3
@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;
}
可以說我有語言工程胚芽,誰講英人,人誰講德語,和人民誰講英,德 我想所有誰講人皆有之英語和德語使用標準。標準多到很多Hibernate
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=?
)
這將列出所有講工程的人,以及所有講GER人,以及所有講這兩種語言的人,我需要的是一個標準,只選擇後來講英語和德語的人。 – Darwly 2010-09-15 11:36:04
啊..比你需要而不是OR:criteria.add(Restrictions.and(languageEN,languageDE)); – Ice 2010-09-15 11:40:13
即使不工作,生病得到sql – Darwly 2010-09-15 11:47:02