Question

我有一个桌面游戏日，其中包含以下列：date_played，winner，loser 具有以下值，

(Jun-03-14, USA, China)
(Jun-05-14, USA, Russia)
(Jun-06-14, France, Germany)
.
.
.
.
(Jun-09-14, USA, Russia)

我需要获得美国在序列中正好赢得3行的所有实例。

我尝试了以下查询。

Select 
    date, winner, loser, 
    RANK() OVER (PARTITION BY winner ORDER BY date rows 2 preceding) as rank 
from playday;

Answer 1

首先你需要找出他们最后一次失败的时间。第二个计算胜利的数量，大于（＆gt;）他们最后一次丢失的日期。第三个返回大于最后一次丢失的所有行，如果count> 3。

抱歉，我前面没有SQL解析器，无法正确地将其放入代码中。

Set @team_name = "USA";
select date, winner, loser
from playday
where (select count(*) as wins_since_loss from playday
       where playday.winner = @team_name
       and playday.date >
    (select max(date) as losing_date from playday where playday.loser = @team_name)) = 3

Answer 2

您可以使用以下查询。

select winner,loser,date,cnt from (select winner, loser, date, date - lag(date,3) over ( order by date) as cnt  from playday) where cnt >=3

Answer 3

查询是拉美国连续三次获胜的行序列，不少于或少于（我使用日期作为date1）

select   date1, winner, loser from
(
  select count (*) over (partition by change) as id, date1,winner,loser from 
  (
    select date1,winner,loser,lag_loser, sum(case when loser <> lag_loser and (loser='USA' or lag_loser='USA') then 1 else 0 end) over (order by date1 rows unbounded preceding) as change from 
    (
      select date1, winner,loser, lag(loser) over (order by date1) as lag_loser from
      (
        select date1, winner, loser from playday
        where winner ='USA' or loser = 'USA'
        ORDER BY date1 ASC
      )
    )
  )
)
where winner ='USA' and id =3

SQL - 行排序

3 个答案: