基于SQL Server

时间:2016-06-09 11:58:57

标签: mysql sql sql-server

我有一张包含用户和日期值的表格。对于每个用户,可以有多个日期值。在下面的脚本中,我根据我在stackoverflow找到的另一个答案,在select,for user期间插入了SincePrevious和sinceFirst列。

SELECT 
 a.user as 'user'
,a.date as 'date'
,ISNULL(DATEDIFF(day,b.date,a.date),0) as 'sincePrevious'
,datediff(day, min(a.date) over (partition by a.user), a.date) as 'sinceFirst'
FROM
(select  *,ROW_NUMBER() OVER(PARTITION BY user ORDER BY date) as Rank from HUT_regels) as a
LEFT JOIN 
(select  *,ROW_NUMBER() OVER(PARTITION BY user ORDER BY date) as Rank from HUT_regels) as b
ON a.user = b.user AND a.Rank = b.Rank + 1
ORDER by 'user', 'date'

我想要做的是以类似的方式插入另一列(在选择期间),根据用户值和两行之间的时间差添加唯一的组ID。在示例中,我添加了一些groupIds。如果同一用户的两个连续日期之间的日期差异大于50天(在此示例中),则必须将其视为出现的新“序列”。 groupIDs 2,3和4反映了这一点。

user      date        sincePrevious  sinceFirst  groupId
100000029 25-05-2012  0              0           1
100002161 08-01-2012  0              0           2
100002161 04-02-2012  27             27          2
100002161 15-02-2012  11             38          2
100002161 28-03-2012  42             80          2
100002161 23-05-2012  56             136         3
100002161 11-07-2012  49             185         3
100002161 29-08-2012  49             234         3
100002161 24-10-2012  56             290         4
100002161 21-11-2012  28             318         4
100005242 07-05-2013  0              0           5
100005242 10-05-2013  3              3           5
100005242 14-05-2013  4              7           5
100005242 17-05-2013  3              10          5
100005242 21-05-2013  4              14          5
100005242 24-05-2013  3              17          5
100005242 28-05-2013  4              21          5
100005242 07-06-2013  10             31          5
...

groupIds应该是唯一的,但不必是连续的或偶数。

我知道CTE可以做到这一点,但我想找到一个类似于SincePrevious和sinceFirst方式的解决方案。

我的用例是SQL Server,但是更通用的解决方案(我提到MySQL,但PostgreSQL也很好)也可以帮助其他人。

1 个答案:

答案 0 :(得分:0)

首先,您应该使用lag()min()来获取值:

select r.*,
       datediff(day, lag(date) over (partition by user order by date), date) as sincePrevious,
       datediff(day, min(date) over (partition by user), date) as sinceFirst
from HUT_regels r;

要添加GroupId,您只需要一个子查询和条件聚合:

select r.*,
       sum(case when sincePrevious < 0 then 0 else 1 end) over
           (partition by user order by date) as groupId
from (select r.*,
             datediff(day, lag(date) over (partition by user order by date), date) as sincePrevious,
             datediff(day, min(date) over (partition by user), date) as sinceFirst
      from HUT_regels r
     ) r;

这是所有ANSI标准功能。但是,直到2012版本才在SQL Server中完全引入它。在早期版本中,您可以使用apply代替。