Question

我想计算两个日期之间的月数。

做：

SELECT TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP '2011-04-30 14:38:40';

返回： 0年0个月409天20小时0分0.00秒

所以：

SELECT extract(month from TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP '2011-04-30 14:38:40');

返回0.

Answer 1

age函数返回interval：

age(timestamp1, timestamp2)

然后我们尝试从间隔中提取年份和月份并相应地添加它们：

select extract(year from age(timestamp1, timestamp2)) * 12 +
extract(month from age(timestamp1, timestamp2))

Answer 2

age函数给出了合理的间隔：

SELECT age(TIMESTAMP '2012-06-13 10:38:40', TIMESTAMP '2011-04-30 14:38:40');

返回1 year 1 mon 12 days 20:00:00，您可以轻松使用EXTRACT来计算月数：

SELECT EXTRACT(YEAR FROM age) * 12 + EXTRACT(MONTH FROM age) AS months_between
FROM age(TIMESTAMP '2012-06-13 10:38:40', TIMESTAMP '2011-04-30 14:38:40') AS t(age);

Answer 3

如果您多次执行此操作，则可以定义以下function：

CREATE FUNCTION months_between (t_start timestamp, t_end timestamp)
    RETURNS integer
    AS $$
        SELECT
            (
                12 * extract('years' from a.i) + extract('months' from a.i)
            )::integer
        from (
            values (justify_interval($2 - $1))
        ) as a (i)
    $$
    LANGUAGE SQL
    IMMUTABLE
    RETURNS NULL ON NULL INPUT;

这样你就可以

了

SELECT months_between('2015-01-01', now());

Answer 4

请注意，当您尝试使用日历月差异时，@ram和@angelin的投票最多是不准确的。

select extract(year from age(timestamp1, timestamp2))*12 + extract(month from age(timestamp1, timestamp2))

例如，如果您尝试这样做：

select extract(year from age('2018-02-02'::date, '2018-03-01'::date))*12 + extract(month from age('2018-02-02'::date , '2018-03-01'::date))

结果将为0 ，但无论日期之间的天数如何，从2月开始的3月之间的月份应为1。

因此公式应类似于以下说法，我们以timestamp1和timestamp2开头：

（（（year2-year1）* 12）-month1 + month2 =两个时间戳之间的日历月

pg中的

会被翻译为：

select ((extract('years' from '2018-03-01 00:00:00'::timestamp)::int -  extract('years' from '2018-02-02 00:00:00'::timestamp)::int) * 12) 
    - extract('month' from '2018-02-02 00:00:00'::timestamp)::int + extract('month' from '2018-03-01 00:00:00'::timestamp)::int;

您可以创建类似的功能

CREATE FUNCTION months_between (t_start timestamp, t_end timestamp)
RETURNS integer
AS $$
select ((extract('years' from $2)::int -  extract('years' from $1)::int) * 12) 
    - extract('month' from $1)::int + extract('month' from $2)::int
$$
LANGUAGE SQL
IMMUTABLE
RETURNS NULL ON NULL INPUT;

Answer 5

给出两个月的两个月的差异

   SELECT ((extract( year FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract( year FROM TIMESTAMP '2011-04-30 14:38:40' )) *12) + extract(MONTH FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract(MONTH FROM TIMESTAMP '2011-04-30 14:38:40' );

结果：14

必须分别提取几个月的日期，然后两个结果的差异

Answer 6

SELECT date_part ('year', f) * 12
      + date_part ('month', f)
FROM age ('2015-06-12', '2014-12-01') f

结果：6个月

Answer 7

我曾经遇到过同样的问题并且写了这个......这很难看：

postgres=>  SELECT floor((extract(EPOCH FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract(EPOCH FROM TIMESTAMP '2005-04-30 14:38:40' ))/30.43/24/3600);
 floor 
-------
    85
(1 row)

在这个解决方案中，“一个月”被定义为30.43天，所以它可能会在较短的时间内产生一些意想不到的结果。

Answer 8

试试这个解决方案：

SELECT extract (MONTH FROM age('2014-03-03 00:00:00'::timestamp, 
'2013-02-03 00:00:00'::timestamp)) + 12 * extract (YEAR FROM age('2014-03-03   
00:00:00'::timestamp, '2013-02-03 00:00:00'::timestamp)) as age_in_month;

Answer 9

SELECT floor(extract(days from TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP
'2011-04-30 14:38:40')/30.43)::integer as months;

给出近似值但避免重复时间戳。这使用来自tobixen's answer的提示除以30来代替30而对于计算月数的长时间间隔不太正确。

Answer 10

按年份和月份提取将取决于月份：

select extract(year from age('2016-11-30'::timestamp, '2015-10-15'::timestamp)); --> 1
select extract(month from age('2016-11-30'::timestamp, '2015-10-15'::timestamp)); --> 1
--> Total 13 months

这种方法可以维持几个月的时间（感谢使用除虫剂Tobixen）

select round(('2016-11-30'::date - '2015-10-15'::date)::numeric /30.43, 1); --> 13.5 months

Answer 11

我做了一个这样的函数：

/* similar to ORACLE's MONTHS_BETWEEN */
CREATE OR REPLACE FUNCTION ORACLE_MONTHS_BETWEEN(date_from DATE, date_to DATE)
RETURNS REAL LANGUAGE plpgsql
AS
$$
DECLARE age INTERVAL;
declare rtn real;
BEGIN
    age := age(date_from, date_to);
    rtn := date_part('year', age) * 12 + date_part('month', age) + date_part('day', age)/31::real;
    return rtn;
END;
$$;

Oracle 示例）

SELECT MONTHS_BETWEEN
(TO_DATE('2015-02-02','YYYY-MM-DD'), TO_DATE('2014-12-01','YYYY-MM-DD') ) 
"Months" FROM DUAL;
--result is: 2.03225806451612903225806451612903225806

我的 PostgreSQL 函数示例)

select ORACLE_MONTHS_BETWEEN('2015-02-02'::date, '2014-12-01'::date) Months;
-- result is: 2.032258

根据结果，您可以使用 CEIL()/FLOOR() 进行四舍五入。

select ceil(2.032258)  --3
select floor(2.032258) --2

Answer 12

尝试;

select extract(month from  age('2012-06-13 10:38:40'::timestamp, '2011-04-30 14:38:40'::timestamp)) as my_months;

计算postgresql上两个时间戳之间的月份？

12 个答案: