2011-12-20 61 views
10

我有一个在EJB中设计ManyToMany的问题,一个连接表怎么可以有一个属性?
下面是一个例子,学生和课程是ManyToMany,每个学生都有很多课程,许多学生选择一门课程。JPA ManyToMany,JoinTable如何拥有一个属性?

@Entity 
    public class Student implements Serializable { 
     @Id 
     @GeneratedValue(strategy = GenerationType.AUTO) 
     Long id; 
     String name; 
     private Collection<Course> courses; 

     @ManyToMany(mappedBy = "students",cascade=CascadeType.ALL)  
     public Collection<Course> getCourses() { 
      return this.courses; 
     } 

     public void setCourses(Collection<Course> courses) { 
      this.courses = courses; 
     } 

    } 


    @Entity 
    public class Course implements Serializable { 
     @Id 
     @GeneratedValue(strategy = GenerationType.AUTO) 
     Long id; 
     String name; 
     private Collection<Student> students; 

     @ManyToMany(cascade=CascadeType.ALL) 
     @JoinTable(name = "Student_Course", 
     joinColumns = {@JoinColumn(name = "Course_ID", referencedColumnName = "id")}, 
     inverseJoinColumns = {@JoinColumn(name = "Student_ID", referencedColumnName = "id")}) 

     public Collection<Student> getStudents() { 
      return this.students; 
     } 

     public void setStudents(Collection<Student> students) { 
      this.students = students; 
     } 
    } 

但是,如果我在JoinTable中有一个属性,例如每个学生在一个课程中都有一个分数。如何使用ManyToMany在EJB中创建它?
非常感谢您的关注!

+1

如果你的关系有属性,那么你应该把它建模为实体。这个问题几乎是一样的:http://stackoverflow.com/questions/7602386/mapping-value-in-junction-table-to-entity/7603036#7603036即使代表关系的实体名称(CourseAssignment)也很适合你案件。 – 2011-12-21 08:16:57

+3

这是不可能的,你不能添加属性关系。如果您需要访问连接表中的属性,那么该属性属于某个实体,因此您需要第三个实体。 – 2011-12-21 09:16:21

+0

对不起,我纠正了这一点。我还有一个问题,当我使用@Embeddable类作为连接表的PK时,它不起作用(编译失败),有人说JPA2.0不支持它!请看看[jpa-eclipselink-manytomany-with-dat](http://stackoverflow.com/questions/4013397/jpa-eclipselink-manytomany-with-data),有什么办法可以使它工作吗? – seaguest 2011-12-21 10:47:45

回答

4

这是不可能的,你不能添加属性关系。如果您需要访问连接表中的属性,那么该属性属于某个实体,因此您需要第三个实体。

3

这是可能的。

您只需要通过第三个实体将一对多和多对一映射的显式组合替换为多对多映射,这将代表两个主要实体(学生和当然在你的例子)。

请仔细阅读细节here

0

实体多对多的关系(商家和服务)。 这可以使用第三个实体如下来实现: -

@Entity 
@Table(name = "merchant") 
public class Merchant implements java.io.Serializable { 

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "pk.merchant",targetEntity = MerchantService.class) 
    private Set<MerchantService> merchantServices = new HashSet<>(); 
} 

@Entity 
@Table(name = "merchant_service") 
@AssociationOverrides({ 
     @AssociationOverride(name = "pk.merchant", 
      joinColumns = @JoinColumn(name = "merchant_id")), 
     @AssociationOverride(name = "pk.service", 
      joinColumns = @JoinColumn(name = "service_id")) }) 
public class MerchantService implements java.io.Serializable { 

    @EmbeddedId 
    private MerchantServiceId pk = new MerchantServiceId(); 

    private boolean isActive; 

    public MerchantServiceId getPk() { 
     return pk; 
    } 

    public void setPk(MerchantServiceId pk) { 
     this.pk = pk; 
    } 

    @Transient 
    public Service getService() { 
     return getPk().getService(); 
    } 


    @Transient 
    public Merchant getMerchant() { 
     return getPk().getMerchant(); 
    } 


    public boolean isActive() { 
     return isActive; 
    } 

    public void setActive(boolean isActive) { 
     this.isActive = isActive; 
    } 

} 

@Embeddable 
public class MerchantServiceId implements java.io.Serializable { 

    private Merchant merchant; 
    private Service service; 

    @ManyToOne 
    public Merchant getMerchant() { 
     return merchant; 
    } 

    @ManyToOne 
    public Service getService() { 
     return service; 
    } 

} 

@Entity 
@Table(name = "service") 
public class Service implements java.io.Serializable { 

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "pk.service",targetEntity = MerchantService.class) 
    private Set<MerchantService> merchantServices = new HashSet<>(); 

}