Excel具有NETWORKDAYS（）函数，可以查找两个日期之间的工作日数。
任何人都有类似MySQL的功能？由于假期增加了复杂性，解决方案不必处理假期。
答案 0 :(得分：60)
此表达式 -
5 * (DATEDIFF(@E, @S) DIV 7) + MID('0123444401233334012222340111123400012345001234550', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)
计算开始日期@S和结束日期@E之间的工作日数。
假设结束日期（@E）不在开始日期之前（@S）。 在相同的开始日期和结束日期与DATEDIFF兼容 给出零工作日。 忽略假期。
数字串的构造如下。创建一个表 开始日期和结束日期，行必须从星期一开始（WEEKDAY 0）并且列也必须以星期一开始。填写 从左上到右下的对角线全部为0（即有0 星期一和星期一，星期二和星期二之间的工作日等）。 每天从对角线开始（必须始终为0）并填写 右边的列，一次一天。如果你降落在 周末日（非营业日）栏，营业日数 不改变，它是从左边进行的。否则，数量 工作日增加一个。当你到达结束时 行循环回到同一行的开头并继续直到你 再次到达对角线。然后继续下一行。
E.g。假设周六和周日不是工作日 -
| M T W T F S S
-|--------------
M| 0 1 2 3 4 4 4
T| 4 0 1 2 3 3 3
W| 3 4 0 1 2 2 2
T| 2 3 4 0 1 1 1
F| 1 2 3 4 0 0 0
S| 1 2 3 4 5 0 0
S| 1 2 3 4 5 5 0
然后将表中的49个值连接成字符串。
如果您发现任何错误，请告诉我。
- 编辑 改进表：
| M T W T F S S
-|--------------
M| 0 1 2 3 4 4 4
T| 4 0 1 2 3 3 3
W| 3 4 0 1 2 2 2
T| 2 3 4 0 1 1 1
F| 1 2 3 4 0 0 0
S| 0 1 2 3 4 0 0
S| 0 1 2 3 4 4 0
改进字符串：'0123444401233334012222340111123400001234000123440'
改进表达：
5 * (DATEDIFF(@E, @S) DIV 7) + MID('0123444401233334012222340111123400001234000123440', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)
答案 1 :(得分：13)
该解决方案使用与罗杰基本相同的方法，除了生成矩阵的方法要复杂得多。注意：此解决方案的此输出与NETWORKDAYS不兼容。
与Rodger的解决方案一样，它计算开始日期（@S）和结束日期（@E）之间的工作日数，而无需定义存储过程。它假定结束日期不在开始日期之前。使用相同的开始和结束日期将产生0.不考虑假期。
这与Rodger的解决方案之间的主要区别在于矩阵和结果的数字串是由一个我没有包含的复杂算法构造的。该算法的输出由单元测试验证（参见下面的测试输入和输出）。在矩阵中，任何给定的x和y值对（WEEKDAY（@S）和WEEKDAY（@E）的交集产生两个值之间的工作日差异。分配顺序实际上并不重要，因为两者被加在一起绘制位置。
营业日为周一至周五
| M T W T F S S
-|--------------
M| 0 1 2 3 4 5 5
T| 5 0 1 2 3 4 4
W| 4 5 0 1 2 3 3
T| 3 4 5 0 1 2 2
F| 2 3 4 5 0 1 1
S| 0 1 2 3 4 0 0
S| 0 1 2 3 4 5 0
表中的49个值连接成以下字符串：
0123455501234445012333450122234501101234000123450
最后，正确的表达是：
5 * (DATEDIFF(@E, @S) DIV 7) + MID('0123455501234445012333450122234501101234000123450', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)
我已使用此解决方案验证了以下输入和输出：
Sunday, 2012-08-26 -> Monday, 2012-08-27 = 0
Sunday, 2012-08-26 -> Sunday, 2012-09-02 = 5
Monday, 2012-08-27 -> Tuesday, 2012-08-28 = 1
Monday, 2012-08-27 -> Monday, 2012-09-10 = 10
Monday, 2012-08-27 -> Monday, 2012-09-17 = 15
Monday, 2012-08-27 -> Tuesday, 2012-09-18 = 16
Monday, 2012-08-27 -> Monday, 2012-09-24 = 20
Monday, 2012-08-27 -> Monday, 2012-10-01 = 25
Tuesday, 2012-08-28 -> Wednesday, 2012-08-29 = 1
Wednesday, 2012-08-29 -> Thursday, 2012-08-30 = 1
Thursday, 2012-08-30 -> Friday, 2012-08-31 = 1
Friday, 2012-08-31 -> Saturday, 2012-09-01 = 1
Saturday, 2012-09-01 -> Sunday, 2012-09-02 = 0
Sunday, 2012-09-02 -> Monday, 2012-09-03 = 0
Monday, 2012-09-03 -> Tuesday, 2012-09-04 = 1
Tuesday, 2012-09-04 -> Wednesday, 2012-09-05 = 1
Wednesday, 2012-09-05 -> Thursday, 2012-09-06 = 1
Thursday, 2012-09-06 -> Friday, 2012-09-07 = 1
Friday, 2012-09-07 -> Saturday, 2012-09-08 = 1
Saturday, 2012-09-08 -> Sunday, 2012-09-09 = 0
Monday, 2012-09-24 -> Sunday, 2012-10-07 = 10
Saturday, 2012-08-25 -> Saturday, 2012-08-25 = 0
Saturday, 2012-08-25 -> Sunday, 2012-08-26 = 0
Saturday, 2012-08-25 -> Monday, 2012-08-27 = 0
Saturday, 2012-08-25 -> Tuesday, 2012-08-28 = 1
Saturday, 2012-08-25 -> Wednesday, 2012-08-29 = 2
Saturday, 2012-08-25 -> Thursday, 2012-08-30 = 3
Saturday, 2012-08-25 -> Friday, 2012-08-31 = 4
Saturday, 2012-08-25 -> Sunday, 2012-09-02 = 0
Monday, 2012-08-27 -> Monday, 2012-08-27 = 0
Monday, 2012-08-27 -> Tuesday, 2012-08-28 = 1
Monday, 2012-08-27 -> Wednesday, 2012-08-29 = 2
Monday, 2012-08-27 -> Thursday, 2012-08-30 = 3
Monday, 2012-08-27 -> Friday, 2012-08-31 = 4
Monday, 2012-08-27 -> Saturday, 2012-09-01 = 5
Monday, 2012-08-27 -> Sunday, 2012-09-02 = 5
答案 2 :(得分：12)
由于您需要在某处跟踪假期，因此日历表似乎合适：
CREATE TABLE Calendar
(
calendar_date DATETIME NOT NULL,
is_holiday BIT NOT NULL,
is_weekend BIT NOT NULL,
CONSTRAINT PK_Calendar PRIMARY KEY CLUSTERED (calendar_date)
)
当然，您需要在应用程序中使用的任何时间段填充所有日期。由于一年只有365天（或366天），从1900年到2100年并不是什么大不了的事。只需确保加载所有日期，而不仅仅是假期。
此时，您需要的查询变得微不足道：
SELECT
COUNT(*)
FROM
Calendar
WHERE
calendar_date BETWEEN '2009-01-01' AND '2009-10-01' AND
is_holiday = 0 AND
is_weekend = 0
警告：我主要使用MS SQL并且长时间没有使用MySQL，因此您可能需要调整上述内容。例如，我甚至不记得MySQL是否具有BIT数据类型。
答案 3 :(得分：8)
仅供进一步参考。以上都不适用于我，但是@Jeff Kooser的修改版本：
SELECT (DATEDIFF(date_end, date_start)) -
((WEEK(date_end) - WEEK(date_start)) * 2) -
(case when weekday(date_end) = 6 then 1 else 0 end) -
(case when weekday(date_start) = 5 then 1 else 0 end) -
(SELECT COUNT(*) FROM holidays WHERE holiday>=date_start and holiday<=data_end)
答案 4 :(得分：8)
建议的字符串是错误的吗？
DATEDIFF（from，to）排除'to'。以同样的方式这个字符串：
星期一 - ＆gt;星期五= {周一，图，周三，周四} = 4
星期一 - ＆gt;周六= {周一，图，周三，周五，周五} = 5
星期二 - ＆gt;周一= {Tu，Wed，Th，Fri，跳过周六，跳过太阳，周一被排除} = 4
等等
拟议矩阵：
| M T W T F S S
-|--------------
M| 0 1 2 3 4 5 5
T| 4 0 1 2 3 4 4
W| 3 4 0 1 2 3 3
T| 2 3 4 0 1 2 2
F| 1 2 3 4 0 1 1
S| 0 1 2 3 4 0 0
S| 0 1 2 3 4 5 0
字符串：'0123455401234434012332340122123401101234000123450'
我在这里遗漏了什么吗？ ：）
答案 5 :(得分：4)
鉴于一个月的第一天，这将返回该月内的工作日数。在MySQL中。没有存储过程。
SELECT (DATEDIFF(LAST_DAY(?),?) + 1) -
((WEEK(LAST_DAY(?)) - WEEK(?)) * 2) -
(case when weekday(?) = 6 then 1 else 0 end) -
(case when weekday(LAST_DAY(?)) = 5 then 1 else 0 end)
答案 6 :(得分：3)
根据Yada的上述功能，这里有一个主题的微小变化，它计算从当前日期（不包括）的工作日，直到目标日期。它还处理以色列不同的周末:-) 请注意，如果目标日期是过去的，这将产生否定结果（这正是我想要的）。
DELIMITER //
DROP FUNCTION IF EXISTS WORKDAYS_LEFT//
CREATE FUNCTION WORKDAYS_LEFT(target_date DATE, location char(2))
RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
DECLARE start_date DATE;
DECLARE end_date DATE;
DECLARE check_date DATE;
DECLARE diff INT;
DECLARE extra_weekend_days INT;
DECLARE weeks_diff INT;
SET start_date = CURDATE();
SET end_date = target_date;
SET diff = DATEDIFF(end_date, start_date);
SET weeks_diff = FLOOR(diff / 7);
SET end_date = DATE_SUB(end_date, INTERVAL (weeks_diff * 7) DAY);
SET check_date = DATE_ADD(start_date, INTERVAL 1 DAY);
SET extra_weekend_days = 0;
WHILE check_date <= end_date DO
SET extra_weekend_days = extra_weekend_days +
IF(DAYNAME(check_date) = 'Saturday', 1, 0) +
IF(DAYNAME(check_date) = IF(location = 'IL','Friday', 'Sunday'), 1, 0);
SET check_date = DATE_ADD(check_date, INTERVAL 1 DAY);
END WHILE;
RETURN diff - weeks_diff*2 - extra_weekend_days;
END//
DELIMITER ;
答案 7 :(得分：2)
Yada的解决方案无法正常工作。我的变化：
DELIMITER $$
DROP FUNCTION IF EXISTS `catalog`.`WORKDAYS` $$
CREATE FUNCTION `catalog`.`WORKDAYS` (first_date DATETIME, second_date DATETIME) RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
DECLARE start_date DATE;
DECLARE end_date DATE;
DECLARE diff INT;
IF (first_date < second_date) THEN
SET start_date = first_date;
SET end_date = second_date;
ELSE
SET start_date = second_date;
SET end_date = first_date;
END IF;
SET diff = DATEDIFF(end_date, start_date);
RETURN (CASE WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) = 'Saturday' THEN diff
WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) = 'Sunday' THEN (diff - 2)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) = 'Sunday' THEN (diff - 1)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) = 'Saturday' THEN (diff + 1)
WHEN DAYNAME(start_date) = 'Sunday' && DAYNAME(end_date) in ('Saturday', 'Sunday') THEN (diff + 1)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) not in ('Saturday', 'Sunday') THEN (diff -1)
WHEN DAYNAME(start_date) = 'Sunday' && DAYNAME(end_date) not in ('Saturday', 'Sunday') THEN (diff + 1)
WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) not in ('Saturday', 'Sunday')
&& WEEKDAY(start_date) > WEEKDAY(end_date) THEN (diff - 2)
ELSE diff END)
- (FLOOR(diff / 7) * 2)
- (CASE WHEN DAYNAME(start_date) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DAYNAME(end_date) = 'Saturday' THEN 1 ELSE 0 END);
END $$
DELIMITER ;
答案 8 :(得分：1)
如果你想真正忽略周末的存在，那么你需要处理源自周六/太阳的东西，好像它起源于星期一;并且在周六/周日结束的事情就好像它真的在星期五结束了。因此，在周末开始和结束的事情，你必须忽略开始和结束。我不认为其他任何答案都是这样做的。
以下功能执行此操作：
CREATE DEFINER=`root`@`localhost` FUNCTION `weekdayDiff`
(
edate datetime,
sdate datetime
)
RETURNS int
DETERMINISTIC
BEGIN
if edate>sdate
then
return 5 * (DATEDIFF(edate, sdate) DIV 7) + MID('+0+1+2+3+4+4+4+4+0+1+2+3+3+3+3+4+0+1+2+2+2+2+3+4+0+1+1+1+1+2+3+4+0+0+0+0+1+2+3+4-1-1+0+1+2+3+4+4-1', 2*(7 * WEEKDAY(sdate) + WEEKDAY(edate)) + 1, 2);
else
return -(5 * (DATEDIFF(sdate, edate) DIV 7) + MID('+0+1+2+3+4+4+4+4+0+1+2+3+3+3+3+4+0+1+2+2+2+2+3+4+0+1+1+1+1+2+3+4+0+0+0+0+1+2+3+4-1-1+0+1+2+3+4+4-1', 2*(7 * WEEKDAY(edate) + WEEKDAY(sdate)) + 1, 2));
end if;
-- The following works unless both start and finish date are on weekends.
-- return 5 * (DATEDIFF(edate, sdate) DIV 7) + MID('0123444401233334012222340111123400001234000123440', 7 * WEEKDAY(sdate) + WEEKDAY(edate) + 1, 1);
END;
用Rodger的答案语言，创建上面字符串的表格在下面（唯一的区别是，如果在星期六/星期日开始和结束时它是-1而不是0）：
| M T W T F S S
-|---------------------
M| +0 +1 +2 +3 +4 +4 +4
T| +4 +0 +1 +2 +3 +3 +3
W| +3 +4 +0 +1 +2 +2 +2
T| +2 +3 +4 +0 +1 +1 +1
F| +1 +2 +3 +4 +0 +0 +0
S| +0 +1 +2 +3 +4 -1 -1
S| +0 +1 +2 +3 +4 +4 -1
答案 9 :(得分：1)
@Rodger Bagnall发布的答案对我不起作用，例如2016-04。它比实际显示的少了1天。
如果谈论按查询计算 - 我用这个：
set
@S = '2016-04-01',
@E = '2016-04-30';
select
case
when WEEKDAY(@S) < 5 then 5 - WEEKDAY(@S)
else 0
end #startweek
+
case
when WEEKDAY(@E) < 5 then WEEKDAY(@E) + 1
else 5
end #endweek
+
(
DATEDIFF(@E, @S) + 1 # plus 1 day cause params is inside 1 month
- (7 - WEEKDAY(@S)) # minus start week
- (WEEKDAY(@E) + 1) # minus end week
) DIV 7 * 5 #rest part
as work_date_count;
查询未优化仅仅是为了显示数字的来源
答案 10 :(得分：1)
非周末天差可以通过这种方式实现：
CREATE FUNCTION `WDDIFF` (d0 DATE, d1 DATE)
RETURNS INT DETERMINISTIC
COMMENT 'Date0, Date1'
BEGIN
RETURN DATEDIFF(d1, d0) - (DATEDIFF(DATE_SUB(d1, INTERVAL WEEKDAY(d1) DAY), DATE_ADD(d0, INTERVAL (7 - WEEKDAY(d0)) DAY))/7+1)*2 + IF(WEEKDAY(d0)>4, 1, 0) + 1;
END
用法：自月初以来的工作日
SELECT ap.WDDIFF(DATE_SUB(CURDATE(), INTERVAL DAYOFMONTH(CURDATE()) - 1 DAY), CURDATE())
注意：该功能计算开始和结束日期
答案 11 :(得分：1)
与“忽略假期”相同的问题是每个国家都有不同的假期。
您必须先为您所在的国家/地区定义假期，然后通过它们查看某个日期是否为假日。
我不知道在mysql中你想要的通用函数
抱歉！
答案 12 :(得分：0)
我有这个要求并编写了完整的功能，可以计算，同时避免给定国家的周末和假期（使用单独的表）。我已将整个功能和详细信息放在我的博客（http://mgw.dumatics.com/mysql-function-to-calculate-elapsed-working-time/）以及解释和流程图以及假期表的创建等等......我很乐意将它放在这里，但它有点太长了......
解决问题的示例：
假设“英国”网站于09:00至16:00之间开放的“2016年6月10日星期五中午12点”发生了事件。该事件随后于“2016年6月14日星期二14:00”结束。
对于上述事件，功能应计算年龄为960分钟= 16小时= [周五4小时（12:00至16:00）+周一7小时（09:00至16:00）+ 5小时周二（09:00至14:00）]
答案 13 :(得分：0)
Below function will give you the Weekdays, Weekends, Date difference with proper results:
You can call the below function like,
select getWorkingday('2014-04-01','2014-05-05','day_diffs');
select getWorkingday('2014-04-01','2014-05-05','work_days');
select getWorkingday('2014-04-01','2014-05-05','weekend_days');
DROP FUNCTION IF EXISTS PREPROCESSOR.getWorkingday;
CREATE FUNCTION PREPROCESSOR.`getWorkingday`(d1 datetime,d2 datetime, retType varchar(20)) RETURNS varchar(255) CHARSET utf8
BEGIN
DECLARE dow1, dow2,daydiff,workdays, weekenddays, retdays,hourdiff INT;
declare newstrt_dt datetime;
SELECT dd.iDiff, dd.iDiff - dd.iWeekEndDays AS iWorkDays, dd.iWeekEndDays into daydiff, workdays, weekenddays
FROM (
SELECT
dd.iDiff,
((dd.iWeeks * 2) +
IF(dd.iSatDiff >= 0 AND dd.iSatDiff < dd.iDays, 1, 0) +
IF (dd.iSunDiff >= 0 AND dd.iSunDiff < dd.iDays, 1, 0)) AS iWeekEndDays
FROM (
SELECT dd.iDiff, FLOOR(dd.iDiff / 7) AS iWeeks, dd.iDiff % 7 iDays, 5 - dd.iStartDay AS iSatDiff, 6 - dd.iStartDay AS iSunDiff
FROM (
SELECT
1 + DATEDIFF(d2, d1) AS iDiff,
WEEKDAY(d1) AS iStartDay
) AS dd
) AS dd
) AS dd ;
if(retType = 'day_diffs') then
set retdays = daydiff;
elseif(retType = 'work_days') then
set retdays = workdays;
elseif(retType = 'weekend_days') then
set retdays = weekenddays;
end if;
RETURN retdays;
END;
Thank You.
Vinod Cyriac.
Bangalore
答案 14 :(得分：0)
SELECT 5* (DATEDIFF(u.EndDate, u.StartDate) DIV 7) + MID('1234555512344445123333451222234511112345001234550', 7 * WEEKDAY(u.StartDate) + WEEKDAY(u.EndDate) + 1, 1)
这是您想要考虑以下情况的时间：
1）如果startdate = enddate，则持续时间= 1 同样..
我使用最多投票答案中提到的逻辑计算了字符串，并根据需要得到了结果。
答案 15 :(得分：0)
我使用此解决方案，最后，请参阅：
DROP FUNCTION IF EXISTS datediff_workdays;
CREATE FUNCTION datediff_workdays(start_date DATE, end_date DATE) RETURNS INTEGER
BEGIN
RETURN 5 * (DATEDIFF(end_date, start_date) DIV 7) + MID('0123455501234445012333450122234501101234000123450', 7 * WEEKDAY(start_date) + WEEKDAY(end_date) + 1, 1);
END
答案 16 :(得分：0)
最高答案计算开始日期和结束日期之间的天数，但不包括结束日期。
同样对于在同一个周末开始和结束的任何日期，例如2018-05-05周六至2018-05-12周六，它还计算了一天。
这是一个适合我的功能！
drop procedure if exists get_duration$$
create procedure get_duration(in data_from date, in data_to date)
begin
if (WEEKDAY(data_from) = 5 AND WEEKDAY(data_to) = 5)
OR (WEEKDAY(data_from) = 6 AND WEEKDAY(data_to) = 6) then
select (5 * (DATEDIFF(data_to, data_from) DIV 7)
+ MID('0123444401233334012222340111123400001234000123440',
7 * WEEKDAY(data_from) + WEEKDAY(data_to) + 1, 1)) dur;
else
select (5 * (DATEDIFF(data_to, data_from) DIV 7)
+ MID('0123444401233334012222340111123400001234000123440',
7 * WEEKDAY(data_from) + WEEKDAY(data_to) + 1, 1))+1 dur;
end if;
end$$
答案 17 :(得分：0)
我在MySQL数据库中添加了一个存储过程，以计算团队的总工作日（我称之为WORKDAYS）：
RETURN ABS(DATEDIFF(date2, date1)) + 1
- ABS(DATEDIFF(ADDDATE(date2, INTERVAL 1 - DAYOFWEEK(date2) DAY),
ADDDATE(date1, INTERVAL 1 - DAYOFWEEK(date1) DAY))) / 7 * 2
- (DAYOFWEEK(IF(date1 < date2, date1, date2)) = 1)
- (DAYOFWEEK(IF(date1 > date2, date1, date2)) = 7)
- (SELECT DISTINCT COUNT(PriKey) FROM holidays WHERE date BETWEEN date1 AND date2)
+ (SELECT DISTINCT COUNT(PriKey) FROM weekenddaysworked WHERE date BETWEEN date1 AND date2)
我在数据库中添加了两个表： 假期和周末工作，都有两列（PriKey（int，11），数据（日期））
在假期中，我添加了需要考虑的假期，在周末工作日中，我添加了我在周末工作的日期。
我将过程添加为函数，结果为INT。 date1和date2定义为DATE。
现在我可以像这样调用MySQL函数：
WORKDAYS（date1，date2）-例如WORKDAYS（'2018-11-01'，'2018-12-01'）
答案 18 :(得分：0)
好吧，男孩和女孩，我显然是最好的解决方案，这是一个简单的select语句，用于获取两个日期之间的工作日数。
select
FLOOR(DATEDIFF(later_date, earlier_date) / 7) * 5 +
least(DATEDIFF(later_date, earlier_date) % 7, 5) +
if(weekday(later_date) < weekday(earlier_date), -2, 0);
简单的解释
答案 19 :(得分：0)
我在@ caveman，@ bryan-geraghty和@ rodger-bagnall的答案上加了一些，我需要一个可以对“工作日之前”查询进行反向计算的版本。当start_date在end_date之前或之后，此调整有效。
SELECT 5 * (DATEDIFF(@E, @S) DIV 7) +
CASE WHEN @E < @S THEN
-1 * MID('0123455401234434012332340122123401101234000123450', 7 * WEEKDAY(@E) + WEEKDAY(@S) + 1, 1)
ELSE
MID('0123455401234434012332340122123401101234000123450', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)
END
两种情况下的抽样结果：
+------------+------------+-----------+
| @S | @E | wday_diff |
+------------+------------+-----------+
| 2019-11-25 | 2019-10-26 | -20 |
| 2019-11-25 | 2019-11-28 | 3 |
+------------+------------+-----------+
如果发现任何错误，请告诉我。
答案 20 :(得分：0)
SELECT FLOOR((DATEDIFF(@E,@S)+1)/7)*5+
LEAST((DATEDIFF(@E,@S)+1)%7,5)+
IF(WEEKDAY(@E)<WEEKDAY(@S),IF(WEEKDAY(@S)<5,-2,WEEKDAY(@S)-7),
IF(WEEKDAY(@E)=WEEKDAY(@S),IF(WEEKDAY(@E) IN (5,6),-1,0),
IF(WEEKDAY(@S)=5,-1,IF(WEEKDAY(@E)=5,-1,IF(WEEKDAY(@E)=6,-2,0)))));
这是从下面的2019年8月起对@jeffery_the_wind的功能答案的更正。
1）周* 5
2）每月增加休息日
3）检查天数并计算其余天的修正值。
答案 21 :(得分：0)
这是我的解决方案
DELIMITER $$
DROP FUNCTION IF EXISTS WORKINGDAYS$$
CREATE DEFINER = 'root'@'localhost'
FUNCTION WORKINGDAYS(DATEFROM DATETIME,
DATETO DATETIME
)
RETURNS INT(11)
BEGIN
DECLARE ACTUALDATE DATETIME;
DECLARE WORKINGDAYS INTEGER;
SET WORKINGDAYS = 0;
SET ACTUALDATE = DATEFROM;
dateloop:
LOOP
IF (ACTUALDATE > DATETO OR DATEFROM > DATETO) THEN
LEAVE dateloop;
END IF;
IF (dayofweek(ACTUALDATE) != 7 AND dayofweek(ACTUALDATE) != 1) THEN
SET WORKINGDAYS = WORKINGDAYS + 1;
END IF;
SET ACTUALDATE = adddate(ACTUALDATE, INTERVAL 1 DAY);
END LOOP dateloop;
RETURN WORKINGDAYS;
END
$$
DELIMITER ;
答案 22 :(得分：0)
此查询可轻松返回两个日期之间的工作日数，不包括周末：
select datediff('2016-06-19','2016-06-01') - (floor(datediff('2016-06-19','2016-06-01')/6) + floor(datediff('2016-06-19','2016-06-01')/7));
答案 23 :(得分：0)
这是DATEDIFF的替代品，可用于+ ve和-ve差异。
DELIMITER $$
DROP FUNCTION IF EXISTS WORKDAYSDIFF$$
CREATE FUNCTION WORKDAYSDIFF(sd DATE, ed DATE)
RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
RETURN IF (sd >= ed,
5 * (DATEDIFF(sd, ed) DIV 7) + MID('0123455501234445012333450122234501101234000123450', 7 * WEEKDAY(ed) + WEEKDAY(sd) + 1, 1),
-(5 * (DATEDIFF(ed, sd) DIV 7) + MID('0123455501234445012333450122234501101234000123450', 7 * WEEKDAY(sd) + WEEKDAY(ed) + 1, 1)) );
END$$
DELIMITER ;
答案 24 :(得分：0)
基于Rodger Bagnall解决方案模拟NETWORKDAYS.INTL的函数https://stackoverflow.com/a/6762805/218418
DELIMITER //
DROP FUNCTION IF EXISTS NETWORKDAYS//
CREATE FUNCTION NETWORKDAYS(sd DATE, ed DATE)
RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
RETURN (5 * (DATEDIFF(ed, sd) DIV 7) + MID('0123444401233334012222340111123400001234000123440', 7 * WEEKDAY(sd) + WEEKDAY(ed) + 1, 1))+1;
END//
DELIMITER ;
并选择
SELECT NETWORKDAYS('2015-01-01 06:00:00', '2015-01-20 06:00:00');
答案 25 :(得分：0)
请Helooo测试。
DELIMITER $$
DROP FUNCTION IF EXISTS `WORKDAYS` $$
CREATE FUNCTION `WORKDAYS` (first_date DATETIME, second_date DATETIME) RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
DECLARE start_date DATE;
DECLARE end_date DATE;
DECLARE diff INT;
DECLARE cnt INT;
IF (first_date < second_date) THEN
SET start_date = first_date;
SET end_date = second_date;
ELSE
SET start_date = second_date;
SET end_date = first_date;
END IF;
SELECT COUNT(*) INTO cnt FROM `holiday` WHERE (hday BETWEEN start_date AND end_date) and (DAYOFWEEK(hday) != 7 and DAYOFWEEK(hday) != 1);
SET diff = DATEDIFF(end_date, start_date) ;
RETURN (CASE WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) = 'Saturday' THEN (diff - cnt)
WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) = 'Sunday' THEN (diff - 2 - cnt)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) = 'Sunday' THEN (diff - 1 - cnt)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) = 'Saturday' THEN (diff + 1 - cnt)
WHEN DAYNAME(start_date) = 'Sunday' && DAYNAME(end_date) in ('Saturday', 'Sunday') THEN (diff + 1 - cnt)
WHEN DAYNAME(start_date) = 'Saturday' && DAYNAME(end_date) not in ('Saturday', 'Sunday') THEN (diff -1 - cnt)
WHEN DAYNAME(start_date) = 'Sunday' && DAYNAME(end_date) not in ('Saturday', 'Sunday') THEN (diff + 1 - cnt)
WHEN DAYNAME(start_date) not in ('Saturday', 'Sunday') && DAYNAME(end_date) not in ('Saturday', 'Sunday')
&& WEEKDAY(start_date) > WEEKDAY(end_date) THEN (diff - 2 - cnt)
ELSE (diff - cnt) END)
- (FLOOR(diff / 7) * 2)
- (CASE WHEN DAYNAME(start_date) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DAYNAME(end_date) = 'Saturday' THEN 1 ELSE 0 END);
END $$
和表假期
DROP TABLE IF EXISTS `holiday`;
CREATE TABLE `holiday` (
`id` bigint(32) unsigned NOT NULL AUTO_INCREMENT,
`hday` date NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
INSERT INTO `holiday` (`id`, `hday`) VALUES
(1, '2012-01-01'),
(2, '2012-05-01'),
(3, '2012-05-08'),
(4, '2012-07-05'),
(5, '2012-07-06'),
(6, '2012-09-28'),
(7, '2012-10-28'),
(8, '2012-11-17'),
(9, '2012-12-24'),
(10, '2012-12-25'),
(11, '2012-12-26');
etc...
答案 26 :(得分：0)
我需要两个功能。一个用于计算两个日期之间的工作日数，另一个用于将x个工作日添加/减去日期。以下是我在互联网上找到的例子。使它们接近标准的DATEDIFF（）和DATE_ADD（）函数以及相互补充计算。例如，DateDiffBusiness（'2014-05-14'，DateAddBusiness（'2014-05-14'，5））将等于5.
DROP FUNCTION IF EXISTS DateDiffBusiness;
DELIMITER &
CREATE FUNCTION DateDiffBusiness( d2 DATE, d1 DATE )
RETURNS INT
DETERMINISTIC
COMMENT 'Calculates the number of bussiness days between two dates'
BEGIN
DECLARE dow1, dow2, days INT;
SET dow1 = DAYOFWEEK(d1);
SET dow2 = DAYOFWEEK(d2);
SET days = FLOOR( DATEDIFF(d2,d1)/7 ) * 5 +
CASE
WHEN dow1=1 AND dow2=7 THEN 5
WHEN dow1 IN(7,1) AND dow2 IN (7,1) THEN 0
WHEN dow1=dow2 THEN 1
WHEN dow1 IN(7,1) AND dow2 NOT IN (7,1) THEN dow2-1
WHEN dow1 NOT IN(7,1) AND dow2 IN(7,1) THEN 7-dow1
WHEN dow1<=dow2 THEN dow2-dow1+1
WHEN dow1>dow2 THEN 5-(dow1-dow2-1)
ELSE 0
END;
RETURN days-1;
END&
DELIMITER ;
DROP FUNCTION IF EXISTS DateAddBusiness;
DELIMITER &
CREATE FUNCTION DateAddBusiness(mydate DATE, numday INT)
RETURNS DATE
DETERMINISTIC
COMMENT 'Adds bussiness days between two dates'
BEGIN
DECLARE num_week INT DEFAULT 0;
DECLARE num_day INT DEFAULT 0;
DECLARE adj INT DEFAULT 0;
DECLARE total INT DEFAULT 0;
SET num_week = numday DIV 5;
SET num_day = MOD(numday, 5);
IF (WEEKDAY(mydate) + num_day >= 5) then
SET adj = 2;
END IF;
SET total = num_week * 7 + adj + num_day;
RETURN DATE_ADD(mydate, INTERVAL total DAY);
END&
DELIMITER ;
答案 27 :(得分：0)
MYSQL函数返回2个日期（含）之间的工作日。 BETWEEN 2和6是周一至周五，可根据您的日历/地区进行调整。
-- Routine DDL
-- Note: comments before and after the routine body will not be stored by the server
-- --------------------------------------------------------------------------------
DELIMITER $$
CREATE DEFINER=`root`@`localhost` FUNCTION `fn_GetBusinessDaysBetweenDates`(d1 DATE, d2 DATE) RETURNS int(11)
BEGIN
DECLARE bDaysInPeriod INT;
SET bDaysInPeriod=0;
WHILE d1<=d2 DO
IF DAYOFWEEK(d1) BETWEEN 2 AND 6 THEN
SET bDaysInPeriod=bDaysInPeriod+1;
END IF;
SET d1=d1+INTERVAL 1 day;
END WHILE;
RETURN bDaysInPeriod;
END
答案 28 :(得分：0)
虽然非常老了但是很有帮助。 按照 @shahcool 提供的解决方案，不会返回确切的天数 e.g。
Workdays('2013-03-26','2013-04-01')返回3天但实际上必须有5天
以下是我测试过的解决方案并且重新确定了确切的工作日
DELIMITER $$
DROP FUNCTION IF EXISTS WORKDAYS $$
CREATE FUNCTION `WORKDAYS` (first_date DATETIME, second_date DATETIME) RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
DECLARE start_date DATE;
DECLARE end_date DATE;
DECLARE diff INT;
DECLARE NumberOfWeeks INT;
DECLARE RemainingDays INT;
DECLARE firstDayOfTheWeek INT;
DECLARE lastDayOfTheWeek INT;
DECLARE WorkingDays INT;
IF (first_date < second_date) THEN
SET start_date = first_date;
SET end_date = second_date;
ELSE
SET start_date = second_date;
SET end_date = first_date;
END IF;
## Add one to include both days in interval
SET diff = DATEDIFF(end_date, start_date)+1;
SET NumberOfWeeks=floor(diff/7);
SET RemainingDays=MOD(diff,7);
SET firstDayOfTheWeek=DAYOFWEEK(start_date);
SET lastDayOfTheWeek=DAYOFWEEK(end_date);
IF(firstDayOfTheWeek <= lastDayOfTheWeek) THEN
IF( firstDayOfTheWeek<=6 AND 6 <=lastDayOfTheWeek) THEN SET RemainingDays=RemainingDays-1; END IF;
IF( firstDayOfTheWeek<=7 AND 7 <=lastDayOfTheWeek) THEN SET RemainingDays=RemainingDays-1; END IF;
ELSE
IF( firstDayOfTheWeek=7) THEN SET RemainingDays=RemainingDays-1;
IF (lastDayOfTheWeek=6) THEN SET RemainingDays=RemainingDays-1; END IF;
ELSE SET RemainingDays=RemainingDays-2;
END IF;
END IF;
SET WorkingDays=NumberOfWeeks*5;
IF(RemainingDays>0) THEN RETURN WorkingDays+RemainingDays;
ELSE RETURN WorkingDays; END IF;
END $$
DELIMITER ;
答案 29 :(得分：0)
对于上面的NETWORKDAYS（）函数，应该添加另外一个条件，以涵盖开始日期到结束日期在7天内以及整个周末的情况。
RETURN (diff + 1)
- (FLOOR(diff / 7) * 2)
- (CASE WHEN DAYNAME(start_date) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DAYNAME(end_date) = 'Saturday' THEN 1 ELSE 0 END)
- (CASE WHEN diff<7 and WEEK(start_date)<>WEEK(end_date) THEN 2 ELSE 0 end);
答案 30 :(得分：0)
我知道这是一个老线程，但我认为我的解决方案可能对某些人有所帮助。这是我在不需要功能的情况下查找商业日的查询。你可以把字段命名为你想要的东西，我只是故意将它们留空。
SELECT
@tmp_s := ept.`date_start`,
@tmp_e := IF(ept.`date_end` IS NULL, NOW(),ept.`date_end`),
@start := IF(DAYOFWEEK(@tmp_s)=1,@tmp_s + INTERVAL 1 DAY,(IF(DAYOFWEEK(@tmp_s)=7,@tmp_s + INTERVAL 2 DAY,@tmp_s)),
@end := IF(DAYOFWEEK(@tmp_e)=1,@tmp_e - INTERVAL 2 DAY,(IF(DAYOFWEEK(@tmp_e)=7,@tmp_e - INTERVAL 1 DAY,@tmp_e)),
@bizdays := CASE
WHEN DATEDIFF(@end,@start)>7 THEN CEIL((DATEDIFF(@end,@start)/7)*5)
WHEN DAYOFWEEK(@end)< DAYOFWEEK(@start) THEN DATEDIFF(@end,@start)-2
ELSE DATEDIFF(@end,@start)
END,
DATE(@start),
DATE(@end),
IF(@bizdays>=10,10,@bizdays)
FROM `employee_points` ept
WHERE ept.`date_start` > '2011-01-01'
答案 31 :(得分：0)
这适用于Sql Server 2005
不知道它是否适合你。
DECLARE @StartDate DATETIME,
@EndDate DATETIME
SELECT @StartDate = '22 Nov 2009',
@EndDate = '28 Nov 2009'
;WITH CTE AS(
SELECT @StartDate DateVal,
DATENAME(dw, @StartDate) DayNameVal
UNION ALL
SELECT DateVal + 1,
DATENAME(dw, DateVal + 1)
FROM CTE
WHERE DateVal < @EndDate
)
SELECT COUNT(1)
FROM (
SELECT *
FROM CTE
WHERE DayNameVal NOT IN ('Sunday','Saturday')
) DayVals
答案 32 :(得分：-1)
DELIMITER //
DROP FUNCTION IF EXISTS NETWORKDAYS//
CREATE FUNCTION NETWORKDAYS(first_date DATE, second_date DATE)
RETURNS INT
LANGUAGE SQL
DETERMINISTIC
BEGIN
DECLARE start_date DATE;
DECLARE end_date DATE;
DECLARE diff INT;
IF (first_date < second_date) THEN
SET start_date = first_date;
SET end_date = second_date;
ELSE
SET start_date = second_date;
SET end_date = first_date;
END IF;
SET diff = DATEDIFF(end_date, start_date);
RETURN (diff + 1)
- (FLOOR(diff / 7) * 2)
- (CASE WHEN DAYNAME(start_date) = 'Sunday' THEN 1 ELSE 0 END)
- (CASE WHEN DAYNAME(end_date) = 'Saturday' THEN 1 ELSE 0 END);
END//
DELIMITER ;
- 测试 SELECT Networkdays（'2009-12-06'，'2009-12-13'）;
答案 33 :(得分：-2)
您需要使用DATEDIFF才能获得MySQL中两个日期之间的天数。 IE：
DATEDIFF(t.date_column_1, t.date_column_2)
但Stephane是正确的 - 假期是联邦和地区定义的。您需要创建一个表来存储日期和时间。在计算中引用它们。