hibernate与spring整合的问题

Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as id0_0_, spring0_.username as username0_0_, spring0_.password as password0_0_ from spring spring0_ where spring0_.id=?
Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as id0_0_, spring0_.username as username0_0_, spring0_.password as password0_0_ from spring spring0_ where spring0_.id=?
Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as id0_0_, spring0_.username as username0_0_, spring0_.password as password0_0_ from spring spring0_ where spring0_.id=?

其中我只用了一次向数据库查询数据怎么这一句

Hibernate: select spring0_.id as col_0_0_ from spring spring0_ where spring0_.username=?

要执行两此
其中查询是在spring的validator中中进行的事物管理使用spring的声明式事物管理

而getUserByName()的定义如下

public class SpringDaoImpl extends HibernateDaoSupport implements SpringDao {

public Spring getUserByName(String name) {
		Session session = this.getSession();
		Query q = session.createQuery("from Spring s where s.username=?");
		q.setString(0, name);
		Spring user = null;
		if(q.iterate().hasNext())
		{
			user = (Spring) q.iterate().next();
		}
		return user;
	}
｝

而现在我有试着重新把这个方法再定义下，其改后如下

public Spring getUserByName(String name) {
		List list = this.getHibernateTemplate().findByNamedParam(
				"from Spring s where s.username=:name", "name", name);
		Iterator it = list.iterator();
		Spring user = null;
		if (it.hasNext()) {
			user = (Spring) it.next();
		}
		return user;
	}

这时我重新运行这个web程序 看下运行的结果，却发现这时没有hibernate所说的lazy-load(懒集合)，而是直接从database里把所有的数据加进来
其运行两次的结果如下：

Hibernate: select spring0_.id as id0_, spring0_.username as username0_, spring0_.password as password0_ from spring spring0_ where spring0_.username=?
Hibernate: select spring0_.id as id0_, spring0_.username as username0_, spring0_.password as password0_ from spring spring0_ where spring0_.username=?

这是否就意味着没有使用到hibernate提供的性能优化的功能呢?或者这背后还有更多的考虑呢？真是迷惑啊，还请各位指点。