下面是代码。你需要使用滞后和求和函数来实现结果：

drop table if exists #a 
drop table if exists #b

CREATE TABLE #a (
Business_Date DATE,
Sales float,
Budget float
);

INSERT INTO #a VALUES
('2023-03-01', 100, 150),
('2023-03-02', 200, 190),
('2023-03-03', 0, 180)

--- Find prevous value of Difference_Budget and prev value of day of Business_Date and subtract it from total days in month
select *,(case when day(Business_Date) = 1 then 0 else lag(Difference_Budget,1) over (order by Business_Date) end )  as New_Budget_previous_day,
isnull(day(dateadd(day,-1,dateadd(month,1,DATEFROMPARTS(year(Business_Date),month(Business_Date),1)))) - day(lag(Business_Date,1) over (order by Business_Date)),0) as total_No_Days
into #b
from
(
select * , 
       abs(case when sales = 0 then 0 else budget - sales end) as Difference_Budget
       
from #a
)as a

--Calculating Running_total_remaining_days by using sum function and then adding its value to budget
select *,sum(remaining_days) over (partition by year(Business_Date),month(Business_Date) 
                                   order by Business_Date 
                                   rows between unbounded preceding and current row
                                ) as Running_total_remaining_days,

        budget + sum(remaining_days) over (partition by year(Business_Date),month(Business_Date) 
                                   order by Business_Date 
                                   rows between unbounded preceding and current row
                                ) as New_budget
from
(
select * , cast(case when total_No_Days = 0 then 0 else New_Budget_previous_day/total_No_Days end as decimal(3,2)) as remaining_days
from #b
) as a