轉載：https://www.cnblogs.com/niniya/p/9046449.html

一、排名

/*普通排名：從1開始，順序往下排*/
SELECT cs.*,@r :=@r + 1 AS rank
FROM cs,(SELECT @r := 0) r
ORDER BY score;

/*並列排名：相同的值是相同的排名*/
SELECT cs.* ,
CASE 
WHEN @p=score THEN @r
WHEN @p:=score THEN @r:=@r+1
END rank
FROM cs,(SELECT @r:=0,@p:=NULL)r
ORDER BY score;

/*並列排名：相同的值名次相同，與上例中的並列排名不同*/
SELECT city,score,rank
FROM
(
SELECT cs.*,
@c:=IF(@p=score,@c,@r) AS rank,
@p:=score,
@r:=@r+1
FROM cs ,(SELECT @p:=NULL,@r:=1,@c:=0)r
ORDER BY score
)c

二、分組後組內排名

/*分組普通排名：順序排名*/
SELECT city,score,rank
FROM
(
SELECT cs.*,IF(@p=city,@r:=@r+1,@r:=1) AS rank,
　　　　@p:=city
FROM cs,(SELECT @p:=NULL,@r:=0)r
ORDER BY city,score
)s;

/* 分組後並列排名：組內相同數值排名相同*/
SELECT city,score,rank
FROM
(
SELECT *,
IF(@p=city,
    CASE 
       WHEN @s=score THEN @r
       WHEN @s:=score THEN @r:=@r+1
    END,
   @r:=1 ) AS rank,
@p:=city,
@s:=score
FROM cs,(SELECT @p:=NULL,@s:=NULL,@r:=0)r
ORDER BY city,score 
)s;

三、分組後取各組的前兩名

/*取每組分數高的前兩個，法一*/
SELECT city,score,rank
FROM
(
SELECT *,
IF(@p=city,
    CASE 
        WHEN @s=score THEN @r
        WHEN @s:=score THEN @r:=@r+1
    END,
  @r:=1 ) AS rank,
@p:=city,
@s:=score
FROM cs,(SELECT @p:=NULL,@s:=NULL,@r:=0)r
ORDER BY city,score DESC 
)s
WHERE rank <3;

/*分組後取前兩個，法二*/
SELECT * FROM cs c
WHERE (
    SELECT count(*) FROM cs
    WHERE city=c.city AND score>c.score )<2
  ORDER BY city,score DESC

參考：

https://www.jianshu.com/p/bb1b72a1623e

http://blog.sina.com.cn/s/blog_4c197d420101e408.html