LeetCode_OJ【178】Rank Scores

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本文介绍了一个SQL查询案例，用于对分数表中的记录进行排名。在出现相同分数时，确保排名一致且连续，不会出现排名空缺的情况。通过使用子查询和DISTINCT关键字，有效地实现了这一目标。

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    |
+-------+------+

这道题就是给得分进行排序，首先记录scores表中的每个元组可以用count()函数找出表中大于该元组score值的元组个数，根据这个即可得到排名

select s1.score,
       (select count(*)+1 from (select DISTINCT score from Scores) s2 where s2.score > s1.score) as rank 
       from Scores s1 
       ORDER BY s1.score DESC;