Question

我有Release表（id，manga_id，chapterization_id）和章节表（id，chapter）。

我们假设我有

推出

id | manga_id | chapterization_id
--------------------------------
1  | 44       | 333
2  | 55       | 222
3  | 66       | 777
4  | 44       | 888
5  | 44       | 999

Chapterization

id  | chapter
------------
111 | 60
222 | 70
333 | 80
444 | 90
777 | 100
888 | 120
999 | 110

我想获取每个manga_id的最大（最新）章节。漫画44有3个章节，但我想获取它的最新章节，即120

所以最终结果将是id为[2,3,4]的版本知道如何在PostgreSQL中编写这样的查询吗？

Answer 1

尝试 -

select r.manga_id, max(c.chapter)
  from release r
  join chapterization c
    on r.chapterization_id = c.id
 group by r.manga_id

根据您的评论，如果您还想要r.id，您可以使用以下任一方法。首先是带内联视图的连接。第二个是子查询。两者都应该有相同的结果，并给你你想要的。

select r.manga_id, r.id, c.chapter
  from release r
  join chapterization c
    on r.chapterization_id = c.id
  join (select r.manga_id, max(c.chapter) as max_chapter
          from release r
          join chapterization c
            on r.chapterization_id = c.id
         group by r.manga_id) x
    on r.manga_id = x.manga_id
   and c.chapter = x.max_chapter

或

select r.manga_id, r.id, c.chapter
  from release r
  join chapterization c
    on r.chapterization_id = c.id
 where c.chapter = (select max(c2.chapter)
                      from release r2
                      join chapterization c2
                        on r2.chapterization_id = c2.id
                     where r2.manga_id = r.manga_id)

Answer 2

您可以使用join和distinct on子句执行此操作：

select distinct on (r.manga_id) r.manga_id, c.id, c.chapter
from release r join
     chapterization c
     on r.chapterization_id = c.id
order by r.manga_id, c.chapter desc;

PostgreSQL：DISTINCT ON另一个表的MAX表

2 个答案: