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

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/

我没有看到任何简单的答案涵盖“PARTITION BY”部分，所以我的答案是:

SELECT
    *
FROM (
    select
        CASE WHEN @partitionBy_1 = l THEN @row_number:=@row_number+1 ELSE @row_number:=1 END AS i
        , @partitionBy_1:=l AS p
        , t.*
    from (
        select @row_number:=0,@partitionBy_1:=null
    ) as x
    cross join (
        select 1 as n, 'a' as l
        union all
        select 1 as n, 'b' as l    
        union all
        select 2 as n, 'b' as l    
        union all
        select 2 as n, 'a' as l
        union all
        select 3 as n, 'a' as l    
        union all    
        select 3 as n, 'b' as l    
    ) as t
    ORDER BY l, n
) AS X
where i > 1

The ORDER BY clause must reflect your ROW_NUMBER need. Thus there's already a clear limitation: you can't have several ROW_NUMBER "emulation" of this form at the same time. The order of the "computed column" matters. If you have mysql compute those column in another order, it might not work. In this simple example I only put one but you can have several "PARTITION BY" parts CASE WHEN @partitionBy_1 = part1 AND @partitionBy_2 = part2 [...] THEN @row_number:=@row_number+1 ELSE @row_number:=1 END AS i , @partitionBy_1:=part1 AS P1 , @partitionBy_2:=part2 AS P2 [...] FROM ( SELECT @row_number:=0,@partitionBy_1:=null,@partitionBy_2:=null[...] ) as x

我想要每对(col1, col2)的col3最高的行。

这是一个分组最大值，是最常被问到的SQL问题之一(因为它看起来应该很简单，但实际上并非如此)。

我经常支持null-self-join:

SELECT t0.col3
FROM table AS t0
LEFT JOIN table AS t1 ON t0.col1=t1.col1 AND t0.col2=t1.col2 AND t1.col3>t0.col3
WHERE t1.col1 IS NULL;

"获取表中没有匹配col1、col2的其他行具有更高的col3的行。(你会注意到，如果不止一行具有相同的col1、col2、col3，那么这个和大多数其他分组最大值解将返回多行。如果这是一个问题，你可能需要一些后期处理。)

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

MySQL从8.0+版本开始支持ROW_NUMBER()。

如果使用MySQL 8.0或更高版本，请检查ROW_NUMBER()函数。 否则，您将使用模拟ROW_NUMBER()函数。

row_number()是一个排序函数，返回一行的顺序编号，第一行从1开始。

对于旧版本，

SELECT t.*, 
       @rowid := @rowid + 1 AS ROWID
  FROM TABLE t, 
       (SELECT @rowid := 0) dummy;