MYSQL 两日期之间的工作日（除去周六日，不考虑节假日）
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最佳答案
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Create proc proc_GetWorkDay @start datetime,@end datetime as declare @weekday char(6) set @start=convert(datetime,convert(varchar(50),@start,23)) set @end=convert(datetime,convert(varchar(50),@end,23)) set nocount on create table #tp ([id] int identity(1,1),[date] varchar(50),[weekday] char(6)) while @start<=@end begin select @weekday=datename(weekday,@start) if @weekday NOT IN ( 'Saturd' ,'Sunday','星期六','星期日') insert #tp([date],[weekday]) values(convert(varchar(50),@start,23),@weekday) select @start=dateadd(day,1,@start) end select * from #tp drop table #tp set nocount off GO 调用： exec proc_GetWorkDay '2015-11-20,'2015-11-23'
奖励园豆：5
求mysql版的啊，太神
@小小高: select (floor(days/7)*5+days%7
-case when 6 between wd and wd+days%7-1 then 1 else 0 end
-case when 7 between wd and wd+days%7-1 then 1 else 0 end ) INTO str
from
(select datediff(e,s) as days,weekday(s)+1 as wd) as a;
ELSE
select (-(floor(days/7)*5+days%7
-case when 6 between wd and wd+days%7-1 then 1 else 0 end
-case when 7 between wd and wd+days%7-1 then 1 else 0 end) ) into str
from
(select datediff(s,e) as days,weekday(e)+1 as wd) as a;
end if;
RETURN str;
END
select first_func('2015-11-17','2015-11-20')
其他回答(1)
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算上2个日期相差的总天数，将总天数放入一个数组，然后判断出为周六和周日的总天数，总天数-周六周日的总天数，就能得到了。
具体代码自行实现。