如果我理解正确，您想计算开始日期和结束日期之间的差异，不包括上午 10 点之前和晚上 7 点之后的时间。
这是示例查询和sql fiddle。
SELECT start_time,
finish_time,
interval_time,
EXTRACT (HOUR FROM interval_time), --extract the hours,mins and seconds from the interval
EXTRACT (MINUTE FROM interval_time),
EXTRACT (SECOND FROM interval_time)
FROM (SELECT start_time,
finish_time,
NUMTODSINTERVAL (
CASE
WHEN finish_time - TRUNC (finish_time) > (19 / 24) --if finish time is after 7pm
THEN
TRUNC (finish_time) + (19 / 24) --set it to 7pm
ELSE
finish_time --else set it to actual finish time
END
- CASE
WHEN start_time - TRUNC (start_time) < (10 / 24) --if start time is before 10 am
THEN
TRUNC (start_time) + (10 / 24) --set it to 10 am.
ELSE
start_time --else set it to the actual start time
END,
'day') --subtract the both and convert the resulting day to interval
interval_time
FROM timings);
我所做的是，
- 检查开始时间是否在上午 10 点之前，结束时间是否在晚上 7 点之后。如果是这样，请将时间设置为上午 10 点和下午 7 点。
- 然后减去日期并将结果天数转换为间隔类型。
- 然后从间隔中提取小时、分钟和秒。
注意：此查询假定两个日期都在同一天，并且都不在上午 10 点之前或晚上 7 点之后。
更新：
要排除假期，查询将变得复杂。我建议编写三个函数并在查询中使用这些函数。
第一个功能：
FUNCTION modify_start_time (p_in_dte DATE) RETURN DATE
----------------------------------
IF p_in_dte - TRUNC (p_in_dte) < (10 / 24)
THEN
RETURN TRUNC (p_in_dte) + (10 / 24);
ELSIF p_in_dte - TRUNC (p_in_dte) > (19 / 24)
THEN
RETURN TRUNC (p_in_dte) + 1 + (10 / 24);
ELSE
RETURN p_in_dte;
END IF;
如果开始时间在工作时间之外，请将开始时间修改为下一个最近的开始时间。
第二个功能：
FUNCTION modify_finish_time (p_in_dte DATE) RETURN DATE
----------------------------------
IF p_in_dte - TRUNC (p_in_dte) > (19 / 24)
THEN
RETURN TRUNC (p_in_dte) + (19 / 24);
ELSIF p_in_dte - TRUNC (p_in_dte) < (10 / 24)
THEN
RETURN TRUNC (p_in_dte) - 1 + (19 / 24);
ELSE
RETURN p_in_dte;
END IF;
如果完成时间在工作时间之外，请将其修改为上一个最近的完成时间。
第三个功能：
FUNCTION get_days_to_exclude (p_in_start_date DATE,
p_in_finish_date DATE) RETURN NUMBER
--------------------------------------------------------
WITH cte --get all days between start and finish date
AS ( SELECT p_in_start_date + LEVEL - 1 dte
FROM DUAL
CONNECT BY LEVEL <= p_in_finish_date + 1 - p_in_starT_date)
SELECT COUNT (1) * 9 / 24 --mutiply the days with work hours in a day
INTO l_num_holidays
FROM cte
WHERE TO_CHAR (dte, 'dy') = 'sun' --find the count of sundays
OR dte IN --fins the count of holidays, assuming leaves are stored in separate table
(SELECT leave_date
FROM leaves
WHERE leave_date BETWEEN p_in_start_date
AND p_in_finish_date);
l_num_holidays :=
l_num_holidays + ( (p_in_finish_date - p_in_start_date) * (15 / 24)); --also, if the dates span more than a day find the non working hours.
RETURN l_num_holidays;
此函数在计算持续时间时查找要排除的天数。
所以，最终的查询应该是这样的，
SELECT start_time,
finish_time,
CASE
WHEN work_duration < 0 THEN NUMTODSINTERVAL (0, 'day')
ELSE NUMTODSINTERVAL (work_duration, 'day')
END
FROM (SELECT start_time, finish_time,
--modify_start_time (start_time), modify_finish_time (finish_time),
modify_finish_time (finish_time)
- modify_start_time (start_time)
- get_days_to_exclude (
TRUNC (modify_start_time (start_time)),
TRUNC (modify_finish_time (finish_time)))
work_duration
FROM timings);
如果持续时间小于 0，则通过将其设置为 0 来忽略它。