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

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

MySQL自版本8以来，支持ROW_NUMBER()，所以你可以像在SQL Server中使用一样使用下面的查询

SELECT 
    col1, col2, 
    ROW_NUMBER() OVER (PARTITION BY col1, col2 ORDER BY col3 DESC) AS intRow
FROM Table1

我还在Maria DB 10.4.21中测试了它。在那里也同样有效。

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。

这允许在MySQL中实现ROW_NUMBER()和PARTITION BY提供的相同功能

SELECT  @row_num := IF(@prev_value=GENDER,@row_num+1,1) AS RowNumber
       FirstName, 
       Age,
       Gender,
       @prev_value := GENDER
  FROM Person,
      (SELECT @row_num := 1) x,
      (SELECT @prev_value := '') y
  ORDER BY Gender, Age DESC

我会定义一个函数:

delimiter $$
DROP FUNCTION IF EXISTS `getFakeId`$$
CREATE FUNCTION `getFakeId`() RETURNS int(11)
    DETERMINISTIC
begin
return if(@fakeId, @fakeId:=@fakeId+1, @fakeId:=1);
end$$

那么我就可以:

select getFakeId() as id, t.* from table t, (select @fakeId:=0) as t2;

现在你没有子查询，视图中没有子查询。