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SQL实现工作日计算_MySQL

程序员文章站 2022-05-22 23:09:16
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CREATE FUNCTION Time_Span_of_minutes(

start_day DATETIME,

end_day DATETIME)

RETURNS FLOAT

BEGIN

-- 返回按分钟计算两段时间的间隔,采用逐日靠近的方法求解,一天按8小时480分钟计算,周末不计,不考虑法定节假日。

-- 如果起始日期在周末,则计算当天时间。

DECLARE minutes FLOAT;

DECLARE next_day DATETIME;

SET minutes=0.0;

SET next_day=start_day;

IF(TIMESTAMPDIFF(DAY,start_day,end_day)

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,start_day,end_day);

ELSE

WHILE TIMESTAMPDIFF(DAY,next_day,end_day)>=1.0 DO

IF ((DAYOFWEEK(next_day)=7) OR (DAYOFWEEK(next_day)=1)) THEN

SET next_day=next_day+INTERVAL 1 DAY;

ELSE

SET next_day=next_day+INTERVAL 1 DAY;

SET minutes=minutes+480.0;

END IF;

END WHILE;

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,next_day,end_day);

IF ((DAYOFWEEK(start_day)=7) OR (DAYOFWEEK(start_day)=1)) THEN

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,start_day,CONVERT(CONCAT(SUBSTRING(start_day FROM 1 FOR 10),' 17:30:00'),DATETIME));

END IF;

END IF;

RETURN minutes;

END;

CREATE FUNCTION Time_Span_of_minutes(

start_day DATETIME,

end_day DATETIME)

RETURNS FLOAT

BEGIN

-- 返回按分钟计算两段时间的间隔,采用逐日靠近的方法求解,一天按8小时480分钟计算,周末不计,不考虑法定节假日。

-- 如果起始日期在周末,则计算当天时间。

DECLARE minutes FLOAT;

DECLARE next_day DATETIME;

SET minutes=0.0;

SET next_day=start_day;

IF(TIMESTAMPDIFF(DAY,start_day,end_day)

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,start_day,end_day);

ELSE

WHILE TIMESTAMPDIFF(DAY,next_day,end_day)>=1.0 DO

IF ((DAYOFWEEK(next_day)=7) OR (DAYOFWEEK(next_day)=1)) THEN

SET next_day=next_day+INTERVAL 1 DAY;

ELSE

SET next_day=next_day+INTERVAL 1 DAY;

SET minutes=minutes+480.0;

END IF;

END WHILE;

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,next_day,end_day);

IF ((DAYOFWEEK(start_day)=7) OR (DAYOFWEEK(start_day)=1)) THEN

SET minutes=minutes+TIMESTAMPDIFF(MINUTE,start_day,CONVERT(CONCAT(SUBSTRING(start_day FROM 1 FOR 10),' 17:30:00'),DATETIME));

END IF;

END IF;

RETURN minutes;

END;

没考虑中午休息时间

摘自 Jasper键盘舞步

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相关标签: 计算 工作日