postgres date difference seconds
snippet in sql

postgres date difference seconds

user7152

-- difference in seconds between two dates

select extract(epoch from ('2020-03-30 09:55:56'::timestamp - '2020-03-30 08:54:55'::timestamp));
-- result : 3661

postgresql get difference between two dates

user6754

select age('2010-04-01', '2012-03-05'),
       date_part('year',age('2010-04-01', '2012-03-05')),
       date_part('month',age('2010-04-01', '2012-03-05')),
       date_part('day',age('2010-04-01', '2012-03-05'));

DATEDIFF minute postgres

user1436

  -- Difference between Dec 30, 2011 08:54:55 and  Dec 30, 2011 08:56:10 in minutes
  SELECT (DATE_PART('day', '2011-12-30 08:56:10'::timestamp - '2011-12-30 08:54:55'::timestamp) * 24 * 60 + 
               DATE_PART('hour', '2011-12-30 08:56:10'::timestamp - '2011-12-30 08:54:55'::timestamp)) * 60 +
               DATE_PART('minute', '2011-12-30 08:56:10'::timestamp - '2011-12-30 08:54:55'::timestamp);
  -- Result: 1
 
  -- Time only
  SELECT DATE_PART('hour', '08:56:10'::time - '08:54:55'::time) * 60 +
              DATE_PART('minute', '08:56:10'::time - '08:54:55'::time);
  -- Result: 1

postgres time difference in minutes

user2034

   CREATE OR REPLACE FUNCTION DateDiff (units VARCHAR(30), start_t TIME, end_t TIME) 
     RETURNS INT AS $$
   DECLARE
     diff_interval INTERVAL; 
     diff INT = 0;
   BEGIN
     -- Minus operator for TIME returns interval 'HH:MI:SS'  
     diff_interval = end_t - start_t;
 
     diff = 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;
   $$ LANGUAGE plpgsql;