找出2个日期之间的月数

Ken*_*han 3 sql postgresql

select 
  (age('2012-11-30 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp)),
  (age('2012-12-31 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp)),
  (age('2013-01-31 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp)),
  (age('2013-02-28 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp))
Run Code Online (Sandbox Code Playgroud)

它给出了以下内容:

  0 years 0 mons 30 days 0 hours 0 mins 0.00 secs
  0 years 2 mons 0 days 0 hours 0 mins 0.00 secs
  0 years 3 mons 0 days 0 hours 0 mins 0.00 secs
  0 years 3 mons 28 days 0 hours 0 mins 0.00 secs
Run Code Online (Sandbox Code Playgroud)

但我希望有下个月的定义,我该怎么做?

  0 years 1 mons 0 days 0 hours 0 mins 0.00 secs
  0 years 2 mons 0 days 0 hours 0 mins 0.00 secs
  0 years 3 mons 0 days 0 hours 0 mins 0.00 secs
  0 years 4 mons 0 days 0 hours 0 mins 0.00 secs
Run Code Online (Sandbox Code Playgroud)

kli*_*lin 5

表达方式

age('2012-11-30 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp) 
Run Code Online (Sandbox Code Playgroud)

30 days.我们期待1 month这两个值都指向一个月的最后几天.如果我们将值增加1天,我们将获得下个月的第一天和

age('2012-12-01 00:00:00'::timestamp, '2012-11-01 00:00:00'::timestamp)
Run Code Online (Sandbox Code Playgroud)

会给我们一个月的预期.因此,让我们检查一下是否有两个月的最后一天,在这种情况下,返回接下来几天的年龄间隔.在其他情况下,我们将返回原始值的年龄间隔:

create or replace function age_m (t1 timestamp, t2 timestamp)
returns interval language plpgsql immutable
as $$
declare
    _t1 timestamp = t1+ interval '1 day';
    _t2 timestamp = t2+ interval '1 day';
begin
    if extract(day from _t1) = 1 and extract(day from _t2) = 1 then
        return age(_t1, _t2);
    else
        return age(t1, t2);
    end if;
end $$;
Run Code Online (Sandbox Code Playgroud)

一些例子:

with my_table(date1, date2) as (
values
    ('2012-11-30 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp),
    ('2012-12-31 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp),
    ('2013-01-31 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp),
    ('2013-02-28 00:00:00'::timestamp, '2012-10-31 00:00:00'::timestamp)
)

select *, age(date1, date2), age_m(date1, date2)
from my_table

        date1        |        date2        |      age       | age_m  
---------------------+---------------------+----------------+--------
 2012-11-30 00:00:00 | 2012-10-31 00:00:00 | 30 days        | 1 mon
 2012-12-31 00:00:00 | 2012-10-31 00:00:00 | 2 mons         | 2 mons
 2013-01-31 00:00:00 | 2012-10-31 00:00:00 | 3 mons         | 3 mons
 2013-02-28 00:00:00 | 2012-10-31 00:00:00 | 3 mons 28 days | 4 mons
(4 rows)
Run Code Online (Sandbox Code Playgroud)