Sql istinct)-功能
我用嵌套聚合函数解决了这个问题:Sql istinct)-功能,sql,hibernate,jpa,hql,jql,Sql,Hibernate,Jpa,Hql,Jql,我用嵌套聚合函数解决了这个问题: select sum(count(distinct u.lastname)) from Personel u group by u.firstname, u.lastname; “u.lastname”必须是非空列 此语句在带有Oracle DB的HQL中工作。如果结果计数为100000条记录,则此解决方案正常且性能良好?您只需要计数,还是在某处显示查询结果?我没有尝试过,但是用这种风格修改了上面的代码,并看到了结果。getMaxResults();我显示分
select sum(count(distinct u.lastname))
from Personel u
group by u.firstname, u.lastname;
“u.lastname”必须是非空列
此语句在带有Oracle DB的HQL中工作。如果结果计数为100000条记录,则此解决方案正常且性能良好?您只需要计数,还是在某处显示查询结果?我没有尝试过,但是用这种风格修改了上面的代码,并看到了结果。getMaxResults();我显示分页结果。例如,pagesize=10,我显示pagenumber=2,所有结果(按组)都是100000这是非常低效的这不是hql查询。在子选择中引用hql不支持hql。请再次检查
select count(*)
from Personel u
where u.lastname='azizkhani'
select u.lastname,count(*)
from Personel u
group by u.lastname;
select count(*)
from (
select u.lastname,count(*)
from tbl_personel u
group by u.lastname
)
public <U> PagingResult<U> getAllGrid(String hql,Map<String, Object> params,PagingRequest searchOption);
String hqlQuery = " select e from Personel e where 1<>2 and e.lastname=:lastname";
HashMap<String, Object> params = new HashMap<String, Object>();
params.put("lastname", 'azizkhani');
return getAllGrid(hqlQuery, params, new PagingRequest( 0/*page*/, 10 /*size*/) );
public class PagingResult<T> {
private int totalElements;
@JsonProperty("rows")
private List<T> items;
public PagingResult() {
}
public PagingResult(int totalElements, List<T> items) {
super();
this.totalElements = totalElements;
this.items = items;
}
public int getTotalElements() {
return totalElements;
}
public void setTotalElements(int totalElements) {
this.totalElements = totalElements;
}
public List<T> getItems() {
return items;
}
public void setItems(List<T> items) {
this.items = items;
}
}
public <U> PagingResult<U> getAllGrid(String hql, Map<String, Object> params, PagingRequest searchOption) {
Session session = getSession();
applyDafaultAuthorizeFilter(session);
Query query = session.createQuery(hql);
if (searchOption != null) {
if (searchOption.getSize() > 0) {
query.setFirstResult(searchOption.getPage() * searchOption.getSize());
query.setMaxResults(searchOption.getSize());
}
}
if (params != null)
HQLUtility.setQueryParameters(query, params);
List<U> list = query.getResultList();
Query countQuery = session.createQuery("select count(*) " + HQLUtility.retriveCountQueryFromHql(hql));
if (params != null)
HQLUtility.setQueryParameters(countQuery, params);
int count = ((Long) countQuery.uniqueResult()).intValue();
if (searchOption != null)
return new PagingResult<U>(searchOption.getPage(), count, searchOption.getSize(), list);
else
return new PagingResult<U>(0, count, 0, list);
}
public static StringBuffer retriveCountQueryFromHql(StringBuffer jql) {
if(jql.indexOf("order by")>=0)
jql.replace(jql.indexOf("order by"), jql.length(),"");
String mainQuery = jql.toString();
jql = new StringBuffer(jql.toString().replace('\t', ' '));
int firstIndexPBas = jql.indexOf(")");
int firstIndexPBaz = jql.lastIndexOf("(", firstIndexPBas);
while (firstIndexPBas > 0) {
for (int i = firstIndexPBaz; i < firstIndexPBas + 1; i++)
jql.replace(i, i + 1, "*");
firstIndexPBas = jql.indexOf(")");
firstIndexPBaz = jql.lastIndexOf("(", firstIndexPBas);
}
int Indexfrom = jql.indexOf(" from ");
return new StringBuffer(" " + mainQuery.substring(Indexfrom, jql.length()));
}
public void applyDafaultAuthorizeFilter(Session session) {
Filter filter = session.enableFilter("defaultFilter");
filter.setParameter("userId", SecurityUtility.getAuthenticatedUserId());
filter.setParameter("orgId", SecurityUtility.getAuthenticatedUserOrganization().getId());
}
Query query = getEntityManager().createQuery("select u.lastname,count(*) from Personel u group by u.lastname;");
List<YourEntity> list = query.getResultList();
String hql = "select count(*) from (select u.lastname,count(*) from tbl_personel u group by u.lastname)";
Query query = session.createQuery(hql);
List listResult = query.list();
select u.lastname, count(u)
from Personel u
group by u.lastname
select count(*)
from (
select u.lastname, count(*)
from tbl_personel u
group by u.lastname
) as tbl_aggr
select u.firstname, u.lastname
from Personel u
group by u.firstname, u.lastname;
select sum(count(distinct u.lastname))
from Personel u
group by u.firstname, u.lastname;