3

我有（简化的）以下模型：

Book
  id
  name

BookCategory
  book_id
  category_id
  rank

Category
  id
  name

使用给定的类别 ID，我希望将具有该类别的书籍作为排名最高的书籍。

我将举一个例子来更清楚地说明它：

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 的结果将是 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排名。我还没有找到合适的解决方案。

注意：我使用的是 postgresql 9.1 版本。

编辑：

DB Schema（取自马丁的 SQL Fiddle 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');
4

3 回答 3

5

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
于 2012-08-08T15:41:07.290 回答
1

建立：

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;

从逻辑上讲，FROM子句中的子查询为每本书生成一个与排名最高的类别的关系，然后您从中选择该类别中的书籍并按该类别中的排名对它们进行排序。

结果：

   姓名    
------------
 悉达多
(1 行)
于 2012-08-08T17:36:33.527 回答
1

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

如果我正确理解您的问题，您需要为 BookCategory 中的每本书获取给定类别的最大值（这是内部选择所做的），然后只需将其加入 book_id 上的 Book 表。

整个例子在SQL Fiddle

编辑：

我看到已经有一个公认的答案，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 小提琴上。

于 2012-08-08T15:59:43.963 回答