在oracle中我可以发现没有:使用MONTHS_BETWEEN函数之间的几个月。
在postgres中,我正在使用提取功能。 eg.like
select
extract(year from age(current_date, '2012-12-09')) * 12
+
extract(month from age(current_date, '2012-12-09'))
postgres中还有其他方法(内置函数)吗?
答案 0 :(得分:10)
这很容易在PostgreSQL中重新实现,只需使用SQL函数来整理你已经拥有的东西:
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 $$
select abs(months_of(age($1, $2)))
$$;
现在select months_between('1978-06-20', '2011-12-09')
生成401。
答案 1 :(得分:3)
不幸的是,似乎没有,因为extract(month ...)
会返回 modulo 12 的月数。
你可以做一个简单的简化;删除age()
的第一个参数 - 默认值为current_date
的年龄,因此这两个参数相同:
age(current_date, '2012-12-09')
age('2012-12-09')
答案 2 :(得分:1)
您可以使用UDF,例如我找到了以下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;
答案 3 :(得分:0)
SELECT date_part ('year', f) * 12
+ date_part ('month', f)
FROM age (CURRENT_DATE, '2014-12-01') f