SQl查询-每个组的2个最大值，第二个必须“年轻”，然后第一个

Question

I have MySQL table. I need select two highest values for each group, but second highest value can`t be "more recent". If the highest value for Project 1 is date '2015-12-01', the second highest value must be earlierl then '2015-12-01'. This date is user input within application, eg.: user input date '2014-01-03' and get 2 highest value order by date.

1. '2013-12-05' - 500 visits

2. '2013-10-11' - 480 viists

    + --------- + ----------- + ---------+ | project | visits | date | + --------- + ----------- + ---------+ | 1 | 750 |2015-12-01| | 1 | 582 |2014-11-01| | 1 | 423 |2013-10-02| | 2 | 666 |2016-01-01| | 2 | 812 |2012-02-02| | 2 | 450 |2014-03-05| + --------- + ----------- + -------- + 

Edit: I try this (it is more complex then example above) + join table due to ProjectName

set @num := 0, @group := '';

select ProjectName,visits, date
                    from 
                    (
                    select date, Projects_ID, visits,
                    @num := if(@group = `Projects_ID`, @num + 1, 1) as row_number,
                    @group := Projects_ID as col
                    from Hits
                    where date < '2016-01-11' and visits > 0
                    order by Projects_ID,date, visits desc
                    ) as x inner join Projects as p on x.Projects_ID = 
                    p.ID
                    where x.row_number <= 2
                    group by ProjectName,Projects_ID, date

I got table where was the highest value, but in some cases second highest value was "most recent" then highest value.

Answer 1

SELECT a.visits, a.project, a.date 
FROM temper1 AS a 
WHERE (SELECT COUNT(*) 
       FROM temper1 AS b 
       WHERE b.project = a.project AND b.date >= a.date
       ) <= 2 
ORDER BY a.visits ASC, a.date DESC