我认为你可以在这里使用DENSE_RANK()函数。 例子:

select `score`, DENSE_RANK() OVER( ORDER BY score desc ) as `rank` from Scores;

https://www.mysqltutorial.org/mysql-window-functions/mysql-dense_rank-function/

有点晚了，但也可能对那些寻找答案的人有帮助……

Between rows/row_number示例-可以在任何SQL中使用的递归查询:

WITH data(row_num, some_val) AS 
(
 SELECT 1 row_num, 1 some_val FROM any_table --dual in Oracle
  UNION ALL
 SELECT row_num+1, some_val+row_num FROM data WHERE row_num < 20 -- any number
)
SELECT * FROM data
 WHERE row_num BETWEEN 5 AND 10
/

ROW_NUM    SOME_VAL
-------------------
5           11
6           16
7           22
8           29
9           37
10          46

我总是遵循这个模式。给定这个表格:

+------+------+
|    i |    j |
+------+------+
|    1 |   11 |
|    1 |   12 |
|    1 |   13 |
|    2 |   21 |
|    2 |   22 |
|    2 |   23 |
|    3 |   31 |
|    3 |   32 |
|    3 |   33 |
|    4 |   14 |
+------+------+

你可以得到这样的结果:

+------+------+------------+
|    i |    j | row_number |
+------+------+------------+
|    1 |   11 |          1 |
|    1 |   12 |          2 |
|    1 |   13 |          3 |
|    2 |   21 |          1 |
|    2 |   22 |          2 |
|    2 |   23 |          3 |
|    3 |   31 |          1 |
|    3 |   32 |          2 |
|    3 |   33 |          3 |
|    4 |   14 |          1 |
+------+------+------------+

通过运行这个不需要定义任何变量的查询:

SELECT a.i, a.j, count(*) as row_number FROM test a
JOIN test b ON a.i = b.i AND a.j >= b.j
GROUP BY a.i, a.j

如果查询有GROUP BY语句，使用交叉连接和逗号的解决方案将不起作用。对于这种情况，您可以使用subselect:

SELECT (@row_number := @row_number + 1) AS rowNumber, res.*
FROM
(
  SELECT SUM(r.amount) 
  FROM Results r 
  WHERE username = 1 
  GROUP BY r.amount
) res
CROSS JOIN (SELECT @row_number := 0) AS dummy

对于另一个列的划分，一种方法是由@abcdn描述的。但是，它的性能很低。我建议使用这段代码，它不需要连接表本身: 考虑同一张桌子。

你可以得到这样的分区:

set @row_num := 0;
set @j:= 0;

select IF(j= @j, @row_num := @row_num + 1, @row_num := 1) as row_num,
       i, @j:= j as j
from tbl fh
order by j, i;

结果是这样的:

这样做的好处是我们不需要连接表本身

SELECT 
    @i:=@i+1 AS iterator, 
    t.*
FROM 
    tablename AS t,
    (SELECT @i:=0) AS foo