SQL 根据日期范围查询获取数据

Question

I have a data which has index and date I need to perform logic in which I need to compare the first date with next date if the date range is lesser than 31 remove the second, again compare the first with third if it's greater than 31 keep the first and third and compare the third with fourth if less remove or if greater add keep on like this. Can anyone help me out on this logic

Simply first record should be compared with prior records with greater of 31

index     DATE
1     2020-01-01
1     2020-01-15
1     2020-01-30
1     2020-02-02
1     2020-02-20
1     2020-03-05
1     2020-03-25
2     2020-04-30

Required output

index     DATE
1     2020-01-01
1     2020-02-02
1     2020-03-05
2     2020-04-30

Answer 1

This answers the original question which was tagged MySQL.

You need to use a recursive CTE for this type of operation:

with recursive cte(ind, date) as (
      select ind, min(date) as date
      from t
      group by ind
      union all
      select cte.ind,
             (select min(t2.date)
              from t t2
              where t2.ind = cte.ind and t2.date > cte.date + interval 30 day
             )
      from cte
      where exists (select 1
                    from t t2
                    where t2.ind = cte.ind and t2.date > cte.date + interval 30 day
                   )
     )
select *
from cte;

Here is a db<>fiddle.

Answer 2

Try this below query

create table indextbl(indexid int, dt date)
insert into indextbl values(1,     '2020-01-01')
,(1,'2020-01-15')
,(1,'2020-01-30')
,(1,'2020-02-02')
,(1,'2020-02-20')
,(1,'2020-03-05')
,(1,'2020-03-25')
,(2,'2020-04-30')

select indexid, MIN(dt) from indextbl
group by indexid, LEFT(dt,7)