Question

我有3张桌子：

publisher (id serial pkey, pub_name text)
book (id serial pkey, id_pub INTEGER NOT NULL REFERENCES publisher (id), book_name TEXT)
rank (id SERIAL PKEY, id_book INTEGER NOT NULL REFERENCES book (id), week DATE, selling INTEGER)

并且只想为每个发布商列出畅销书。我现在所拥有的东西归还所有书籍的销售：

SELECT publisher.pub_name, book.book_name, SUM(rank.selling)
 FROM rank, publisher, book
 WHERE rank.id_book = book.id AND book.id_pub = publisher.id
 GROUP BY publisher.pub_name, book.book_name;

这样返回：

BigPublisher, Book1, 100300
GoodPublisher, BeBook, 10003
BigPublisher, Book2, 50200
OldPublisher, N-Book, 20009
GoodPublisher, CeBook, 4000
GoodPublisher, DeBook, 3001

我正在寻找那种结果：

BigPublisher, Book1, 100300
GoodPublisher, BeBook, 10003
OldPublisher, N-Book, 20009

Answer 1

使用窗口功能。请参阅the window functions tutorial in the PostgreSQL documentation。

您正在寻找以下内容：

SELECT
  rank() OVER (PARTITION BY publisher ORDER BY sum_of_sold DESC) AS pos
-- ... FROM clause etc here ...
WHERE pos = 1;

Answer 2

select
    distinct on (pub_name),
    pub_name,
    book_name,
    selling
from (
    select
        publisher.pub_name,
        book.book_name,
        sum(rank.selling) selling
    from
        publisher
        inner join 
        book on book.id_pub = publisher.id
        inner join
        rank on rank.id_book = book.id
    group by 1, 2
) s
order by pub_name, s.selling desc

Postgres SELECT GROUP BY出版商最畅销的商品

2 个答案: