SQL 查询 - 基于多列组合行

Question

On the image above, I'd like to combine rows with the same value on consecutive days. Combined rows will have the earliest date on From column and the latest date on To column. Looking at the example, even if Rows 3 and 4 have the same value, they were not combined because of the date gap.

I've tried using LAG and LEAD functions but no luck.

Answer 1

You can try below way -

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

OUTPUT:

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

Answer 2

I am going to recommend the following approach:

1. Find where each new group begins. You can do this by comparing the previous maximum todate with the fromdate in this row.
2. Do a cumulative sum of the starts to define a group.
3. Aggregate the results.

This can be handled using window functions and aggregation:

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 is a db<>fiddle.