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 |
+-------+------+此题，其本质就是赋值行号(需要注意分数相同的情景).
在实践过程中，初版答案如下所示：
# Write your MySQL query statement below
SELECT
a.Score AS Score,
(SELECT COUNT(DISTINCT b.Score) FROM Scores b where b.Score >= a.Score) AS Rank
FROM Scores a ORDER BY a.Score DESC;此处，使用select count来统计行号，注意使用distinct来区分相同分数.
但是，此解题方案的效率较差，sql运行肯定是越快越好.
因此，在sql中引入变量来赋值行号，以替代耗时的select count操作.
# Write your MySQL query statement below
SELECT
Score,
@row := @row + (@pre <> (@pre := Score)) AS Rank
FROM Scores, (SELECT @row := 0, @pre := -1) t
ORDER BY Score DESC;此处，查询是在Scores与临时表之间进行cross join.
此外，使用临时变量(@row,@pre)记录行号.Tips:
- 通过
@pre与当前Score的比较，确定是否+1,需要注意mysql中的true/false为0/1); - 在
sql中,set/update语句中,=表示赋值；在set/update/select中,:=表示赋值，因此使用:=.
通过以上改进，mysql的运行效率得到了较大的提高.
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