SUM Value Based On Day Near Each Other By A Day

Question

So, I wanted to get the SUM of New Ink (ml) and Ink Used (ml) grouped by Date, Paper Code, and Ink Code but I also wanted the Date group the difference of the Date only by a day

Main Table

Date	Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
10-1-2022	911	C21	10	8
10-1-2022	911	C29	9	3
10-2-2022	911	C21	8	3
10-5-2022	911	C21	25	15
10-12-2022	911	C21	10	8
10-13-2022	911	C21	8	6
10-15-2022	911	C21	6	6
10-15-2022	911	C29	9	9

Expected result should be :

For Date 10-1-2022 and 10-2-2022

Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
911	C21	18	11
911	C29	9	3

For Date 10-5-2022

Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
911	C21	25	15

For Date 10-12-2022 and 10-13-2022

Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
911	C21	18	14

For Date 10-15-2022

Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
911	C21	6	6
911	C29	9	9

and all combined would be

Paper Code	Ink Code	New Ink (ml)	Ink Used (ml)
911	C21	18	11
911	C29	9	3
911	C21	25	15
911	C21	18	14
911	C21	6	6
911	C29	9	9

I've tried

SELECT 
    [Paper Code],
    [Ink Code],
    SUM([New Ink (ml)]) AS [New Ink (ml)],
    SUM([Ink Used (ml)]) AS [Ink Used (ml)]
FROM [Table Ink]
GROUP BY [Paper Code], [Ink Code]

Answer 1

As suggested in comments, you can use LAG() or LEAD() to identify the group of rows. After that just use GROUP BY as per normal

with cte as
(
  select *,
         g = case when datediff(day, lag([Date]) over (order by [Date]),
                                     [Date]) > 1
                  then 1
                  else 0
                  end
  from   [Table Ink]
),
cte2 as
(
  select *, grp = sum(g) over (order by [Date])
  from   cte
)
select [Paper Code],  [Ink Code],
       sum([New Ink (ml)])  as [New Ink (ml)],
       sum([Ink Used (ml)]) as [Ink Used (ml)]
from   cte2
group by grp, [Paper Code],  [Ink Code]