问题
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
extract(year from age(current_date, '2012-12-09')) * 12
+
extract(month from age(current_date, '2012-12-09'))
Is there any other ways(built in functions) in postgres??
回答1:
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.
回答2:
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')
回答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;
回答4:
SELECT date_part ('year', f) * 12
+ date_part ('month', f)
FROM age (CURRENT_DATE, '2014-12-01') f
来源:https://stackoverflow.com/questions/14251456/months-between-two-dates-function