Calculate exact month-difference between two dates - sql

Question

DB-Fiddle
CREATE TABLE inventory
(
id SERIAL PRIMARY KEY,
inventory_date DATE,
product_name VARCHAR(255),
product_value VARCHAR(255)
);
INSERT INTO inventory (inventory_date, product_name, product_value)
VALUES ('2020-10-19', 'Product_A', '400'),
('2020-10-22', 'Product_B', '400'),
('2020-11-20', 'Product_C', '900'),
('2020-11-25', 'Product_D', '300');
Expected result:
product_name | months_in_inventory
-------------+--------------------
Product_A | 2
Product_B | 1
Product_C | 1
Product_D | 0
I want to calculate the months_in_inventory by calculating the difference between a fixed_date and the inventory_date.
In the example the fixed_date is '2020-12-20' and I am using it my query.
So far I am able to calculate the difference in days:
SELECT
iv.product_name,
'2020-12-20'::date - MAX(iv.inventory_date::date) AS days_in_inventory
FROM
inventory iv
GROUP BY
1
ORDER BY
1;
However, I could not figure out how to change it to a difference in month. Do you have any idea?
NOTE
I know that one way to approach this would be extracting the month from the fixed_date and inventory_date and subtract both numbers. However, this would not give me the correct result because I need it exactly based on the dates.
For example Product_B is only 1 month in inventory because 2020-10-22 is not two months compared to 2020-12-20.

Answer 1

You can use age(). If the value is always less than 12 months, then one method is:
SELECT iv.product_name,
extract(month form age('2020-12-20'::date, MAX(iv.inventory_date::date))) AS months_in_inventory
FROM inventory iv
GROUP BY 1
ORDER BY 1;
A more accurate calculation takes the year into account:
SELECT iv.product_name,
(extract(year from age('2020-12-20'::date, MAX(iv.inventory_date::date))) * 12 +
extract(month from age('2020-12-20'::date, MAX(iv.inventory_date::date)))
) AS months_in_inventory
FROM inventory iv
GROUP BY 1
ORDER BY 1;
Here is a db<>fiddle.