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  • 0x0 题目详情
  • 0x1 解题思路
  • 0x2 代码实现
  • 0x3 课后总结

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  1. Mysql

178-Rank-Scores

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Last updated 4 years ago

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0x0 题目详情

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      | +-------+------+

0x1 解题思路

这道题采用局部变量的方法,我也是第一次采用的,高级高级。我们先给每行初始化一个Rank变量。给rank赋值的逻辑就比较简单了。

如果前一行的分数等于当前行,那么rank不变,否则rank加1,当然这样操作的前提是需要先对Score字段排序。当时对于这段sql是如何正确运行的困扰了我很长时间。

我的问题:首先order by是在select后面直面,但是实现正确的rank又需要先排序,所以这段代码到底是怎么执行的?

答案:之前我一直以为在order之前Rank字段的值就已经计算好了,但是事实并不是这样,以下面的代码为例作为讲解(与原题无关):

SELECT socre, cast(CASE
WHEN @pre = socre THEN @rk + 0
WHEN (@pre := socre) is not null THEN @rk := @rk + 1
END as signed)AS `Rank`
FROM Score, (SELECT @pre := NULL, @rk := 0) AS init
ORDER BY socre DESC;

结果如下:

如果在最后加上limit,因为limit是最后执行的,我们再来看看结果:

SELECT socre, cast(CASE
WHEN @pre = socre THEN @rk + 0
WHEN (@pre := socre) is not null THEN @rk := @rk + 1
END as signed)AS `Rank`
FROM Score, (SELECT @pre := NULL, @rk := 0) AS init
ORDER BY socre DESC
limit 1,3;

结果如下:

神奇的结果出现了,我们发现是先limit再计算Rank的。因为如果是先计算Rank,那么limit后的Rank应该从2开始,而结果是从1开始。所以说明:变量是在limit之后赋值的。

0x2 代码实现

# Write your MySQL query statement below
SELECT Score, cast(CASE
WHEN @pre = Score THEN @rk + 0
WHEN @pre := Score THEN @rk := @rk + 1
ELSE @rk := @rk + 1
END as signed)AS `Rank`
FROM Scores, (SELECT @pre := NULL, @rk := 0) AS init
ORDER BY Score DESC;

这里首先简单说明以下语法规则:MySQL中变量符号以@开头,:=表示对局部变量赋值。首先会执行from语句,这里会将Score表中的每一行加上一个新字段,也就是Rank。

下面这一句可以看做对每一行新添加的局部变量做初始化。

(SELECT @pre := NULL, @rk := 0) AS init

还有一点需要注意:

WHEN @pre := Score THEN @rk := @rk + 1

这句比较特殊,按道理来说,when中应该执行的是条件语句,而这里却给了一个赋值语句,那么赋值结果如果是0的话,再进行逻辑判断的结果就是false,自然不会执行后面的then语句。这样如果分数中出现零分,就不会实现正确的Rank排名。

这里比较笨拙的办法就是加一句else,在后面加一句else,把刚才没有执行then部分补上(因为上一句并没有成功执行,自然会走到else)。ok,这样就修复了0分的情况。

还有一点需要注意:

这里因为Score是浮点数,所以计算出来的Rank也是浮点数,所以需要强制将Rank向下转型。

下面是一种不用加else的方法,就是在赋值完成,就直接判断局部变量是否为null,因为null和0还是不等的,所以如果为0分还是会执行后面的then语句。

# Write your MySQL query statement below
SELECT Score, cast(CASE
WHEN @pre = Score THEN @rk + 0
WHEN (@pre := Score) is not null THEN @rk := @rk + 1
END as signed)AS `Rank`
FROM Scores, (SELECT @pre := NULL, @rk := 0) AS init
ORDER BY Score DESC;

0x3 课后总结

嗯,sql语句里的局部变量还是第一次用到。

原题链接
178-no-limit
178-with-limit