题目4：MySQL----------Rank Scores

最新推荐文章于 2025-01-21 21:04:43 发布

原创最新推荐文章于 2025-01-21 21:04:43 发布 · 1.5k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#database #MySQL #Rank Scores

数据库MySQL 专栏收录该内容

18 篇文章

订阅专栏

本文介绍如何使用SQL查询实现动态排名，解决分数相同情况下的并列排名问题，确保排名序列连续无空缺。

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

For example, given the above Scores table, your query should generate the following report (order by highest score):

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

题目解答：

一、

SELECT
  Score,
  (SELECT count(*) FROM (SELECT distinct Score s FROM Scores) tem WHERE s >= Score) Rank
FROM Scores
ORDER BY Score desc;

二、

# Write your MySQL query statement below
SELECT Scores.Score, COUNT(Ranking.Score) AS RANK
  FROM Scores
     , (
       SELECT DISTINCT Score
         FROM Scores
       ) Ranking
 WHERE Scores.Score <= Ranking.Score
 GROUP BY Scores.Id, Scores.Score
 ORDER BY Scores.Score DESC;