如何消除Oracle的非工作时间 [英] How to eliminate non-working hours in Oracle
问题描述
我有两列DateTime类型.第一个在进程启动时存储DateTime，另一个在该进程完成时存储DateTime.我想计算完成工作所需的总工作时间.工作时间是上午10点至晚上7点，周日不营业.
I have two columns of DateTime type. The first one stores the DateTime when a process it started, the other one stores the DateTime when that process is finished. I want to calculate the total working hours put in to complete the work. The working hours are 10 am to 7 pm and Sundays are off.
以下是我简要介绍的一些摘要:
Here are a few snippets that I've jotted down:
SELECT col1, col2, floor(((date2-date1)*24*60*60)/3600)
|| ' HOURS ' ||
floor((((date2-date1)*24*60*60) -
floor(((date2-date1)*24*60*60)/3600)*3600)/60)
|| ' MINUTES ' ||
round((((date2-date1)*24*60*60) -
Floor(((date2-date1)*24*60*60)/3600)*3600 -
(floor((((date2-date1)*24*60*60) -
floor(((date2-date1)*24*60*60)/3600)*3600)/60)*60) ))
|| ' SECS ' Time_Difference
From Table_Name;
然后
Select To_Number(To_Char(date1, 'HH24')) || ':'
|| to_number(to_char(date1, 'MI')) || ':'||
to_number(to_char(date1, 'SS')) from Table_Name
请帮助.
对于您提到的上述解决方案，这就是结果！
To the above solution you have mentioned, this is the result!
START DATE DAY FINISH DATE DAY Date Diff Total Hours Work Hours
07-AUG-12 21:55:21 TUE 08-AUG-12 11:09:10 WED 0 13:13:49.0 13 13
13-NOV-12 15:45:25 TUE 14-NOV-12 10:41:42 WED 0 18:56:17.0 18 18
20-DEC-12 20:31:03 THU 21-DEC-12 11:03:36 FRI 0 14:32:33.0 14 14
14-MAR-13 20:39:00 THU 15-MAR-13 11:00:04 FRI 0 14:21:4.0 14 14
07-JUN-12 21:17:36 THU 08-JUN-12 11:02:23 FRI 0 13:44:47.0 13 13
18-SEP-12 20:48:27 TUE 19-SEP-12 11:07:35 WED 0 14:19:8.0 14 14
新方案: https://stackoverflow. com/questions/17248741/如何在Oracle中两次约会之间获取业务时间
推荐答案
如果我理解正确，则要计算开始日期和结束日期之间的时差，不包括上午10点之前和下午7点之后的时间.
If i understand correctly, you want to calculate the difference between the start and finish date excluding the time before 10 am and after 7 pm.
这是示例查询和 sql小提琴.
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，结束时间是否在下午7点之后.如果是这样，请将时间设置为上午10点和晚上7点.
- 然后减去日期并将结果转换为间隔类型.
- 然后从时间间隔中提取小时，分钟和秒.
注意::该查询假设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.
Note: This query assumes that both dates fall on same day and both are not before 10am or after 7 pm.
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.
UPDATE: To exclude holidays, the query will become complicated. I suggest writing three functions and use these functions in the query.
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接近的开始时间.
If the start time is outside the work hours, modify the start time to next nearest start time.
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;
如果结束时间不在工作时间之内，请将其修改为最近的结束时间.
If the finish time is outside the work hours, modify it to the previous nearest finish time.
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;
此功能可在计算持续时间时找到要排除的天数.
This function finds the no of days to be excluded while calculating the duration.
因此，最终查询应该是这样的
So, the final query should be something like this,
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，则将其设置为0，以将其忽略.
If the duration is less than 0, ignore it by setting it to 0.
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