我正在使用PostgreSQL 11，我创建了一个函数来获得2个不同时间戳之间的时间

create function analysis.calcHours(datetime1 timestamp, datetime2 timestamp)
    returns integer
    language plpgsql as $$
    declare
        diff interval;
    begin
        diff = datetime2 - datetime1;
        return (abs(extract(days from diff))*24 + abs(extract(hours from diff)))::integer;
    end; $$;

如果你想要整数，即天数:

SELECT (EXTRACT(epoch FROM (SELECT (NOW() - '2014-08-02 08:10:56')))/86400)::int

如果你转换表字段:

定义字段，使其包含秒: 创建表如果不存在测试( .．. 字段INTERVAL秒(0) ）; 提取值。记住，如果转换为int，你可能会在间隔变大时得到一个不愉快的惊喜: EXTRACT(EPOCH FROM field)::int

我正在使用PostgreSQL 11，我创建了一个函数来获得2个不同时间戳之间的时间

create function analysis.calcHours(datetime1 timestamp, datetime2 timestamp)
    returns integer
    language plpgsql as $$
    declare
        diff interval;
    begin
        diff = datetime2 - datetime1;
        return (abs(extract(days from diff))*24 + abs(extract(hours from diff)))::integer;
    end; $$;

select floor((date_part('epoch', order_time - '2016-09-05 00:00:00') / 3600)), count(*)
from od_a_week
group by floor((date_part('epoch', order_time - '2016-09-05 00:00:00') / 3600));

::int转换遵循舍入原则。 如果你想要一个不同的结果，比如四舍五入，你可以使用相应的数学函数，比如floor。

可能最简单的方法是:

SELECT EXTRACT(epoch FROM my_interval)/3600