SQL Query to return all rows with the earliest date available for each year

I am trying to design a SQL query that can return the records for the earliest date in each year. Suppose I have a table as follows:

Date       | Category | Value
========== | ======== | =====
2019-01-03 | Apple    | 5
2019-01-03 | Orange   | 2
2019-01-20 | Apple    | 5
2019-01-20 | Orange   | 8
2019-02-05 | Apple    | 1
2019-02-05 | Peach    | 5
2018-01-02 | Apple    | 2
2018-01-02 | Orange   | 9
2018-05-10 | Apple    | 3
2018-05-10 | Orange   | 5
2018-07-20 | Apple    | 6
2018-07-20 | Orange   | 1

I am trying to generate a table shown below:

Date       | Category | Value
========== | ======== | =====
2019-01-03 | Apple    | 5
2019-01-03 | Orange   | 2
2018-01-02 | Apple    | 2
2018-01-02 | Orange   | 9

The earliest date for each year will change, which means I cannot simply query by day & month. I have tried using:

SELECT MIN(Date), * 
FROM mytable 
GROUPBY YEAR(Date)

But this results in an aggregate error: ‘Category’ is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause

What would be the best way to achieve this?

Answer

A simple way is a correlated subquery:

select t.*
from mytable t
where t.date = (select min(t2.date)
                from mytable t2
                where t2.category = t.category and year(t2.date) = year(t.date)
               );

You can also use row_number():

select t.*
from (select t.*,
             row_number() over (partition by category, year(date) order by date) as seqnum
      from mytable t
     ) t
where seqnum = 1