Question

我想要做的是查找某人在星期日活动的连续周数并为其分配值。他们必须每天参加至少2场比赛，才能算上一周的比赛。

如果他们连续2周活跃，我想指定100，连续3周，值200，连续4周，值300，并持续连续9周。

我的困难不在于确定连续几周，而是在连续日期之间中断。假设以下数据集：

CustomerID  RaceDate    Races
1           2/2/2014    2
1           2/9/2014    5
1           2/16/2014   3
1           2/23/2014   3
1           3/2/2014    4
1           3/9/2014    3
1           3/16/2014   3
2           2/2/2014    2
2           2/9/2014    3
2           3/2/2014    2
2           3/9/2014    4  
2           3/16/2014   3

CustomerID 1连续7周的价值为600。

对我来说困难的部分是CustomerID 2.他们将连续2周和连续3周。所以他们的总价值将是100 + 200 = 300。

我希望能够连续几周的任何不同组合来做到这一点。

请帮忙吗？

编辑：我正在使用SQL Server 2008 R2。

Answer 1

在寻找顺序值时，有一个简单的观察有助于。如果从日期中减去序列，则该值为常量。您可以将其用作分组机制

select CustomerId, min(RaceDate) as seqStart, max(RaceDate) as seqEnd,
       count(*) as NumDaysRaced
from (select t.*,
              dateadd(week, - row_number() over (partition by customerID, RaceDate),
                      RaceDate) as grp
      from table t
      where races >= 2
     ) t
group by CustomerId, grp;

然后您可以使用它来获得最终的“积分”：

select CustomerId,
       sum(case when NumDaysRaced > 1 then (NumDaysRaced - 1) * 100 else 0 end) as Points
from (select CustomerId, min(RaceDate) as seqStart, max(RaceDate) as seqEnd,
             count(*) as NumDaysRaced
      from (select t.*,
                    dateadd(week, - row_number() over (partition by customerID, RaceDate),
                            RaceDate) as grp
            from table t
            where races >= 2
           ) t
      group by CustomerId, grp
     ) c
group by CustomerId;

如何在SQL中找到连续的活动周？

1 个答案: