我有两张桌子:
CATEGORY (id)
POSTING (id, categoryId)
我正在尝试编写HQL或SQL查询来查找排名最多的前10个类别。
非常感谢帮助。
答案 0 :(得分:5)
SQL查询:
SELECT c.Id, sub.POSTINGCOUNT
FROM CATEGORY c where c.Id IN
(
SELECT TOP 10 p.categoryId
FROM POSTING p
GROUP BY p.categoryId
order by count(1) desc
)
HQL:
Session.CreateQuery("select c.Id
FROM CATEGORY c where c.Id IN
(
SELECT p.categoryId
FROM POSTING p
GROUP BY p.categoryId
order by count(1) desc
)").SetMaxResults(10).List();
http://sqlinthewild.co.za/index.php/2010/01/12/in-vs-inner-join/
答案 1 :(得分:1)
在SQL中你可以这样做:
SELECT c.Id, sub.POSTINGCOUNT
FROM CATEGORY c
INNER JOIN
(
SELECT p.categoryId, COUNT(id) AS 'POSTINGCOUNT'
FROM POSTING p
GROUP BY p.categoryId
) sub ON c.Id = sub.categoryId
ORDER BY POSTINGCOUNT DESC
LIMIT 10
答案 2 :(得分:0)
SQL可以是:
SELECT c.* from CATEGORY c, (SELECT count(id) as postings_count,categoryId
FROM POSTING
GROUP BY categoryId ORDER BY postings_count
LIMIT 10) d where c.id=d.categoryId
此输出可以映射到类别实体。
答案 3 :(得分:0)
我知道这是一个老问题,但我得到了满意的答案。
JPQL:
//the join clause is necessary, because you cannot use p.category in group by clause directly
@NamedQuery(name="Category.topN",
query="select c, count(p.id) as uses
from Posting p
join p.category c
group by c order by uses desc ")
爪哇:
List<Object[]> list = getEntityManager().createNamedQuery("Category.topN", Object[].class)
.setMaxResults(10)
.getResultList();
//here we must made a conversion, because the JPA cannot order using a non select field (used stream API, but you can do it in old way)
List<Category> cats = list.stream().map(oa -> (Category) oa[0]).collect(Collectors.toList());