分享程序员开发的那些事...
更新时间：2022-04-14 09:11:47
根据您的问题 工作时间 : Mo–Fr，08 :00–15:00 .
According to your question working hours are: Mo–Fr, 08:00–15:00.
以 1小时为单位运行.分数被忽略，因此不是 precise 而是简单的:
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';
函数> 如果结尾大于起点，则generate_series() 会生成一行，而在给定间隔(1小时)内，每个 full 都会生成另一行.每小时输入 的次数.要忽略小数小时，请从末尾减去1小时.而且不要计算从14:00开始的小时数.
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.
对time进行简单(且非常便宜)的转换可以轻松确定排位赛时间.
A simple (and very cheap) cast to time makes it easy to identify qualifying hours.
一个小时的分数将被忽略，即使该间隔开始和结尾的分数加起来等于一个小时或更多.
Fractions of an hour are ignored, even if fractions at begin and end of the interval would add up to an hour or more.
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
查询:
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;
要获得更高的精度，可以使用较小的时间单位.例如5分钟的片:
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;
单位越小，成本越高..
The smaller the unit the higher the cost.
与新的 LATERAL 功能结合使用对于Postgres 9.3，上面的查询可以这样写:
In combination with the new LATERAL feature in Postgres 9.3, the above query can then be written as:
1小时精度:
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分钟精度:
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;
这具有附加优势，与上述版本一样，结果中也不排除包含零工作时间的间隔.
This has the additional advantage that intervals containing zero working hours are not excluded from the result like in the above versions.
有关LATERAL的更多信息:
或者您分别处理时间范围的开始和结束，以使 exact 的结果精确到微秒.使查询更复杂，但更便宜且更精确:
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;
新的范围类型结合简单的功能，适用于仅跨越一天的时间范围:
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;
如果您的范围从未跨越多天，那么这就是您所需要的.
否则，使用此包装器函数来处理 any 间隔:
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;
致电:
SELECT t_id, f_worktime(t_start, t_end) AS worktime
FROM t
ORDER BY 1;