mysql自定義函式計算時間段內的工作日(支援跨年)
阿新 • • 發佈:2020-07-27
①
CREATE DEFINER=`root`@`%` FUNCTION `WORKDAYSONEYEAR`(`datefrom` datetime,`dateto` datetime) RETURNS int(20) NO SQL BEGIN declare days int default 1; # 如果起始時間大於結束時間或者日期跨年那麼直接返回-1,表示不支援 if (datefrom > dateto or year(datefrom) != year(dateto)) then return -1; end if; set days = case # 同一周的情況:計算時間間隔再減去週六週日的天數 # 每週開始時間為星期日,1是星期日 7是星期六 when week(dateto)-week(datefrom) = 0 then dayofweek(dateto) - dayofweek(datefrom) + 1 - case when (dayofweek(datefrom) > 1 and dayofweek(dateto) < 7) then 0 when (dayofweek(datefrom) = 1 and dayofweek(dateto) =7) then 2 else 1 end #不是同一周的情況:間隔週數 * 5 加上同一周的工作日 else (week(dateto)-week(datefrom)-1) * 5 + case when dayofweek(datefrom) = 1 then 5 when dayofweek(datefrom) = 7 then 0 else 7 - dayofweek(datefrom) end + case when dayofweek(dateto) = 1 then 0 when dayofweek(dateto) = 7 then 5 else dayofweek(dateto) - 1 end end; return days; end
②
CREATE DEFINER=`root`@`%` FUNCTION `WORKDAYSTWOYEARS`(`startdate` datetime,`enddate` datetime) RETURNS int(20) BEGIN #起始時間大於結束時間,直接返回-1,表示不支援 if (startdate > enddate) then return -1; #同一年的情況下,直接使用上面的WORKDAYSONEYEAR()函式計算 ELSEIF (year(startdate) = year(enddate)) then set @days = WORKDAYSONEYEAR(startdate,enddate); return @days; #年份相差一年,分兩段進行處理 ELSEIF (year(startdate) < year(enddate)) then set @yearofstartdate = year(startdate); set @yearofenddate = year(enddate); set @lastdayofstartdate = CONCAT(@yearofstartdate,'-12-31'); set @intervelone = workdaynum(startdate,@lastdayofstartdate); set @days = @intervelone; set @firstdayofenddate = CONCAT(@yearofenddate,'-01-01'); set @interveltwo = workdaynum(@firstdayofenddate,enddate); set @days = @intervelone + @interveltwo; end if; return @days; end
測試:
select WORKDAYS('2019-12-15','2020-01-05');