Question

到目前为止，我有以下查询，它可以完美地根据我们的内部分数对书籍进行排名。

UPDATE CER
SET CER.book_rank = Ranker.ranc
FROM book_ranks CER
INNER JOIN
(SELECT Rank() over (Order by book_score desc, book_id) as ranc, book_id
FROM book_ranks
WHERE Category = 'Fiction'
GROUP BY book_id, book_score 
) Ranker
ON
CER.book_id = Ranker.book_id

代码工作正常，但没有处理案例。

输入：

bookName  book_score
--------  ----------
book2     45
book3     35
book5     35
book7     35
book9     30

当前输出：

bookName  book_score  book_rank
--------  ----------  ---------
book2     45          1
book3     35          2
book5     35          3
book7     35          4
book9     30          5

必需输出：

bookName  book_score  book_rank
--------  ----------  ---------
book2     45          1
book3     35          2
book5     35          2
book7     35          2
book9     30          5

作为book3，book5，book7具有相同的分数，它们的等级应该相同，book9应该具有5的等级。

Answer 1

尝试：

SELECT * FROM book_ranks CER
INNER JOIN
(SELECT Rank() over (Order by book_score desc  ) 
  -- only difference is here ----------------^^
  as ranc, book_id
  FROM book_ranks
  WHERE Category = 'Fiction'
GROUP BY book_id, book_score 
) Ranker
ON CER.book_id = Ranker.book_id;

Answer 2

而不是

SELECT Rank() over (Order by book_score desc, book_id)

你应该使用

SELECT Rank() over (Order by book_score desc)

这可以解决您的问题。

排列有关系的书籍

2 个答案: