8

我几乎看不到与 Hibernate 相关的以下问题的任何指针。这与使用与其自身具有父子关系的单个数据库表实现继承有关。例如:

CREATE TABLE Employee (
  empId BIGINT NOT NULL AUTO_INCREMENT,
  empName VARCHAR(100) NOT NULL,
  managerId BIGINT,
  CONSTRAINT pk_employee PRIMARY KEY (empId)
)

这里,managerId列可能为空,也可能指向Employee表的另一行。业务规则要求员工了解他的所有报告人,并让他了解他/她的经理。业务规则还允许行的managerId为空(组织的 CEO 没有经理)。

我们如何在 Hibernate 中映射这种关系,标准的多对一关系在这里不起作用?特别是,如果我不仅想将我的实体实现为相应的“员工”实体类,还想实现多个类,例如“经理”、“助理经理”、“工程师”等,每个类都继承自“员工”超级实体类,某些实体的属性实际上并不适用于所有人,例如“经理”获得 Perks,其他实体则没有(相应的表列当然会接受 null)。

示例代码将不胜感激(我打算使用 Hibernate 3 注释)。

4

3 回答 3

9

您在这里表达了两个概念:

  1. 继承,并且您希望将继承层次结构映射到单个表中。
  2. 父/子关系。

要实现 1.,您需要使用 Hibernate 的每个类层次结构的单表策略:

@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
    name="emptype",
    discriminatorType=DiscriminatorType.STRING
)
public abstract class Employee { ... }

@Entity
@DiscriminatorValue("MGR")
public class Manager extends Employee { ... }

要实现 2.,您需要在 上添加两个自引用关联Employee

  • 许多员工有零个或一个经理(这也是一个Employee
  • 一名员工有零个或多个报告人

结果Employee可能如下所示:

@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
    name="emptype",
    discriminatorType=DiscriminatorType.STRING
)
public abstract class Employee { 

    ... 

    private Employee manager;
    private Set<Employee> reportees = new HashSet<Employee>();

    @ManyToOne(optional = true)
    public Employee getManager() {
        return manager;
    }

    @OneToMany
    public Set<Employee> getReportees() {
        return reportees;
    }

    ...
}

这将导致下表:

CREATE TABLE EMPLOYEE_EMPLOYEE (
        EMPLOYEE_ID BIGINT NOT NULL,
        REPORTEES_ID BIGINT NOT NULL
    );

CREATE TABLE EMPLOYEE (
        EMPTYPE VARCHAR(31) NOT NULL,
        ID BIGINT NOT NULL,
        NAME VARCHAR(255),
        MANAGER_ID BIGINT
    );

ALTER TABLE EMPLOYEE ADD CONSTRAINT SQL100311183749050 PRIMARY KEY (ID);

ALTER TABLE EMPLOYEE_EMPLOYEE ADD CONSTRAINT SQL100311183356150 PRIMARY KEY (EMPLOYEE_ID, REPORTEES_ID);

ALTER TABLE EMPLOYEE ADD CONSTRAINT FK4AFD4ACE7887BF92 FOREIGN KEY (MANAGER_ID)
    REFERENCES EMPLOYEE (ID);

ALTER TABLE EMPLOYEE_EMPLOYEE ADD CONSTRAINT FKDFD1791F25AA2BE0 FOREIGN KEY (REPORTEES_ID)
    REFERENCES EMPLOYEE (ID);

ALTER TABLE EMPLOYEE_EMPLOYEE ADD CONSTRAINT FKDFD1791F1A4AFCF1 FOREIGN KEY (EMPLOYEE_ID)
    REFERENCES EMPLOYEE (ID);
于 2010-03-11T17:46:34.240 回答
3

非常感谢你们。我创建了我的员工实体如下:

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
    name="EMPLOYEE_TYPE", 
    discriminatorType = DiscriminatorType.STRING
)
@DiscriminatorValue("Employee")
public abstract class Employee {

    @Id 
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="EMPLOYEE_ID") 
    private Integer empId = null;

    @Column(name="EMPLOYEE_NAME") 
    private String empName = null;

