Question

我有（简化）以下模型：

Book
  id
  name

BookCategory
  book_id
  category_id
  rank

Category
  id
  name

对于给定的类别ID，我希望将该类别的图书列为排名最高的图书。

我将举一个例子来说明它：

Book   

id    name   
---  -------
1     On Writing
2     Zen teachings
3     Siddharta

BookCategory   

book_id category_id      rank   
---       -------        -----
1        2               34.32
1        5               24.23
1        9               54.65
2        5               27.33
2        9               28.32
3        2               30.43
3        5               27.87

Category   

id    name   
---  -------
2     Writing
5     Spiritual
9     Buddism

category_id = 2的结果是id = 3的书。

这是我正在运行的查询：

SELECT book."name" AS bookname
FROM bookcategory AS bookcat
LEFT JOIN book ON bookcat."book_id" = book."id" 
LEFT JOIN category cat ON bookcat."category_id" = cat."id" 
WHERE cat."id" = 2
ORDER BY bookcat."rank"

这不是正确的方法，因为它没有选择每本书的最高等级。我还没有找到合适的解决方案。

注意：我正在使用postgresql 9.1版本。

编辑：

DB Schema（取自马丁的SQL小提琴答案）：

create table Book (
  id int,
  name varchar(16)
  );

insert into Book values(1, 'On Writing');
insert into Book values(2, 'Zen teachings');
insert into Book values(3, 'Siddharta');

create table BookCategory (
  book_id int,
  category_id int,
  rank real
  );

insert into BookCategory values(1,2,34.32);
insert into BookCategory values(1,5,24.23);
insert into BookCategory values(1,9,54.65);
insert into BookCategory values(2,5,27.33);
insert into BookCategory values(2,9,28.32);
insert into BookCategory values(3,2,30.43);
insert into BookCategory values(3,5,27.87);

create table Category (
  id int,
  name varchar(16)
  );

insert into Category values(2, 'Writing');
insert into Category values(5,'Spiritual');
insert into Category values(9,    'Buddism');

Answer 1

添加另一列来计算排名：

dense_rank() OVER (PARTITION BY book."name" ORDER BY bookcat."rank"
s ASC) AS rank

Answer 2

这是你想要的吗？

SELECT 
  book.name, mx.max_rank
FROM
  (SELECT 
     max(rank) AS max_rank , book_id 
   FROM BookCategory WHERE category_id = 2 
   GROUP BY 
     book_id
  ) mx
JOIN Book ON 
  mx.book_id = Book.id

如果我理解你的问题，你需要获得BookCategory中每本书的给定类别的最大值（这是内部选择所做的），然后简单地将它加入book_id上的Book表。

整个示例位于SQL Fiddle

编辑：

我看到已经有一个已接受的答案，但为了完整起见，这是我在澄清问题后的答案：

SELECT 
  Book.name 
FROM
  (SELECT max(rank) AS max_rank, book_id AS bid
   FROM BookCategory GROUP BY book_id
  ) mx
JOIN BookCategory ON
  rank = max_rank
  AND book_id = bid
JOIN Book
  ON book_id = Book.id
WHERE category_id = 2

在SQL Fiddle。

Answer 3

设置：

CREATE TABLE Book
(
  id int PRIMARY KEY,
  name text not null
);

CREATE TABLE Category
(
  id int PRIMARY KEY,
  name text not null
);

CREATE TABLE BookCategory
(
  book_id int,
  category_id int,
  rank numeric not null,
  primary key (book_id, category_id)
);

INSERT INTO Book VALUES
  (1, 'On Writing'),
  (2, 'Zen teachings'),
  (3, 'Siddharta');

INSERT INTO Category VALUES
  (2, 'Writing'),
  (5, 'Spiritual'),
  (9, 'Buddism');

INSERT INTO BookCategory VALUES
  (1, 2, 34.32),
  (1, 5, 24.23),
  (1, 9, 54.65),
  (2, 5, 27.33),
  (2, 9, 28.32),
  (3, 2, 30.43),
  (3, 5, 27.87);

解决方案：

SELECT Book.name
  FROM (
         SELECT DISTINCT ON (book_id)
             *
           FROM BookCategory
           ORDER BY book_id, rank DESC
       ) t
  JOIN Book ON Book.id = t.book_id
  WHERE t.category_id = 2
  ORDER BY t.rank;

逻辑上，FROM子句中的子查询会为每本书生成与排名最高类别的关系，然后您可以从中选择该类别中的书籍，并按该类别中的排名进行排序。

结果：

   name    
-----------
 Siddharta
(1 row)

查询以获得每个项目的最高排名

3 个答案: