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

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

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

也有点晚了，但今天我有同样的需要，所以我在谷歌上搜索了一下，最后在Pinal Dave的文章http://blog.sqlauthority.com/2014/03/09/mysql-reset-row-number-for-each-group-partition-by-row-number/中找到了一个简单的一般方法

我想把重点放在Paul最初的问题上(这也是我的问题)，所以我把我的解决方案总结为一个工作示例。

因为我们想要在两个列上进行分区，我将在迭代期间创建一个SET变量来识别是否启动了一个新组。

SELECT col1, col2, col3 FROM (
  SELECT col1, col2, col3,
         @n := CASE WHEN @v = MAKE_SET(3, col1, col2)
                    THEN @n + 1 -- if we are in the same group
                    ELSE 1 -- next group starts so we reset the counter
                END AS row_number,
         @v := MAKE_SET(3, col1, col2) -- we store the current value for next iteration
    FROM Table1, (SELECT @n := 0, @v := NULL) r -- helper table for iteration with startup values
   ORDER BY col1, col2, col3 DESC -- because we want the row with maximum value
) x WHERE row_number = 1 -- and here we select exactly the wanted row from each group

3意味着在MAKE_SET的第一个参数，我想在SET中的值(3=1|2)。 当然，如果没有两个或更多列来构造组，则可以取消MAKE_SET操作。结构完全相同。这对我来说是必要的。非常感谢Pinal Dave的清晰演示。

我也会投票给Mosty Mostacho的解决方案，对他的查询代码进行了轻微的修改:

SELECT a.i, a.j, (
    SELECT count(*) from test b where a.j >= b.j AND a.i = b.i
) AS row_number FROM test a

会得到相同的结果:

+------+------+------------+
|    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 |
+------+------+------------+

对于表格:

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

唯一的区别是查询不使用JOIN和GROUP BY，而是依赖于嵌套选择。