Fill missing gaps in data using a date column - sql

Question

I have a temp table that returns this output
PRICE | DATE
1.491500 | 2019-02-01
1.494000 | 2019-02-04
1.486500 | 2019-02-06
I want to fill in the missing gaps in data by duplicating the last known record prior to the gaps in data using the date. Is their a way to update the existing temp table or create a new temp table with this desired output dynamically:
PRICE | DATE
1.491500 | 2019-02-01
1.491500 | 2019-02-02
1.491500 | 2019-02-03
1.494000 | 2019-02-04
1.494000 | 2019-02-05
1.486500 | 2019-02-06
I am working on sql server 2008r2

Answer 1

Because SQL Server does not support IGNORE NULLS in LAG() this is a bit tricky. I would go for a recursive subquery of the form:
with cte as (
select price, date, dateadd(day, -1, lead(date) over (order by date)) as last_date
from t
union all
select price, dateadd(day, 1, date), last_date
from cte
where date < last_date
)
select price, date
from cte
order by date;
Here is a db<>fiddle.
In SQL Server 2008, you can replace the lead() with:
with cte as (
select price, date,
(select min(date)
from t t2
where t2.date > t.date
) as last_date
from t
union all
select price, dateadd(day, 1, date), last_date
from cte
where date < last_date
)
select price, date
from cte
order by date;

Answer 2

Assuming there is a dates table (if not you can easily make one), you can do this by left joining the existing table to the dates table. Thereafter assign groups per dates found using a running sum. The max value per group is what would be needed to fill in the missing values.
select dt,max(price) over(partition by grp) as price
from (select p.price,d.dt,sum(case when p.dt is null then 0 else 1 end) over(order by d.dt) as grp
from dates d
left join prices p on p.dt = d.dt
) t
Sample Demo
Making a dates table with a recursive cte. Persist it as needed.
--Generate dates in 2019
with dates(dt) as (select cast('2019-01-01' as date)
union all
select dateadd(day,1,dt)
from dates
where dt < '2019-12-31'
)
select * from dates
option(maxrecursion 0)