hibernate查询时Criteria怎么根据级联的两个类里的属性进行查询？
0
悬赏园豆：20
[已解决问题]
解决于 2015-11-12 09:35
有3个表，A，B，C，
关系如下：
A-->List<B>，B属性有time等;
A-->List<C>，C属性有time等
Java代码：
Entity：
@OneToMany(fetch = FetchType.EAGER)
@JoinColumn(name = "axxx_id")
@Fetch(FetchMode.SELECT)
public List<B> getB() {
return b;
}
public void setB(List<B> b) {
this.b = b;
}
@OneToMany(fetch = FetchType.EAGER)
@JoinColumn(name = "axxx_id")
@Fetch(FetchMode.SELECT)
public List<C> getC() {
return c;
}
public void setC(List<C> c) {
this.c = c;
}
Server:
Criteria countCriteria = entityDao.getCriteria();
Criteria listCriteria = entityDao.getCriteria();
...
//*********************start***************************
//加条件 当类为一时可以这样
{
countCriteria.createAlias("B", "b");
listCriteria.createAlias("B", "b");
countCriteria.add(Restrictions.le("b.time", time));
listCriteria.add(Restrictions.le("b.time", time));
}
//********************end***************************
//现在类为一个List型的，该怎么加条件??
listCriteria.setFirstResult((pageNo-1)*pageSize);
listCriteria.setMaxResults(pageSize);
List<A> items = listCriteria.list();
hibernate查询时Criteria怎么根据两个time进行查询？
最佳答案
0
问题已解决：
底层代码：
@Autowired @Qualifier("sessionFactory") protected SessionFactory sessionFactory; protected Session getSession() { return sessionFactory.getCurrentSession(); } @Override public Criteria getExecutableCriteria(){ DetachedCriteria dc = DetachedCriteria .forClass(this.entityClass); return dc.getExecutableCriteria(getSession()); }
service调用代码：
Criteria listCriteria = entityDao.getExecutableCriteria(); listCriteria.createAlias("A","a",JoinType.LEFT_OUTER_JOIN); listCriteria.createAlias("B","b",JoinType.LEFT_OUTER_JOIN); listCriteria.add(Restrictions.or(Restrictions.lt("a.time", time, MatchMode.ANYWHERE),Restrictions.lt("b.time", time, MatchMode.ANYWHERE))); ... ...
