我的问题与this question非常相似。
我从table1中选择了table2中field3和field4的所有匹配唯一组合的所有数据。
这是我剥离的SQL:
select *
from table1 as t1
where (t1.field1, t1.field2) in (select distinct field3, field4
from table2 as t2
where t2.id=12345);
我需要将我的SQL翻译成Hibernate Criteria。我让我的实体对象正确映射到表并将响应转换为正确的结果实体,但我无法正确翻译我的where子句。
我有什么
Criteria criteria = getSession().createCriteria(Table1.class);
DetachedCriteria subquery = DetachedCriteria.forClass(Table2.class);
ProjectionList projectionList = Projections.projectionList();
projectionList.add(Projections.property("field3"), "field3");
projectionList.add(Projections.property("field4"), "field4");
subquery.setProjection(Projections.distinct(projectionList));
subquery.add(Restrictions.eq("id", 12345));
我希望我的where子句类似于:
criteria.add(Subqueries.in("field1, field2", subquery));
但是Hibernate不允许这样做。
我试过推出where子句以获得两个子查询,并根据结果检查field1和field2,但看起来子查询总是必须返回多个列。我使用group by做了这个,但是Hibernate会自动将组中的列添加到投影列表中,我找不到删除它们的方法。
以下是使用group by的相同查询:
select *
from table1 as t1
where t1.field1 in (select field3
from table2 as t2
where t2.id=12345
group by field3, field4)
and t1.field2 in (select field4
from table2 as t2
where t2.id=12345
group by field3, field4);
是否可以使用Hibernate Criteria执行where子句?
如果无法使用Hibernate Criteria,是否可以使用HQL执行where子句?
修改
@ Larry.Z使用HQL回答我的问题。
我能用Hibernate Criteria解决我的问题,但我不得不将查询修改为:
select *
from table1 as t1
where exists (select 1
table2 as t2
where t2.id=12345
and t2.field3=t1.field1
and t2.field4=t1.field2);
转换为Hibernate标准:
Criteria criteria = getSession().createCriteria(Table1.class, "t1");
DetachedCriteria subquery = DetachedCriteria.forClass(Table2.class, "t2");
subquery.add(Restrictions.eq("t2.id", 12345));
subquery.add(Restrictions.eqProperty("t2.field3", "t1.field1"));
subquery.add(Restrictions.eqProperty("t2.field4", "t1.field2"));
subquery.setProjection(Projections.property("t2.id")); // select the ID rather than 1
我仍然很好奇是否可以使用原始SQL编写Hibernate Criteria。
答案 0 :(得分:5)
尝试像这样编写HQL查询
String hql = "from Table1 t1 where (t1.field1, t1.field2) in (
select distinct t2.field3, t2.field4
from Table2 t2
where t2.id=12345)";
sessionFactory.getCurrentSession().createQuery(hql).list()
答案 1 :(得分:4)
Subqueries.propertiesIn就是您所需要的:
criteria.add(Subqueries.propertiesIn(
new String[] { "field1", "field2" },
detachedCriteria));