Calculate working hours between 2 dates in PostgreSQL

According to your question working hours are: Mo–Fr, 08:00–15:00.

Rounded results

For just two given timestamps

Operating on units of 1 hour. Fractions are ignored, therefore not precise but simple:

SELECT count(*) AS work_hours
FROM   generate_series (timestamp '2013-06-24 13:30'
                      , timestamp '2013-06-24 15:29' - interval '1h'
                      , interval '1h') h
WHERE  EXTRACT(ISODOW FROM h) < 6
AND    h::time >= '08:00'
AND    h::time <= '14:00';

• The function generate_series() generates one row if the end is greater than the start and another row for every full given interval (1 hour). This wold count every hour entered into. To ignore fractional hours, subtract 1 hour from the end. And don't count hours starting before 14:00.

• Use the field pattern ISODOW instead of DOW for EXTRACT() to simplify expressions. Returns 7 instead of 0 for Sundays.

• A simple (and very cheap) cast to time makes it easy to identify qualifying hours.

• Fractions of an hour are ignored, even if fractions at begin and end of the interval would add up to an hour or more.

For a whole table

CREATE TEMP TABLE t (t_id int PRIMARY KEY, t_start timestamp, t_end timestamp);
INSERT INTO t VALUES 
  (1, '2009-12-03 14:00', '2009-12-04 09:00')
 ,(2, '2009-12-03 15:00', '2009-12-07 08:00')  -- examples in question
 ,(3, '2013-06-24 07:00', '2013-06-24 12:00')
 ,(4, '2013-06-24 12:00', '2013-06-24 23:00')
 ,(5, '2013-06-23 13:00', '2013-06-25 11:00')
 ,(6, '2013-06-23 14:01', '2013-06-24 08:59');  -- max. fractions at begin and end

Query:

SELECT t_id, count(*) AS work_hours
FROM  (
   SELECT t_id, generate_series (t_start, t_end - interval '1h', interval '1h') AS h
   FROM   t
   ) sub
WHERE  EXTRACT(ISODOW FROM h) < 6
AND    h::time >= '08:00'
AND    h::time <= '14:00'
GROUP  BY 1
ORDER  BY 1;

SQL Fiddle.

More precision

To get more precision you can use smaller time units. 5-minute slices for instance:

SELECT t_id, count(*) * interval '5 min' AS work_interval
FROM  (
   SELECT t_id, generate_series (t_start, t_end - interval '5 min', interval '5 min') AS h
   FROM   t
   ) sub
WHERE  EXTRACT(ISODOW FROM h) < 6
AND    h::time >= '08:00'
AND    h::time <= '14:55'  -- 15.00 - interval '5 min'
GROUP  BY 1
ORDER  BY 1;

The smaller the unit the higher the cost.

Cleaner with LATERAL in Postgres 9.3+

In combination with the new LATERAL feature in Postgres 9.3, the above query can then be written as:

1-hour precision:

SELECT t.t_id, h.work_hours
FROM   t
LEFT   JOIN LATERAL (
   SELECT count(*) AS work_hours
   FROM   generate_series (t.t_start, t.t_end - interval '1h', interval '1h') h
   WHERE  EXTRACT(ISODOW FROM h) < 6
   AND    h::time >= '08:00'
   AND    h::time <= '14:00'
   ) h ON TRUE
ORDER  BY 1;

5-minute precision:

SELECT t.t_id, h.work_interval
FROM   t
LEFT   JOIN LATERAL (
   SELECT count(*) * interval '5 min' AS work_interval
   FROM   generate_series (t.t_start, t.t_end - interval '5 min', interval '5 min') h
   WHERE  EXTRACT(ISODOW FROM h) < 6
   AND    h::time >= '08:00'
   AND    h::time <= '14:55'
   ) h ON TRUE
ORDER  BY 1;

This has the additional advantage that intervals containing zero working hours are not excluded from the result like in the above versions.

More about LATERAL:

• Find most common elements in array with a group by
• Insert multiple rows in one table based on number in another table

Exact results

Postgres 8.4+

Or you deal with start and end of the time frame separately to get exact results to the microsecond. Makes the query more complex, but cheaper and exact:

WITH var AS (SELECT '08:00'::time  AS v_start
                  , '15:00'::time  AS v_end)
SELECT t_id
     , COALESCE(h.h, '0')  -- add / subtract fractions
       - CASE WHEN EXTRACT(ISODOW FROM t_start) < 6
               AND t_start::time > v_start
               AND t_start::time < v_end
         THEN t_start - date_trunc('hour', t_start)
         ELSE '0'::interval END
       + CASE WHEN EXTRACT(ISODOW FROM t_end) < 6
               AND t_end::time > v_start
               AND t_end::time < v_end
         THEN t_end - date_trunc('hour', t_end)
         ELSE '0'::interval END                 AS work_interval
FROM   t CROSS JOIN var
LEFT   JOIN (  -- count full hours, similar to above solutions
   SELECT t_id, count(*)::int * interval '1h' AS h
   FROM  (
      SELECT t_id, v_start, v_end
           , generate_series (date_trunc('hour', t_start)
                            , date_trunc('hour', t_end) - interval '1h'
                            , interval '1h') AS h
      FROM   t, var
      ) sub
   WHERE  EXTRACT(ISODOW FROM h) < 6
   AND    h::time >= v_start
   AND    h::time <= v_end - interval '1h'
   GROUP  BY 1
   ) h USING (t_id)
ORDER  BY 1;

SQL Fiddle.

Postgres 9.2+ with tsrange

The new range types offer a more elegant solution for exact results in combination with the intersection operator *:

Simple function for time ranges spanning only one day:

CREATE OR REPLACE FUNCTION f_worktime_1day(_start timestamp, _end timestamp)
  RETURNS interval AS
$func$  -- _start & _end within one calendar day! - you may want to check ...
SELECT CASE WHEN extract(ISODOW from _start) < 6 THEN (
   SELECT COALESCE(upper(h) - lower(h), '0')
   FROM  (
      SELECT tsrange '[2000-1-1 08:00, 2000-1-1 15:00)' -- hours hard coded
           * tsrange( '2000-1-1'::date + _start::time
                    , '2000-1-1'::date + _end::time ) AS h
      ) sub
   ) ELSE '0' END
$func$  LANGUAGE sql IMMUTABLE;

If your ranges never span multiple days, that's all you need.
Else, use this wrapper function to deal with any interval:

CREATE OR REPLACE FUNCTION f_worktime(_start timestamp
                                    , _end timestamp
                                    , OUT work_time interval) AS
$func$
BEGIN
   CASE _end::date - _start::date  -- spanning how many days?
   WHEN 0 THEN                     -- all in one calendar day
      work_time := f_worktime_1day(_start, _end);
   WHEN 1 THEN                     -- wrap around midnight once
      work_time := f_worktime_1day(_start, NULL)
                +  f_worktime_1day(_end::date, _end);
   ELSE                            -- multiple days
      work_time := f_worktime_1day(_start, NULL)
                +  f_worktime_1day(_end::date, _end)
                + (SELECT count(*) * interval '7:00'  -- workday hard coded!
                   FROM   generate_series(_start::date + 1
                                        , _end::date   - 1, '1 day') AS t
                   WHERE  extract(ISODOW from t) < 6);
   END CASE;
END
$func$  LANGUAGE plpgsql IMMUTABLE;

Call:

SELECT t_id, f_worktime(t_start, t_end) AS worktime
FROM   t
ORDER  BY 1;

SQL Fiddle.

How about this: create a small table with 24*7 rows, one row for each hour in a week.

CREATE TABLE hours (
  hour timestamp not null,
  is_working boolean not null
);

INSERT INTO hours (hour, is_working) VALUES
 ('2009-11-2 00:00:00', false),
 ('2009-11-2 01:00:00', false),
 . . .
 ('2009-11-2 08:00:00', true),
 . . .
 ('2009-11-2 15:00:00', true),
 ('2009-11-2 16:00:00', false),
 . . .
 ('2009-11-2 23:00:00', false);

Likewise add 24 rows for each of the other days. It doesn't matter what year or month you give, as you'll see in a moment. You just need to represent all seven days of the week.

SELECT t.id, t.start, t.end, SUM(CASE WHEN h.is_working THEN 1 ELSE 0 END) AS hours_worked
FROM mytable t JOIN hours h 
ON (EXTRACT(DOW FROM TIMESTAMP h.hour) BETWEEN EXTRACT(DOW FROM TIMESTAMP t.start) 
      AND EXTRACT(DOW FROM TIMESTAMP t.end))
  AND (EXTRACT(DOW FROM TIMESTAMP h.hour) > EXTRACT(DOW FROM TIMESTAMP t.start)
      OR EXTRACT(HOUR FROM TIMESTAMP h.hour) >= EXTRACT(HOUR FROM TIMESTAMP t.start))
  AND (EXTRACT(DOW FROM TIMESTAMP h.hour) < EXTRACT(DOW FROM TIMESTAMP t.end)
      OR EXTRACT(HOUR FROM TIMESTAMP h.hour) <= EXTRACT(HOUR FROM TIMESTAMP t.end))
GROUP BY t.id, t.start, t.end;