@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=?
)
最佳答案
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();
关于java - 标准多对多 Hibernate,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3711374/