Calculate the days to reach a certain date - PostgreSQL - sql

Question

I need to create a query to calculate the difference in days until a date reach another date. Something like the "how many days until my birthday".
Current_date | Reach_date
2000-01-01 | 2015-01-03
-- Should Return: 2
2015-03-01 | 2021-03-05
-- Should Return: 4
The most similar built-in function I found to solve this problem, was using "age()", but it returns me "year, month and days":
select age(current_date,reach_date) from sample_table;
age
-------------------------
3 years 10 mons 1 day
I also tried to use "extract()" trying to get the difference in days, but it just returns me the part of the age function of the days. At my last sample, instead of it returns me more than 1000 days, it returns me just 1.

Answer 1

Try if this works for you. It checks where it's a leap year to calculate the difference correctly, and then uses different logic to calculate the difference between the dates depending on whether the dates are in the same year or not.
with cte as
(
SELECT *,
CASE WHEN extract(year from CurrentDate)::INT % 4 = 0
and (extract(year from CurrentDate)::INT % 100 <> 0
or extract(year from CurrentDate)::INT % 400 = 0)
THEN TRUE
ELSE FALSE
END AS isLeapYear,
Extract(day from (Reach_date - CurrentDate)) AS diff_in_days
FRoM test
)
SELECT CurrentDate,
Reach_date,
CASE WHEN isLeapYear
THEN
CASE WHEN diff_in_days < 366
THEN diff_in_days
ELSE Extract(day from AGE(Reach_date, CurrentDate))
END
ELSE CASE WHEN diff_in_days < 365
THEN diff_in_days
ELSE Extract(day from AGE(Reach_date, CurrentDate))
END
END AS diff
FROM cte
Test here: SQL Fiddle

Answer 2

SELECT
d_date,
'2021-01-01'::date - '2020-01-01'::date AS diff_2021_minus_2020,
CASE WHEN (date_part('month', d_date)::integer) = 1
AND (date_part('day', d_date)::integer) = 1 THEN
(date_trunc('year', d_date) + interval '1 year')::date - date_trunc('year', d_date)::date
WHEN ((d_date - (date_trunc('year', d_date))::date)) <= 182 THEN
(d_date - (date_trunc('year', d_date))::date)
ELSE
365 - (d_date - (date_trunc('year', d_date))::date)
END AS till_to_birthday
FROM (
VALUES ('2021-12-01'::date),
('2021-06-01'::date),
('2020-01-01'::date),
('2021-01-01'::date),
('2021-09-01'::date),
('2021-11-01'::date),
('2020-06-01'::date)) s (d_date);
returns:
+------------+----------------------+------------------+
| d_date | diff_2021_minus_2020 | till_to_birthday |
+------------+----------------------+------------------+
| 2021-12-01 | 366 | 31 |
| 2021-06-01 | 366 | 151 |
| 2020-01-01 | 366 | 366 |
| 2021-01-01 | 366 | 365 |
| 2021-09-01 | 366 | 122 |
| 2021-11-01 | 366 | 61 |
| 2020-06-01 | 366 | 152 |
+------------+----------------------+------------------+

Answer 3

The behavior that you've got with using age() is because extract() only extract the amount of days but it won't convert months and years into days for you before extraction.
On a SQL Server you could use DATEDIFF() but in Postgre you have to compute it yourself by substracting dates, as shown in this answer.
There's also few examples with all the time units here.