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

    @Column(name="EMPLOYEE_PERKS")
    private int perks;       

    @ManyToOne(targetEntity = Employee.class, optional=true)
@JoinColumn(name="MANAGER_ID", nullable=true)
private Employee manager = null;

    @OneToMany  
    private Set<Employee> reportees = new HashSet<Employee>();

    ...        

    public Set<Employee> getReportees() {
        return reportees;
    } 
}

然后,我添加了其他没有正文但只有鉴别器列值的实体类,例如 Manager、CEO 和 AsstManager。我选择让 Hibernate 为我创建表。以下是主程序:

SessionFactory sessionFactory = HibernateUtil.sessionFactory;
Session session = sessionFactory.openSession();
Transaction newTrans = session.beginTransaction();

CEO empCeo = new CEO();
empCeo.setEmpName("Mr CEO");
empCeo.setSecretary("Ms Lily");

Manager empMgr = new Manager();
empMgr.setEmpName("Mr Manager1");
empMgr.setPerks(1000);
empMgr.setManager(empCeo);

Manager empMgr1 = new Manager();
empMgr1.setEmpName("Mr Manager2");
empMgr1.setPerks(2000);
empMgr1.setManager(empCeo);

AsstManager asstMgr = new AsstManager();
asstMgr.setEmpName("Mr Asst Manager");
asstMgr.setManager(empMgr);

session.save(empCeo);
session.save(empMgr);
session.save(empMgr1);
session.save(asstMgr);
newTrans.commit();

System.out.println("Mr Manager1's manager is : "
        + empMgr.getManager().getEmpName());
System.out.println("CEO's manager is : " + empCeo.getManager());
System.out.println("Asst Manager's manager is : " + asstMgr.getManager());
System.out.println("Persons Reporting to CEO: " + empCeo.getReportees());

session.clear();
session.close();

代码运行良好,Hibernate 自己创建了一个列“MANAGER_EMPLOYEE_ID”,用于存储 FK。我指定了 JoinColumn 名称以使其成为“MANAGER_ID”。Hibernate 还创建了一个表 EMPLOYEE_EMPLOYED,但是数据并没有保存在那里。

方法 getReportees() 方法返回 null,而 getManager() 工作正常,正如预期的那样。

于 2010-03-12T14:51:27.350 回答
1

我不确定你是否真的想要,但我认为你想要每个类层次结构的表

在这种情况下,每个实体都按 DISCRIMINATOR_COLUMN 排序,如下所示

@Entity
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(
    name="EMPLOYEE_TYPE", 
    discriminatorType = DiscriminatorType.STRING
)
@DiscriminatorValue("EMPLOYEE")
public class Employee {

    @Id @GeneratedValue
    @Column(name="EMPLOYEE_ID") 
    private Integer id = null;

}

它的孩子是根据

@Entity
@DiscriminatorValue("MANAGER")
public class Manager extends Employee {

    // Manager properties goes here        
     ...
}

为了测试,让我们执行以下操作

SessionFactory sessionFactory = HibernateUtil.getSessionFactory();
Session session = sessionFactory.openSession();

/*
insert 
into
    Employee
    (EMPLOYEE_TYPE) 
values
    ('EMPLOYEE')
*/
session.save(new Employee());

/*
insert 
into
    Employee
    (EMPLOYEE_TYPE) 
values
    ('MANAGER')
*/
session.save(new Manager());

session.clear();
session.close();

但是,不是继承(由于多个实体共享同一个表,您可以看到很多 NULL 列 - 使用 InheritanceType.SINGLE_TABLE 策略时),您的模型会更好,如下所示

@Entity
public class Employee { 

    private Employee manager;
    private List<Employee> reporteeList = new ArrayList<Employee>();

    /**
    * optional=true
    * because of an Employee could not have a Manager
    * CEO, for instance, do not have a Manager
    */  
    @ManyToOne(optional=true)
    public Employee getManager() {
        return manager;
    }

    @OneToMany
    public List<Employee> getReporteeList() {
        return reporteeList;
    }

}

随意选择满足您需求的最佳方法。

问候,

于 2010-03-11T16:42:47.290 回答