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

这也可以是一个解决方案:

SET @row_number = 0;

SELECT 
    (@row_number:=@row_number + 1) AS num, firstName, lastName
FROM
    employees

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。

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

在MySQL中没有像rownum, row_num()这样的函数，但方法如下:

select 
      @s:=@s+1 serial_no, 
      tbl.* 
from my_table tbl, (select @s:=0) as s;

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