How do I convert an interval into a number of hours with postgres?

Say I have an interval like

4 days 10:00:00

in postgres. How do I convert that to a number of hours (106 in this case?) Is there a function or should I bite the bullet and do something like

extract(days, my_interval) * 24 + extract(hours, my_interval)

Probably the easiest way is:

SELECT EXTRACT(epoch FROM my_interval)/3600

If you want integer i.e. number of days:

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

To get the number of days the easiest way would be:

SELECT EXTRACT(DAY FROM NOW() - '2014-08-02 08:10:56');

As far as I know it would return the same as:

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

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));

The ::int conversion follows the principle of rounding. If you want a different result such as rounding down, you can use the corresponding math function such as floor.

If you convert table field:

1. Define the field so it contains seconds:

    CREATE TABLE IF NOT EXISTS test (
        ...
        field        INTERVAL SECOND(0)
    );
   

2. Extract the value. Remember to cast to int other wise you can get an unpleasant surprise once the intervals are big:

EXTRACT(EPOCH FROM field)::int

I'm working with PostgreSQL 11, and I created a function to get the hours betweeen 2 differents timestamps

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 date 'now()' - date '1955-12-15';

Here is the simple query which calculates total no of days.