加入无关儿童的Hibernate Criteria？-Java 学习之路

我有三个实体，一个父母有两个孩子，但我在这里只列出了两个实体 .

@Entity
@Table(name = "person")
public class Person{
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", unique = true, nullable = false) 
    private Long id;

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

    @OneToMany(fetch=FetchType.EAGER)
    private Set<Phone> phones = new HashSet<Phone>();


    @OneToMany(fetch=FetchType.EAGER)
    @JoinColumn(name="person_id", referencedColumnName="id")        
    private List<Book> books = new ArrayList<Book>();

}


@Entity
@Table(name = "book")   
public class Book{
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", unique = true, nullable = false) 
    private Long id;

    @Column(name="person_id")
    private Long personId;

}

当我按照Hibernate标准运行并注意到SQL生成的日志时，它正在加入两个孩子 - 电话和书籍 . 我和这两个孩子没有任何关系 . 为什么从Hibernate标准生成的SQL查询尝试加入无关的两个孩子？我希望父母和孩子之间能够加入，而不是孩子之间 .

Criteria criteria = session.createCriteria(Person.class)
                .add(Restrictions.in("id", ids))
                .setFetchMode("phones", FetchMode.JOIN).setFetchMode("books", FetchMode.JOIN)
                .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);               
        result = criteria.list();

1 回答

如果要使用条件，则需要为每个关系发出两个条件：

result = session.createCriteria(Person.class)
                .add(Restrictions.in("id", ids))
                .setFetchMode("phones", FetchMode.JOIN)
                .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list();

   and

result =         session.createCriteria(Person.class)
                .add(Restrictions.in("id", ids))
                .setFetchMode("books", FetchMode.JOIN)
                .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list();

否则你可以使用hql而不是标准，并具有：

String hql=  "select new list(person.phones, person.books) from Person person";
session.createQuery(hql).list();

EDIT:

String hqlQuery = "select p from Person p left outer join p.phones as ph left outer join p.books as b";

回复于 2024-04-28T18:49:46+08:00

加入无关儿童的Hibernate Criteria？

1 回答

相关问题