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Java 联接表的休眠条件_Java_Hibernate_Many To Many - Fatal编程技术网

Java 联接表的休眠条件

Java 联接表的休眠条件,java,hibernate,many-to-many,Java,Hibernate,Many To Many,帮我冬眠大师。。 我有两个关系类,我们称之为A类和B类 @Entity @Table(name="A") public class A extends Serializable{ @Id @Column(name="a_id") private int id; @Column(name="a_name") private String name; /* *.....Setter and Getter

帮我冬眠大师。。 我有两个关系类,我们称之为A类和B类

@Entity
@Table(name="A")
public class A extends Serializable{

      @Id
      @Column(name="a_id")
      private int id;
      @Column(name="a_name")
      private String name;

     /*
      *.....Setter and Getter
      */
}



@Entity
@Table(name="B")
public class B extends Serializable{

      @Id
      @Column(name="b_id")
      private int id;
      @ManyToMany(
          fetch= FetchType.EAGER,
          targetEntity=package.A.class,
          cascade={CascadeType.ALL}
      )
      @JoinTable(
          name="B_A",
          joinColumns=@JoinColumn(name="b_id"),
          inverseJoinColumns=@JoinColumn(name="a_id")
      )
      @Fetch(FetchMode.SUBSELECT)
      private List<A> list;

     /*
      *.....Setter and Getter
      */
}
那么我必须如何创建标准代码呢????对于预期的列表结果列表,我想使用Transformer

谢谢

试试这个:

criteria.createCriteria(A.class)
.createCriteria("id", "join_between_a_b")
.add(Restrictions.eq("some_field_of_A", someValue));

记住,A和B必须有
id
作为它们的标识符才能使它们加入。

我所说的
A的某个\u字段
是指你在A类中喜欢的任何东西,比如
name
。您可以对这些类的任何属性进行限制。

您尝试了什么?你看过文件了吗?
criteria.createCriteria(A.class)
.createCriteria("id", "join_between_a_b")
.add(Restrictions.eq("some_field_of_A", someValue));