PostgreSQL: days/months/years between two dates

Question

I am looking for a way to implement the SQLServer-function datediff in PostgreSQL. That is,

This function returns the count (as a signed integer valu

Answer 1

SELECT
  AGE('2012-03-05', '2010-04-01'),
  DATE_PART('year', AGE('2012-03-05', '2010-04-01')) AS years,
  DATE_PART('month', AGE('2012-03-05', '2010-04-01')) AS months,
  DATE_PART('day', AGE('2012-03-05', '2010-04-01')) AS days;

This will give you full years, month, days ... between two dates:

          age          | years | months | days
-----------------------+-------+--------+------
 1 year 11 mons 4 days |     1 |     11 |    4

More detailed datediff information.

Answer 2

This question is full of misunderstandings. First lets understand the question fully. The asker wants to get the same result as for when running the MS SQL Server function DATEDIFF ( datepart , startdate , enddate ) where datepart takes dd, mm, or yy.

This function is defined by:

This function returns the count (as a signed integer value) of the specified datepart boundaries crossed between the specified startdate and enddate.

That means how many day boundaries, month boundaries, or year boundaries, are crossed. Not how many days, months, or years it is between them. That's why datediff(yy, '2010-04-01', '2012-03-05') is 2, and not 1. There is less than 2 years between those dates, meaning only 1 whole year has passed, but 2 year boundaries have crossed, from 2010 to 2011, and from 2011 to 2012.

The following are my best attempt at replicating the logic correctly.

-- datediff(dd`, '2010-04-01', '2012-03-05') = 704 // 704 changes of day in this interval
select ('2012-03-05'::date - '2010-04-01'::date );
-- 704 changes of day

-- datediff(mm, '2010-04-01', '2012-03-05') = 23  // 23 changes of month
select (date_part('year', '2012-03-05'::date) - date_part('year', '2010-04-01'::date)) * 12 + date_part('month', '2012-03-05'::date) - date_part('month', '2010-04-01'::date)
-- 23 changes of month

-- datediff(yy, '2010-04-01', '2012-03-05') = 2   // 2 changes of year
select date_part('year', '2012-03-05'::date) - date_part('year', '2010-04-01'::date);
-- 2 changes of year

Answer 3

SELECT date_part ('year', f) * 12
     + date_part ('month', f)
FROM age ('2015-06-12'::DATE, '2014-12-01'::DATE) f

Result: 6

Answer 4

I would like to expand on Riki_tiki_tavi's answer and get the data out there. I have created a datediff function that does almost everything sql server does. So that way we can take into account any unit.

create function datediff(units character varying, start_t timestamp without time zone, end_t timestamp without time zone) returns integer
language plpgsql
 as
 $$
DECLARE
 diff_interval INTERVAL; 
 diff INT = 0;
 years_diff INT = 0;
BEGIN
 IF units IN ('yy', 'yyyy', 'year', 'mm', 'm', 'month') THEN
   years_diff = DATE_PART('year', end_t) - DATE_PART('year', start_t);

   IF units IN ('yy', 'yyyy', 'year') THEN
     -- SQL Server does not count full years passed (only difference between year parts)
     RETURN years_diff;
   ELSE
     -- If end month is less than start month it will subtracted
     RETURN years_diff * 12 + (DATE_PART('month', end_t) - DATE_PART('month', start_t)); 
   END IF;
 END IF;

 -- Minus operator returns interval 'DDD days HH:MI:SS'  
 diff_interval = end_t - start_t;

 diff = diff + DATE_PART('day', diff_interval);

 IF units IN ('wk', 'ww', 'week') THEN
   diff = diff/7;
   RETURN diff;
 END IF;

 IF units IN ('dd', 'd', 'day') THEN
   RETURN diff;
 END IF;

 diff = diff * 24 + DATE_PART('hour', diff_interval); 

 IF units IN ('hh', 'hour') THEN
    RETURN diff;
 END IF;

 diff = diff * 60 + DATE_PART('minute', diff_interval);

 IF units IN ('mi', 'n', 'minute') THEN
    RETURN diff;
 END IF;

 diff = diff * 60 + DATE_PART('second', diff_interval);

 RETURN diff;
END;
$$;