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  • [SQL]LeetCode178. 分数排名 | Rank Scores

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    ➤原文地址:https://www.cnblogs.com/strengthen/p/10145554.html 
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    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    |
    +-------+------+

    编写一个 SQL 查询来实现分数排名。如果两个分数相同,则两个分数排名(Rank)相同。请注意,平分后的下一个名次应该是下一个连续的整数值。换句话说,名次之间不应该有“间隔”。

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

    例如,根据上述给定的 Scores 表,你的查询应该返回(按分数从高到低排列):

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

    137ms
    1 # Write your MySQL query statement below
    2 Select Score,
    3 @rank:=@rank + (@pre<>(@pre:=Score)) as Rank
    4 From Scores, (Select @rank:=0, @pre:=-1) INIT
    5 Order by Score DESC;

    139ms

    1 # Write your MySQL query statement below
    2 
    3 select f.score Score, f.Rank from
    4 (select 
    5 @rnk:=@rnk + case when @score > s.score then 1 else 0 end as Rank
    6 ,@score:= s.score as score
    7 from 
    8 (select score from scores order by score desc) s,(select @score:=0,@rnk:=1) t) f

    142ms

    1 # Write your MySQL query statement below
    2 Select Score,
    3 @rank:=@rank + (@pre<>(@pre:=Score)) as Rank
    4 From Scores, (Select @rank:=0, @pre:=-2) INIT
    5 Order by Score DESC;

    143ms

    1 # Write your MySQL query statement below
    2 SELECT
    3   Score,
    4   @rank := @rank + (@prev <> (@prev := Score)) Rank
    5 FROM
    6   Scores,
    7   (SELECT @rank := 0, @prev := -1) init
    8 ORDER BY Score desc

    146ms

    1 # Write your MySQL query statement below
    2 select Score, CONVERT(Rank, SIGNED) AS Rank from (
    3 select Score,
    4 if(Score = @previous_score, @rank := @rank, @rank := @rank +1) as Rank,
    5 @previous_score := score
    6 from Scores, (select @rank := 0, @previous_score:= null) r
    7 order by Score desc)v

    552ms

    1 # Write your MySQL query statement below
    2 
    3 SELECT s.Score, count(distinct t.score) Rank
    4 FROM Scores s JOIN Scores t ON s.Score <= t.score
    5 GROUP BY s.Id
    6 ORDER BY s.Score desc

    553ms

    1 # Write your MySQL query statement below
    2 SELECT s1.Score, count(distinct s2.Score) as Rank
    3 FROM Scores s1 
    4     join Scores s2 on s1.Score <= S2.Score
    5 GROUP BY s1.Id
    6 ORDER BY 2
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  • 原文地址:https://www.cnblogs.com/strengthen/p/10145554.html
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