Question

数据看起来像下面-我想要的输出是，当一天中的下一个条目是相同的值时，这些时间将相加以获得总时间。值更改后，将第一个条目（该值）的日期更改为新值，以便获得该值持续的总时间。

    Value            Date
60                  1/5/2019 8:00
60                  1/5/2019 9:00
60                  1/5/2019 10:00
75                  1/5/2019 10:30
60                  1/5/2019  11:00
40                  1/5/2019  12:00
40                  1/5/2019  13:00

所需的输出

    Value            Total Time
60                  1/5/2019 8:00 - 10:30 = 2 and a half hours
75                  1/5/2019 10:30 - 11:00 = half hour
60                  1/5/2019  11:00 - 12:00 = 1 hour
40                  1/5/2019  12:00 - 13:00 = 1 hour

Answer 1

这是一个空白和孤岛的问题。对于此版本，我认为行号之间的差异是最简单的解决方案。因此，这几乎可以解决您的问题：

select value, min(date), max(date)
from (select t.*,
             row_number() over (order by date) as seqnum,
             row_number() over (partition by value order by date) as seqnum_v
      from t
     ) t
group by (seqnum - seqnum_v), value;

但是您要 next 开始，所以我们也需要lead()：

select value, min(date) as startdate,
       lead(min(date), 1, max(date)) over (order by min(date)) as enddate
from (select t.*,
             row_number() over (order by date) as seqnum,
             row_number() over (partition by value order by date) as seqnum_v
      from t
     ) t
group by (seqnum - seqnum_v), value;

并获取总时间：

select value,
       datediff(minute,
                min(date),
                lead(min(date), 1, max(date)) over (order by min(date))
               ) as dur_minutes
from (select t.*,
             row_number() over (order by date) as seqnum,
             row_number() over (partition by value order by date) as seqnum_v
      from t
     ) t
group by (seqnum - seqnum_v), value;

合并具有相同值的行的日期范围

1 个答案: