我搜索了很多例子,我得到了很好的例子:
Count days between two dates, excluding weekends (MySQL only)
How to count date difference excluding weekend and holidays in MySQL
Calculate diffference between 2 dates in SQL, excluding weekend days
但是没有得到最有希望的解决方案,所以我可以在我的mysql函数中使用来查询行的十万行。
这个是一个非常新的概念,但是对于像@start_date = '2013-08-03' , @end_date = '2013-08-21'
预期的答案:13,它仅给出12,
SELECT 5 * (DATEDIFF(@end_date, @start_date) DIV 7) + MID('0123444401233334012222340111123400012345001234550', 7 * WEEKDAY(@start_date) + WEEKDAY(@end_date) + 1, 1);
所以我试图自己做 -
Concept :
Input : 1. period_from_date - from date
2. period_to_date - to date
3. days_to_exclude - mapping : S M T W TH F Sat => 2^0 + 2^6
(sat and sun to exclude) ^ ^ ^ ^ ^ ^ ^
0 1 2 3 4 5 6
DELIMITER $$
USE `db_name`$$
DROP FUNCTION IF EXISTS `FUNC_CALC_TOTAL_WEEKDAYS`$$
CREATE DEFINER=`name`@`%` FUNCTION `FUNC_CALC_TOTAL_WEEKDAYS`( period_from_date DATE, period_to_date DATE, days_to_exclude INT ) RETURNS INT(11)
BEGIN
DECLARE period_total_num_days INT DEFAULT 0;
DECLARE period_total_working_days INT DEFAULT 0;
DECLARE period_extra_days INT DEFAULT 0;
DECLARE period_complete_weeks INT DEFAULT 0;
DECLARE extra_days_start_date DATE DEFAULT '0000-00-00';
DECLARE num_days_to_exclude INT DEFAULT 0;
DECLARE start_counter_frm INT DEFAULT 0;
DECLARE end_counter_to INT DEFAULT 6;
DECLARE temp_var INT DEFAULT 0;
# if no day to exclude return date-diff only
IF days_to_exclude = 0 THEN
RETURN DATEDIFF( period_to_date, period_from_date ) + 1 ;
END IF;
# get total no of days to exclude
WHILE start_counter_frm <= end_counter_to DO
SET temp_var = POW(2,start_counter_frm) ;
IF (temp_var & days_to_exclude) = temp_var THEN
SET num_days_to_exclude = num_days_to_exclude + 1;
END IF;
SET start_counter_frm = start_counter_frm + 1;
END WHILE;
# Get period days count
SET period_total_num_days = DATEDIFF( period_to_date, period_from_date ) + 1 ;
SET period_complete_weeks = FLOOR( period_total_num_days /7 );
SET period_extra_days = period_total_num_days - ( period_complete_weeks * 7 );
SET period_total_working_days = period_complete_weeks * (7 - num_days_to_exclude);
SET extra_days_start_date = DATE_SUB(period_to_date,INTERVAL period_extra_days DAY);
# get total working days from the left days
WHILE period_extra_days > 0 DO
SET temp_var = DAYOFWEEK(period_to_date) -1;
IF POW(2,temp_var) & days_to_exclude != POW(2,temp_var) THEN
SET period_total_working_days = period_total_working_days +1;
END IF;
SET period_to_date = DATE_SUB(period_to_date,INTERVAL 1 DAY);
SET period_extra_days = period_extra_days -1;
END WHILE;
RETURN period_total_working_days;
END$$
DELIMITER ;
请告诉我这会失败的漏洞。打开任何建议和评论。
答案 0 :(得分:15)
更新:如果你需要两个日期之间的工作日,你可以像这样得到它
CREATE FUNCTION TOTAL_WEEKDAYS(date1 DATE, date2 DATE)
RETURNS INT
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);
注意:如果您切换开始date1
和结束date2
日期,该功能仍然有用。
样本用法:
SELECT TOTAL_WEEKDAYS('2013-08-03', '2013-08-21') weekdays1,
TOTAL_WEEKDAYS('2013-08-21', '2013-08-03') weekdays2;
输出:
| WEEKDAYS1 | WEEKDAYS2 | ------------------------- | 13 | 13 |
这是 SQLFiddle 演示
答案 1 :(得分:6)
此查询将正常运行,上述所有查询都无法正常运行。试试这个:
SELECT ((DATEDIFF(date2, date1)) -
((WEEK(date2) - WEEK(date1)) * 2) -
(case when weekday(date2) = 6 then 1 else 0 end) -
(case when weekday(date1) = 5 then 1 else 0 end)) as DifD
像这样测试:
SELECT ((DATEDIFF('2014-10-25', '2014-10-15')) -
((WEEK('2014-10-25') - WEEK('2014-10-15')) * 2) -
(case when weekday('2014-10-25') = 6 then 1 else 0 end) -
(case when weekday('2014-10-15') = 5 then 1 else 0 end)) as DifD
结果:
DifD
8
答案 2 :(得分:1)
我用这个。意味着没有功能,因此可以在视图中使用:
select
datediff(@dateto, @datefrom) +
datediff(@datefrom,
date_add(@datefrom, INTERVAL
floor(datediff(@dateto, @datefrom) / 7) day)) * 2
- case
when weekday(@dateto) = 6 then 2
when weekday(@dateto) = 5 then 1
when weekday(@dateto) < weekday(@datefrom) then 2
else 0
end;
答案 3 :(得分:0)
有一个类似的问题,我使用PHP删除周末,需要知道开始日期和天数:
EG SQL:
SELECT DAYOFWEEK(`date1`) AS `startday`, TIMESTAMPDIFF(DAY, `date1`, `date2`) AS `interval` FROM `table`
然后通过PHP函数运行结果:
function noweekends($startday, $interval) {
//Remove weekends from an interval
$wecount = 0; $tmp = $interval;
while($interval/7 > 1) { $interval-=7; $wecount++; }
if($interval+$startday > 5) $wecount++;
$interval = $tmp-($wecount*2);
return $interval;
}
答案 4 :(得分:0)
仅排除星期日:
CREATE FUNCTION TOTAL_WEEKDAYS(date1 DATE, date2 DATE)
RETURNS INT
RETURN ABS(DATEDIFF(date2, date1)) + 1
- ABS(DATEDIFF(ADDDATE(date2, INTERVAL 1 - DAYOFWEEK(date2) DAY),
ADDDATE(date1, INTERVAL 1 - DAYOFWEEK(date1) DAY))) / 7
- (DAYOFWEEK(IF(date1 < date2, date1, date2)) = 1);