Question

我有一个Entity FooBar，它作为Foo和Bar实体的@ManyToMany连接表，包括一些附加信息。

@Entity
@Table(name = "foo_bar")
public class FooBar
{
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id", unique = true, nullable = false)
    protected Long id;

    @Column(name = "someInfo", nullable = true)
    private String someInfo;

    @ManyToOne(optional = false)
    private Foo foo;

    @ManyToOne(optional = false)
    private Bar bar;


    //getters, setters, and toString()
}

@Entity
@Table(name = "foo")
public class Foo
{
   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   @Column(name = "id", unique = true, nullable = false)
   protected Long id;

   @OneToMany(mappedBy = "foo", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
   private Set<FooBar> fooBars;

   //Foo has a number of other fields
   @Column(name = "orderIndex", nullable = false)
   private int orderIndex;

   @Column(name = "upgradeDirection", nullable = false)
   @Enumerated(EnumType.STRING)
   private Order direction;

   @ManyToOne(optional = false)
   private SomeEntity e;


   //getters, setters, and toString()
}


@Entity
@Table(name = "bar")
public class Bar
{
   @Id
   @GeneratedValue(strategy = GenerationType.AUTO)
   @Column(name = "id", unique = true, nullable = false)
   protected Long id;

   @OneToMany(mappedBy = "bar") //TODO specify a cascade and fetch attribute
   private Set<FooBar> fooBars;

   //Bar contains a number of other fields
   @Column(name = "value", nullable = false)
   private String value;


   //getters, setters, and toString()
}

当Hibernate创建表时，其中包含列＆＃39; id＆＃39;，＆＃39; someInfo＆＃39;，＆＃39; foo_id＆＃39;和＆＃39; bar_id＆＃39;。＆＃39; foo_id＆＃39;和＆＃39; bar_id＆＃39;用作复合键而不是使用＆＃39; id＆＃39;领域，任何想法为什么？

Hibernate创建Composite Key而不是使用Identifier

0 个答案: