标准多对多休眠

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

从可以访问会话对象的 DAO 对象（可能是扩展 HibernateDaoSupport 的对象）：

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();