Hibernate使用distinct返回不重复的数据，使用group by 进行分组

distinct使用

 
public list<String> distinctDutyDate() {
        String hql="select distinct(dutyDate) from TestInfo";
        Query query=getSession().createQuery(hql);
        List list= query.list();
        Iterator it= list.iterator();
        List<String> list1=new ArrayList<String>();
        while(it.hasNext()){
            String dutyDate=it.next()+"";
            list1.add(dutyDate);
        }
        return list1;
    }
 

group by使用

public List<YearMonthDTO> getYearMonthByUserId(Integer userId, String submitType) {
        String hql="select submitYear,submitMonth from TotalInfoSubmit
              where userId=:userId and submitType=:submitType group by submitYear,submitMonth ";
        Query query = getSession().createQuery(hql)
                .setParameter("userId",userId)
                .setParameter("submitType",submitType);
        List list= query.list();
        Iterator it= list.iterator();
        List<YearMonthDTO> list1=new ArrayList<>();
        while(it.hasNext()){
            Object[] res=(Object[]) it.next();
            YearMonthDTO dto=new YearMonthDTO();
            String year=res[0]+"";
            String month=res[1]+"";
            dto.setYear(year);
            dto.setMonth(month);
            list1.add(dto);
        }
        return list1;
    }