mysql - HQL/SQL 根据计数选择前 10 条记录

Question

我有 2 张桌子：

CATEGORY (id)
POSTING (id, categoryId)

我正在尝试编写 HQL 或 SQL 查询来查找帖子数量最多的前 10 个类别。

帮助表示赞赏。

Answer 1

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
)

总部：

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/

Answer 2

在 SQL 中，您可以这样做：

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

Answer 3

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 Category 实体。

Answer 4

我知道这是一个老问题，但我得到了一个令人满意的答案。

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());