SQL first_value() by specific datepart

Question

I have data where it has specific date (yyyy-mm-dd) with its city and value

-------------------------------------
   city   |  date       | value
-------------------------------------
   city A |  2018-05-01 |  10
   city A |  2018-04-03 |  15
   city A |  2019-01-03 |  9
   city A |  2019-05-01 |  13
   city B |  2019-05-01 |  12
   city B |  2019-02-12 |  11

and I would like to get the first value for each year partitioned by city in the next column :

--------------------------------------------------
   city   |  date       | value | first_value_year_by_city
--------------------------------------------------
   city A |  2018-05-01 |  10   |    15
   city A |  2018-04-03 |  15   |    15
   city A |  2019-01-03 |  9    |    9
   city A |  2019-05-01 |  13   |    9
   city B |  2019-05-01 |  12   |    11
   city B |  2019-02-12 |  11   |    11

For that I used SQL :

select *,
,first_value(value) over(partition by city order by date rows between unbounded preceding and unbounded following) as first_value_year_by_city
 from table

But I couldn't put datepart(year,date) = 2018 and datepart(year,date) = 2019 in the first_value(). How we can do this?

Answer 1

You can put the year in the partition by:

select t.*,
       first_value(value) over (partition by city, year(date)
                                order by date
                               ) as first_value_year_by_city
from table t;