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 DoctorDuty";
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 TotalBranchSubmit 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;
}
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