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

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

查询mysql中的row_number

set @row_number=0;
select (@row_number := @row_number +1) as num,id,name from sbs

MySQL中没有排名功能。最接近的方法是使用变量:

SELECT t.*, 
       @rownum := @rownum + 1 AS rank
  FROM YOUR_TABLE t, 
       (SELECT @rownum := 0) r

那么在我的案例中这是如何起作用的呢?我需要两个变量，分别代表col1和col2?当col1改变时，Col2需要以某种方式重置…?

是的。如果它是Oracle，您可以使用LEAD函数在下一个值处达到峰值。值得庆幸的是，Quassnoi涵盖了在MySQL中需要实现的逻辑。

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。