Question

我们说我有以下两个班级：

@Entity
class Article {

    @Id
    int id;

    @ManyToMany(mappedBy="articles")
    List<Tag> tags;
}

@Entity
@NamedQueries({
    @NamedQuery(name = "Tag.trends", query = "SELECT t FROM Tag t") // Some magic stuff here..
})
class Tag {

    @Id
    String name;

    @ManyToMany
    List<Articles> articles;
}

现在我想创建一个命名查询来选择最常用的标签。

因此，这需要加入，计数和订单。但我似乎无法解决这个问题。

问题： 如何选择最常用的标签？

例如，如果以下文章具有以下标记：

Article: Tags
0: A, B
1: A
2: C
3: A, C

输出

A // 3 times
C // 2 times
B // 1 time

Answer 1

您可以使用

之类的东西来实现您想要的效果

SELECT t FROM Tag t ORDER BY SIZE(t.articles) ASC

这将为您提供按文章数量排序的标签列表。

Answer 2

在这些问题上向后退一步。从SQL查询开始：

select 
  count(t.name), t.name 
from article a 
  join article_tags ats 
  join tag t 
where 
  a.id=ats.article_id 
  and ats.tag_id = t.id 
group by 
  t.name 
order by 
  count(t.name) 
desc;

然后JPQL应该或多或少地遵循：

List<Tuple> t = em.createQuery("select t.name as name, count(t.name) as count from Article a join a.tags t group by (t.name) order by count(t.name) desc", Tuple.class).getResultList();
t.stream().forEach((tp)-> {
    System.out.println("Count of " + tp.get("name") + " = " + tp.get("count"));
});

这给了我以下输出：

Hibernate: select tag2_.name as col_0_0_, count(tag2_.name) as col_1_0_ from Article article0_ inner join Tag_Article tags1_ on article0_.id=tags1_.articles_id inner join Tag tag2_ on tags1_.tags_id=tag2_.id group by tag2_.name order by count(tag2_.name) desc
Count of A = 2
Count of B = 1
Count of C = 1

更新：如果你想要一个标签列表，只需将元组的第一个元素更改为标签。当然，如果你愿意，你可以忽略计数：

List<Tag> t = em.createQuery("select t from Article a join a.tags t group by (t.name) order by count(t.name) desc", Tag.class).getResultList();
t.stream().forEach((tag)-> {
    System.out.println("Tag = " + tag.getName());
});

获得以下输出：

Hibernate: select tag2_.id as id1_1_, tag2_.name as name2_1_ from Article article0_ inner join Tag_Article tags1_ on article0_.id=tags1_.articles_id inner join Tag tag2_ on tags1_.tags_id=tag2_.id group by tag2_.name order by count(tag2_.name) desc
Tag = A
Tag = B
Tag = C

最常用标签的NamedQuery

2 个答案: