我正在尝试计算Oracle select中两个日期之间的工作日。我得到的结论是，我的计算给出了给定日期的大多数结果是正确的（我将它与excel中的NETWORKDAYS进行比较）但有时它在2天到-2天之间变化 - 我不知道为什么......
这是我的代码：
SELECT
((to_char(CompleteDate,'J') - to_char(InstallDate,'J'))+1) - (((to_char(CompleteDate,'WW')+ (52 * ((to_char(CompleteDate,'YYYY') - to_char(InstallDate,'YYYY'))))) - to_char(InstallDate,'WW'))*2) as BusinessDays
FROM TABLE
谢谢！
答案 0 :(得分：23)
解决方案，最后：
SELECT OrderNumber, InstallDate, CompleteDate,
(TRUNC(CompleteDate) - TRUNC(InstallDate) ) +1 -
((((TRUNC(CompleteDate,'D'))-(TRUNC(InstallDate,'D')))/7)*2) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END) as BusinessDays
FROM Orders
ORDER BY OrderNumber;
感谢您的所有回复！
答案 1 :(得分：5)
我考虑了上面讨论的所有不同方法，并提出了一个简单的查询，它给出了两个日期之间一年中每个月的工作日数：
WITH test_data AS
(
SELECT TO_DATE('01-JAN-14') AS start_date,
TO_DATE('31-DEC-14') AS end_date
FROM dual
),
all_dates AS
(
SELECT td.start_date, td.end_date, td.start_date + LEVEL-1 as week_day
FROM test_data td
CONNECT BY td.start_date + LEVEL-1 <= td.end_date)
SELECT TO_CHAR(week_day, 'MON'), COUNT(*)
FROM all_dates
WHERE to_char(week_day, 'dy', 'nls_date_language=AMERICAN') NOT IN ('sun' , 'sat')
GROUP BY TO_CHAR(week_day, 'MON');
请随时根据需要修改查询。
答案 2 :(得分：1)
试试这个：
with holidays as
(
select d from (
select minDate + level -1 d
from (select min(submitDate) minDate, max (completeDate) maxDate
from t)
connect by level <= maxDate - mindate + 1)
where to_char(d, 'dy', 'nls_date_language=AMERICAN') not in ('sun' , 'sat')
)
select t.OrderNo, t.submitDate, t.completeDate, count(*) businessDays
from t join holidays h on h.d between t.submitDate and t.completeDate
group by t.OrderNo, t.submitDate, t.completeDate
order by orderno
答案 3 :(得分：1)
我看到标记的最终解决方案总是不正确。假设，InstallDate是本月的第1天（如果是星期六），而CompleteDate是月份的第16天（如果是星期日）
在这种情况下，实际营业日数是10，但标记的查询结果会给出答案为12.所以，我们也必须处理这类案例，我用过
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SAT' AND TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SUN' THEN 2 ELSE 0 END
来处理它。
SELECT OrderNumber, InstallDate, CompleteDate,
(TRUNC(CompleteDate) - TRUNC(InstallDate) ) +1 -
((((TRUNC(CompleteDate,'D'))-(TRUNC(InstallDate,'D')))/7)*2) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SAT' AND TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SUN' THEN 2 ELSE 0 END)as BusinessDays
FROM Orders
ORDER BY OrderNumber;
答案 4 :(得分：0)
我将我的示例更改为更具可读性并返回总线计数。几天之间。我不知道为什么你需要'J'-朱利安格式。所需要的只是开始/安装和结束/完成日期。您将使用此日期在两个日期之间获得正确的天数。用你的日期替换我的日期，如果需要，添加NLS ......：
SELECT Count(*) BusDaysBtwn
FROM
(
SELECT TO_DATE('2013-02-18', 'YYYY-MM-DD') + LEVEL-1 InstallDate -- MON or any other day
, TO_DATE('2013-02-25', 'YYYY-MM-DD') CompleteDate -- MON or any other day
, TO_CHAR(TO_DATE('2013-02-18', 'YYYY-MM-DD') + LEVEL-1, 'DY') InstallDay -- day of week
FROM dual
CONNECT BY LEVEL <= (TO_DATE('2013-02-25', 'YYYY-MM-DD') - TO_DATE('2013-02-18', 'YYYY-MM-DD')) -- end_date - start_date
)
WHERE InstallDay NOT IN ('SAT', 'SUN')
/
SQL> 5
答案 5 :(得分：0)
接受的解决方案非常接近但在某些情况下似乎是错误的（例如，2015年2月1日至2015年2月28日或2015年5月1日至2015年5月31日）。这是一个精致的版本......
end_date-begin_date+1 /* total days */
- TRUNC(2*(end_date-begin_date+1)/7) /* weekend days in whole weeks */
- (CASE
WHEN TO_CHAR(begin_date,'D') = 1 AND REMAINDER(end_date-begin_date+1,7) > 0 THEN 1
WHEN TO_CHAR(begin_date,'D') = 8 - REMAINDER(end_date-begin_date+1,7) THEN 1
WHEN TO_CHAR(begin_date,'D') > 8 - REMAINDER(end_date-begin_date+1,7) THEN 2
ELSE 0
END) /* weekend days in partial week */
AS business_days
处理7（整周）的倍数的部分是好的。但是，在考虑部分周部分时，它取决于星期偏移量和部分部分中的天数，根据以下矩阵...
654321
1N 111111
2M 100000
3T 210000
4W 221000
5R 222100
6F 222210
7S 222221
答案 6 :(得分：0)
要删除星期日和星期六，您可以使用此
SELECT Base_DateDiff
- (floor((Base_DateDiff + 0 + Start_WeekDay) / 7))
- (floor((Base_DateDiff + 1 + Start_WeekDay) / 7))
FROM (SELECT 1 + TRUNC(InstallDate) - TRUNC(InstallDate, 'IW') Start_WeekDay
, CompleteDate - InstallDate + 1 Base_DateDiff
FROM TABLE) a
Base_DateDiff计算两个日期之间的天数
(floor((Base_DateDiff + 0 + Start_WeekDay) / 7))计算星期日的数量
(floor((Base_DateDiff + 1 + Start_WeekDay) / 7))计算星期六的数量
1 + TRUNC(InstallDate) - TRUNC(InstallDate, 'IW')星期一获得1，星期日获得7
答案 7 :(得分：0)
此查询可用于从指定日期（仅限工作日）起N天后退
例如，从2017-05-17后退15天：
select date_point, closest_saturday - (15 - offset + floor((15 - offset) / 6) * 2) from(
select date_point,
closest_saturday,
(case
when weekday_num > 1 then
weekday_num - 2
else
0
end) offset
from (
select to_date('2017-05-17', 'yyyy-mm-dd') date_point,
to_date('2017-05-17', 'yyyy-mm-dd') - to_char(to_date('2017-05-17', 'yyyy-mm-dd'), 'D') closest_saturday,
to_char(to_date('2017-05-17', 'yyyy-mm-dd'), 'D') weekday_num
from dual
))
一些简短的解释：假设我们想要在给定日期的N天后退 - 找到小于或等于给定日期的最接近的星期六。 - 从最近的星期六开始，回到病房（N - 偏移）天。 offset是最接近的星期六和给定日期（不包括给定日期）之间的工作日数。
*要从星期六（仅限工作日）返回M天，请使用此公式DateOfMonthOfTheSaturday - [M + Floor（M / 6）* 2]
答案 8 :(得分：-1)
这是一个快速而灵活的功能。您可以计算日期范围内的任何工作日。
CREATE OR REPLACE FUNCTION wfportal.cx_count_specific_weekdays( p_week_days VARCHAR2 DEFAULT 'MON,TUE,WED,THU,FRI'
, p_start_date DATE
, p_end_date DATE)
RETURN NUMBER
IS
/***************************************************************************************************************
*
* FUNCTION DESCRIPTION:
*
* This function calculates the total required week days in a date range.
*
* PARAMETERS:
*
* p_week_days VARCHAR2 The week days that need to be counted, comma seperated e.g. MON,TUE,WED,THU,FRU,SAT,SUN
* p_start_date DATE The start date
* p_end_date DATE The end date
*
* CHANGE history
*
* No. Date Changed by Change Description
* ---- ----------- ------------- -------------------------------------------------------------------------
* 0 07-May-2013 yourname Created
*
***************************************************************************************************************/
v_date_end_first_date_range DATE;
v_date_start_last_date_range DATE;
v_total_days_in_the_weeks NUMBER;
v_total_days_first_date_range NUMBER;
v_total_days_last_date_range NUMBER;
v_output NUMBER;
v_error_text CX_ERROR_CODES.ERROR_MESSAGE%TYPE;
--Count the required days in a specific date ranges by using a list of all the weekdays in that range.
CURSOR c_total_days ( v_start_date DATE
, v_end_date DATE ) IS
SELECT COUNT(*) total_days
FROM ( SELECT ( v_start_date + level - 1) days
FROM dual
CONNECT BY LEVEL <= ( v_end_date - v_start_date ) + 1
)
WHERE INSTR( ',' || p_week_days || ',', ',' || TO_CHAR( days, 'DY', 'NLS_DATE_LANGUAGE=english') || ',', 1 ) > 0
;
--Calculate the first and last date range by retrieving the first Sunday after the start date and the last Monday before the end date.
