Months between two dates function

Question

In oracle i can find out no:of months between using MONTHS_BETWEEN function.

In postgres i am using extract function for this. eg.like

select 
    ex         


        

Answer 1

SELECT date_part ('year', f) * 12
      + date_part ('month', f)
FROM age (CURRENT_DATE, '2014-12-01') f

Answer 2

This is easy to re-implement in PostgreSQL just using SQL functions to tidy up what you've already got:

create function months_of(interval)
 returns int strict immutable language sql as $$
  select extract(years from $1)::int * 12 + extract(month from $1)::int
$$;

create function months_between(date, date)
 returns int strict immutable language sql as $$
   select abs(months_of(age($1, $2)))
$$;

And now select months_between('1978-06-20', '2011-12-09') produces 401.

Answer 3

You can use UDF, e.g. I've found the following here:

   CREATE OR REPLACE FUNCTION DateDiff (units VARCHAR(30), start_t TIMESTAMP, end_t TIMESTAMP) 
     RETURNS INT 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;
   $$ LANGUAGE plpgsql;

Answer 4

Unfortunately it seems not, because extract(month ...) returns the number of months modulo 12.

There is one small simplification you can make; remove the first parameter of age() - the default is age from current_date, so these two are equivalent:

age(current_date, '2012-12-09')
age('2012-12-09')