当前位置: 代码迷 >> SQL >> SQL兑现工作日计算
  详细解决方案

SQL兑现工作日计算

热度:61   发布时间:2016-05-05 14:03:04.0
SQL实现工作日计算
CREATE FUNCTION Time_Span_of_minutes(start_day DATETIME,end_day DATETIME)RETURNS FLOATBEGIN			-- 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)<1.0) THEN					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
  相关解决方案