SQL 查询 - 基于多列组合行

Question

在上图中，我想在连续几天合并具有相同值的行。 合并的行将在 From 列上具有最早日期，在 To 列上具有最晚日期。 看示例，即使第 3 行和第 4 行具有相同的值，它们也不会因为日期间隔而合并。

我尝试过使用 LAG 和 LEAD 函数，但没有成功。

Answer 1

我将推荐以下方法：

1. 查找每个新组的开始位置。为此，您可以将之前的最大值 todate 与此行中的 fromdate 进行比较。
2. 对开始次数进行累积求和以定义一个组。
3. 汇总结果。

这可以使用窗口函数和聚合来处理：

select value, min(fromdate) as fromdate, max(todate) as todate
from (select t.*,
             sum(case when prev_todate >= dateadd(day, -1, fromdate)
                      then 0   -- overlap, so this does not begin a new group
                      else 1   -- no overlap, so this does begin a new group
                 end) over
                 (partition by value order by fromdate) as grp
      from (select t.*,
                   max(todate) over (partition by value
                                     order by fromdate
                                     rows between unbounded preceding and 1 preceding
                                    ) as prev_todate
            from t
           ) t
     ) t
 group by value, grp
 order by value, min(fromdate);

Here 是一个 dbfiddle。

Answer 2

你可以试试下面的方法-

DEMO

    with c as
    (
    select *, datediff(dd,todate,laedval) as leaddiff,
    datediff(dd,todate,lagval) as lagdiff
    from
    (
    select *,lead(todate) over(partition by value order by todate) laedval,
    lag(todate) over(partition by value order by todate) lagval
    from t1
    )A
    ) 



select * from
(
select value,min(todate) as fromdate,max(todate) as todate from c
    where coalesce(leaddiff,0)+coalesce(lagdiff,0) in (1,-1)
    group by value
    union all
    select value,fromdate,todate from c
    where coalesce(leaddiff,0)+coalesce(lagdiff,0)>1 or coalesce(leaddiff,0)+coalesce(lagdiff,0)<-1
)A order by value

输出：

value   fromdate            todate
1       16/07/2019 00:00:00 17/07/2019 00:00:00
3       21/07/2019 00:00:00 26/07/2019 00:00:00
2       18/07/2019 00:00:00 18/07/2019 00:00:00
2       20/07/2019 00:00:00 20/07/2019 00:00:00