Question

我试图在我的数据中找到客户的最大天数。我想了解特定客户制作的最大天数。如果有人在25/8/16 AND 26/08/16和27/08/16以及01/09/16和02/09/16进入我的应用程序 - 最大顺序为3天（25,26， 27）。

最后（输出）我想得到两个字段：custid | MaxDaySequence

我的数据表中有以下字段： custid |订购日期（timestemp）

例如：custid orderdate 1 25/08/2007 1 03/10/2007 1 13/10/2007 1 15/01/2008 1 16/03/2008 1 09/04/2008 2 18/09/2006 2 08/08/2007 2 28/11/2007 2 04/03/2008 3 27/11/2006 3 15/04/2007 3 13/05/2007 3 19/06/2007 3 22/09/2007 3 25/09/2007 3 28/01/2008

我正在使用SQL Server 2014。

由于

Answer 1

有一个技巧，如果你的日期有一个递增的数字，那么如果它们是连续的，那么减去你日期的天数就会相同。像这样：

SELECT custid, 
       min(orderdate) as start_of_group,
       max(orderdate) as end_of_group,
       count(*) as num_days
FROM (
   SELECT custid, orderdate
          ROW_NUMBER() OVER (PARTITION  BY custid ORDER BY orderdate) as rn
) x
GROUP BY custid, dateadd(day, - rn, orderdate);

您可以获取此结果并提取解决问题的最长天数：

SELECT custid, max(num_days) as longest
FROM (
  SELECT custid, 
         count(*) as num_days
  FROM (
     SELECT custid, orderdate
            ROW_NUMBER() OVER (PARTITION  BY custid ORDER BY orderdate) as rn
  ) x
  GROUP BY custid, dateadd(day, - rn, orderdate)
) y
GROUP BY custid

Answer 2

如果你想用MySQL解决它：

select user_id,max(num_days) as longest
from(
select user_id, count(*) as num_days
from
(
SELECT    (CASE a1.user_id 
           WHEN @curType 
           THEN @curRow := @curRow + 1 
           ELSE @curRow := 1 AND @curType := a1.user_id END
          ) AS rank,
          a1.user_id,
          a1.last_update as dat
FROM      (select a2.user_id,left(FROM_UNIXTIME(a2.last_update),10) as 'last_update' 
                from visits as a2 group by 1,2) as a1 ,
            (SELECT @curRow := 0, @curType := '') r

ORDER BY  a1.user_id DESC, dat) x
group by user_id, DATE_ADD(dat,INTERVAL -rank day)
) y
group by 1
order by longest desc

在SQL Server中查找日期序列

2 个答案: