Answers for "postgres get date difference in days"

SQL

postgresql difference between two dates in days

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

Posted by: Guest on March-11-2021

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-1

postgresql get difference between two dates

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'));

Posted by: Guest on April-08-2020

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postgres between dates

SELECT user_id 
FROM user_logs 
WHERE login_date BETWEEN '2014-02-01' AND '2014-03-01'

Posted by: Guest on September-03-2021

Source

-1

postgres time difference in minutes

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;

Posted by: Guest on February-07-2020

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