--Calculate the total amount of weeks in between and multiply that with the total required days.
CURSOR c_calculate_new_dates ( v_start_date DATE
, v_end_date DATE ) IS
SELECT date_end_first_date_range
, date_start_last_date_range
, (
(
( date_start_last_date_range - ( date_end_first_date_range + 1 ) )
) / 7
) * total_required_days total_days_in_the_weeks --The total amount of required days
FROM ( SELECT v_start_date + DECODE( TO_CHAR( v_start_date, 'DY', 'NLS_DATE_LANGUAGE=english')
, 'MON', 6
, 'TUE', 5
, 'WED', 4
, 'THU', 3
, 'FRI', 2
, 'SAT', 1
, 'SUN', 0
, 0 ) date_end_first_date_range
, v_end_date - DECODE( TO_CHAR( v_end_date, 'DY', 'NLS_DATE_LANGUAGE=english')
, 'MON', 0
, 'TUE', 1
, 'WED', 2
, 'THU', 3
, 'FRI', 4
, 'SAT', 5
, 'SUN', 6
, 0 ) date_start_last_date_range
, REGEXP_COUNT( p_week_days, ',' ) + 1 total_required_days --Count the commas + 1 to get the total required weekdays
FROM dual
)
;
BEGIN
--Verify that the start date is before the end date
IF p_start_date < p_end_date THEN
--Get the new calculated days.
OPEN c_calculate_new_dates( p_start_date, p_end_date );
FETCH c_calculate_new_dates INTO v_date_end_first_date_range
, v_date_start_last_date_range
, v_total_days_in_the_weeks;
CLOSE c_calculate_new_dates;
--Calculate the days in the first date range
OPEN c_total_days( p_start_date, v_date_end_first_date_range );
FETCH c_total_days INTO v_total_days_first_date_range;
CLOSE c_total_days;
--Calculate the days in the last date range
OPEN c_total_days( v_date_start_last_date_range, p_end_date );
FETCH c_total_days INTO v_total_days_last_date_range;
CLOSE c_total_days;
--Sum the total required days
v_output := v_total_days_first_date_range + v_total_days_last_date_range + v_total_days_in_the_weeks;
ELSE
v_output := 0;
END IF;
RETURN v_output;
EXCEPTION
WHEN OTHERS
THEN
RETURN NULL;
END cx_count_specific_weekdays;
/
答案 9 :(得分：-1)
你去......
在两个日期之间获取工作日（MON到FRI），然后减去假期。
create or replace
FUNCTION calculate_business_days (p_start_date IN DATE, p_end_date IN DATE)
RETURN NUMBER IS
v_holidays NUMBER;
v_start_date DATE := TRUNC (p_start_date);
v_end_date DATE := TRUNC (p_end_date);
BEGIN
IF v_end_date >= v_start_date
THEN
SELECT COUNT (*)
INTO v_holidays
FROM holidays
WHERE day BETWEEN v_start_date AND v_end_date
AND day NOT IN (
SELECT hol.day
FROM holidays hol
WHERE MOD(TO_CHAR(hol.day, 'J'), 7) + 1 IN (6, 7)
);
RETURN GREATEST (NEXT_DAY (v_start_date, 'MON') - v_start_date - 2, 0)
+ ( ( NEXT_DAY (v_end_date, 'MON')
- NEXT_DAY (v_start_date, 'MON')
)
/ 7
)
* 5
- GREATEST (NEXT_DAY (v_end_date, 'MON') - v_end_date - 3, 0)
- v_holidays;
ELSE
RETURN NULL;
END IF;
END calculate_business_days;
之后你可以测试一下，如：
select
calculate_business_days('21-AUG-2013','28-AUG-2013') as business_days
from dual;
答案 10 :(得分：-1)
还有另一种更简单的方法，使用connect by和dual ...
with t as (select to_date('30-sep-2013') end_date, trunc(sysdate) start_date from dual)select count(1) from dual, t where to_char(t.start_date + level, 'D') not in (1,7) connect by t.start_date + level <= t.end_date;
通过connect获取从start_date到end_date的所有日期。然后，您可以排除您不需要的日期，并仅计算所需的日期。
答案 11 :(得分：-1)
这将返回工作日：
(CompleteDate-InstallDate)-2*FLOOR((CompleteDate-InstallDate)/7)-
DECODE(SIGN(TO_CHAR(CompleteDate,'D')-
TO_CHAR(InstallDate,'D')),-1,2,0)+DECODE(TO_CHAR(CompleteDate,'D'),7,1,0)-
DECODE(TO_CHAR(InstallDate,'D'),7,1,0) as BusinessDays,