn(name = "employee_id") )
    @ManyToMany(fetch = FetchType.LAZY)
    private List<Employee> employees;
}


@Entity
public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "employee_id") , inverseJoinColumns = @JoinColumn(name = "company_id") )
    @ManyToMany(fetch = FetchType.LAZY)
    private List<Company> companies;
}

更新1

以下是保存和获取结果的方法。

  Employee e1 = new Employee();
  Company c1 = new Company();
  c1.employees.add(e1);

  e1.companies.add(c1);

  ScheduleOwner so = new ScheduleOwner();
  so.c = c1;
  so.e = e1;

  Schedule s = new Schedule();
  s.owner = so;

  session.save(c1);
  session.save(e1);
  session.save(s);

  // below query will fetch from schedule, where company id = 9
  Schedule ss = (Schedule) session.createQuery("From Schedule sh where sh.owner.c.id = 9").uniqueResult();

更新2

@Entity
public class Company {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "company_id", referencedColumnName="id") 
                                        , inverseJoinColumns = @JoinColumn(name = "employee_id", referencedColumnName="id"))
    @ManyToMany(fetch = FetchType.LAZY)
    List<Employee> employees = new ArrayList<>();

    String name;
}

@Entity
public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

    @ManyToMany(fetch = FetchType.LAZY, mappedBy = "employees")
    List<Company> companies = new ArrayList<>();

    String name;
}

@Entity
public class Schedule {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    int schedule_id;

    @ManyToOne
    @JoinColumn(name = "company_id", insertable = false, updatable = false)
    private Company company;
    @ManyToOne
    @JoinColumn(name = "employee_id", insertable = false, updatable = false)
    private Employee employee;

    String description;

    @OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, orphanRemoval = true, mappedBy = "schedule")
    List<Vacation> vacations;

}

@Entity
public class Vacation {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int vacation_id;

    @ManyToOne
    @JoinColumn(name = "schedule_id" ) 
    Schedule schedule;

    @OneToMany(mappedBy = "vacation")
    List<Day> days;
}

Day实体与Vacation直接相关。不要Schedule。

@Entity
public class Day {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;


    @ManyToOne
    @JoinColumn(name = "vacation_id")
    Vacation vacation;
}

希望这有帮助。

Hibernate两个父母一个孩子映射

1 个答案: