当我们用HQL进行子查询的时候,如select * from Tree where pid in (select id from Tree,此时HIBERANTE就会报错,说什么*号错误之类的。但如果将*改为Tree类里的所有子段时就不会有问题了。就会像平时一样第一行数据返回一个Object[],然后你再根据Tree类里字段对Object[]数组里的值进行转换。这样一来比较麻烦。今天发现如果我SQL来查有一个方法可以返回一个对象的。
Configuration config = new Configuration().configure();
SessionFactory sf = config.buildSessionFactory();
Session session = sf.openSession();
Transaction ts = session.beginTransaction();
Query query = session.createSQLQuery("select * from Tree t where pid in (select id from Tree) ").addEntity(Tree.class); //返回对象
List list = query.list();
此时在遍历list时就可以(Tree)list.get[i];将每一行的内容变换为一个对象了。
另还可以返回一个Map对象,也就是说在在list里包含多个Map,代码如下
Query query = session.createSQLQuery("select id,name from Tree t where pid in (select id from Tree) ").setResultTransformer(Transformers.ALIAS_TO_ENTITY_MAP); //返回一个map,KEY:为DB中名称一致(大小写一致)遍历list时就可以
Map map = (Map)list.get[i];
map.get("id");map.get("name");来取值。按你的SQL语句select后的字段名来作为map的Key,但这个key必须与数据库中的字段名一模一样。
还可以用作函数方面的。如
Query query = session.createSQLQuery("select sum(id) SUMID from Tree t where pid in (select id from Tree)
.addScalar("SUMID",Hibernate.INTEGER) //转换类型,按DB中的type转
.setResultTransformer(Transformers.ALIAS_TO_ENTITY_MAP); //返回一个map,KEY:为DB中名称一致(大小写一致)
直接就map.get("SUMID")可以取值了
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