How to use date_part function to split value per month to each day and country - postgresql

Question

DB-Fiddle
CREATE TABLE sales (
id SERIAL PRIMARY KEY,
country VARCHAR(255),
sales_date DATE,
sales_volume DECIMAL,
fix_costs DECIMAL
);
INSERT INTO sales
(country, sales_date, sales_volume, fix_costs
)
VALUES
('DE', '2020-01-03', '500', '2000'),
('FR', '2020-01-03', '350', '2000'),
('None', '2020-01-31', '0', '2000'),
('DE', '2020-02-15', '0', '5000'),
('FR', '2020-02-15', '0', '5000'),
('None', '2020-02-29', '0', '5000'),
('DE', '2020-03-27', '180', '4000'),
('FR', '2020-03-27', '970', '4000'),
('None', '2020-03-31', '0', '4000');
Expected Result:
sales_date | country | sales_volume | fix_costs
-------------|--------------|------------------|------------------------------------------
2020-01-03 | DE | 500 | 37.95 (= 2000/31 = 64.5 x 0.59)
2020-01-03 | FR | 350 | 26.57 (= 2000/31 = 64.5 x 0.41)
-------------|--------------|------------------|------------------------------------------
2020-02-15 | DE | 0 | 86.21 (= 5000/28 = 172.4 x 0.50)
2020-02-15 | FR | 0 | 86.21 (= 5000/28 = 172.4 x 0.50)
-------------|--------------|------------------|------------------------------------------
2020-03-27 | DE | 180 | 20.20 (= 4000/31 = 129.0 x 0.16)
2020-03-27 | FR | 970 | 108.84 (= 4000/31 = 129.0 x 0.84)
-------------|--------------|------------------|-------------------------------------------
The column fix_costs in the expected result is calculated as the following:
Step 1) Get the daily rate of the fix_costs per month.(2000/31 = 64.5; 5000/29 = 172.4; 4000/31 = 129.0)
Step 2) Split the daily value to the countries DE and FR based on their share in the sales_volume. (500/850 = 0.59; 350/850 = 0.41; 180/1150 = 0.16; 970/1150 = 0.84)
Step 3) In case the sales_volume is 0 the daily rate gets split 50/50 to DE and FR as you can see for 2020-02-15.
In MariaDB I was able to this with the below query:
SELECT
s.sales_date,
s.country,
s.sales_volume,
(CASE WHEN SUM(sales_volume) OVER (PARTITION BY sales_date) > 0
THEN ((s.fix_costs/ DAY(LAST_DAY(sales_date))) *
sales_volume / NULLIF(SUM(sales_volume) OVER (PARTITION BY sales_date), 0)
)
ELSE (s.fix_costs / DAY(LAST_DAY(sales_date))) * 1 / SUM(country <> 'None') OVER (PARTITION by sales_date)
END) AS imputed_fix_costs
FROM sales s
WHERE country <> 'None'
GROUP BY 1,2,3
ORDER BY 1;
However, in PostgresSQL I get an error on DAY(LAST_DAY(sales_date)).
I tried to replace this part with (date_part('DAY', ((date_trunc('MONTH', s.sales_date) + INTERVAL '1 MONTH - 1 DAY')::date)))
However, this is causing another error.
How do I need to modify the query to get the expected result?

Answer 1

The Postgresql equivalent of DAY(LAST_DAY(sales_date)) would be:
extract(day from (date_trunc('month', sales_date + interval '1 month') - interval '1 day'))
The expression SUM(country <> 'None') also needs to be fixed as
SUM(case when country <> 'None' then 1 else 0 end)
It might be a good idea to define this compatibility function:
create function last_day(d date) returns date as
$$
select date_trunc('month', d + interval '1 month') - interval '1 day';
$$ language sql immutable;
Then the first expression becomes simply
extract(day from last_day(sales_date))

Answer 2

I would create a function to return the last day (number) for a given date - which is actually the "length" of the month.
create function month_length(p_input date)
returns integer
as
$$
select extract(day from (date_trunc('month', p_input) + interval '1 month - 1 day'));
$$
language sql
immutable;
Then the query can be written as:
select sales_date, country,
sum(sales_volume),
sum(fix_costs_per_day * cost_factor)
from (
select id, country, sales_date, sales_volume, fix_costs,
fix_costs / month_length(sales_date) as fix_costs_per_day,
case
when sum(sales_volume) over (partition by sales_date) > 0
then sales_volume::numeric / sum(sales_volume) over (partition by sales_date)
else sales_volume::numeric / 2
end as cost_factor
from sales
where country <> 'None'
) t
group by sales_date, country
order by sales_date, country