Find all instances of a date in a date range - SQL Server - date

Question

I need to find the price for an item for each financial year end date in a date range. In this case the financial year is e.g. 31 March
The table I have for example:
ItemID
Value
DateFrom
DateTo
1
10
'2019/01/01'
'2021/02/28'
1
11
'2021/03/01'
'2021/05/01'
SQL Fiddle
The SQL would thus result in the above table to be:
ItemID
Value
DateFrom
DateTo
1
10
'2019/01/01'
'2019/03/30'
1
10
'2020/03/31'
'2021/02/28'
1
11
'2020/03/01'
'2021/03/30'
1
11
'2020/03/31'
'2021/05/01'

Answer 1

You can solve it, but a prerequisite is the creation of a table called financial_years and filling it with data. This would be the structure of the table:
financial_years(id, DateFrom, DateTo)
Now that you have this table, you can do something like this:
select ItemID, Value, financial_years.DateFrom, financial_years.DateTo
from items
join financial_years
on (items.DateFrom between financial_years.DateFrom and financial_years.DateTo) or
(items.DateTo between financial_years.DateFrom and financial_years.DateTo)
order by financial_years.DateFrom;

Answer 2

The accepted answer is not correct, as it does not split out different parts of the year which have different values.
You also do not need a Year table, although it can be beneficial. You can generate it on the fly using a VALUES table.
Note also a better way to check the intervals overlap, using AND not OR
WITH Years AS (
SELECT
YearStart = DATEFROMPARTS(v.yr, 3, 31),
YearEnd = DATEFROMPARTS(v.yr + 1, 3, 31)
FROM (VALUES
(2015),(2016),(2017),(2018),(2019),(2020),(2021),(2022),(2023),(2024),(2025),(2026),(2027),(2028),(2029),(2030),(2031),(2032),(2033),(2034),(2035),(2036),(2037),(2038),(2039)
) v(yr)
)
SELECT
i.ItemID,
i.Value,
DateFrom = CASE WHEN i.DateFrom > y.YearStart THEN i.DateFrom ELSE y.YearStart END,
DateTo = CASE WHEN i.DateTo > y.YearEnd THEN y.YearEnd ELSE i.DateTo END
FROM items i
JOIN Years y ON i.DateFrom <= y.YearEnd
AND i.DateTo >= y.YearStart;