java - 标准多对多 Hibernate

Question

@Entity
public class Person implements Serializable {
    private int id;
           ...........
    private Set<Languages> languages = new HashSet<Languages>();
       ...............
    @ManyToMany
    @JoinTable(name = "link_person_languages")
    public Set<Languages> getLanguages() {
       return languages;
    }
}

@Entity
public class Languages implements Serializable {
    private int id;
    private String name;
    @Id
    @GeneratedValue
    public int getId() {
        return id;
    }
    @Column(nullable = false, length = 40, unique = true)
    public String getName() {
        return name;
    }

假设我有语言 Eng Germ、说英语的人、说德语的人以及说英语和德语的人我想让所有会说英语和德语的人都使用 Criteria。

crit.createAlias("languages", "l");
Conjunction con = Restrictions.conjunction();
 for (int j = 0; j < o.length; j++) {
          Criterion tmp = Restrictions.eq("l.id", ((Languages)o[j]).getId());
         con.add(tmp);
 }
 crit.add(con);


 select
 this_.id as y0_,
    this_.lastName as y1_,
    this_.firstName as y2_,
    this_.socialNumber as y3_
from
    Person this_ 
inner join
    link_person_languages languages3_ 
        on this_.id=languages3_.Person_id 
inner join
    Languages l1_ 
        on languages3_.languages_id=l1_.id 
where
    (
        l1_.id=? 
        and l1_.id=?
    )

Answer 1

从可以访问 session 对象(可能是扩展 HibernateDaoSupport 的对象)的 DAO 对象中:

Criteria criteria = getSession().createCriteria(Person.class);
criteria = criteria.createCriteria("languages");

Criterion languageEN = Restrictions.eq("name", "en");
Criterion languageDE = Restrictions.eq("name", "de");
criteria.add(Restrictions.and(languageEN, languageDE));

List<Person> result = criteria.list();