如何把时间相减,求助中?
比如网吧的时间差,我去上网的时候它给我记下一个时间,走的时候又获得一个时间.这中间要如何计算时间差来获得金额啊.用C#设计的话.学习啊...希望大虾指点一下.谢先. 问题点数:20、回复次数:5Top
1 楼zhongkeruanjian(编程亮子)回复于 2006-03-04 19:06:24 得分 5
dt1-dt2Top
2 楼syhan(藏书人)回复于 2006-03-04 20:04:19 得分 5
直接减啊,又没有进位什么的,获取时间减就是了Top
3 楼lkcllll(为了工作没办法)回复于 2006-03-04 20:10:47 得分 5
DateTime dt1;
DateTime dt2;
TimeSpan ts=dt1-dt2;
ts里就有属性了,要获取分钟就ts.minute总之里面分,秒,那些都在属性里了Top
4 楼singlepine(小山)回复于 2006-03-04 20:34:32 得分 5
//在asp.net中怎么样计算两个日期相差的年、月份、日期、小时、分钟 、妙等
// 调用
// DateTime a=Convert.ToDateTime("2005-09-03 20:15");
// DateTime b=Convert.ToDateTime("2005-09-04 09:09 ");
// double d=Bll.Common.DateDiff(Bll.Common.EnumDateCompare.day,a,b);
// Response.Write(d.ToString("f0"));//四舍五入
public enum EnumDateCompare
{
year =1,
month =2,
day =3,
hour =4,
minute =5,
second =6
}
public static double DateDiff(EnumDateCompare howtocompare, System.DateTime startDate, System.DateTime endDate)
{
double diff=0;
System.TimeSpan TS = new System.TimeSpan(endDate.Ticks-startDate.Ticks);
switch (howtocompare)
{
case EnumDateCompare.year:
diff = Convert.ToDouble(TS.TotalDays/365);
break;
case EnumDateCompare.month:
diff = Convert.ToDouble((TS.TotalDays/365)*12);
break;
case EnumDateCompare.day:
diff = Convert.ToDouble(TS.TotalDays);
break;
case EnumDateCompare.hour:
diff = Convert.ToDouble(TS.TotalHours);
break;
case EnumDateCompare.minute:
diff = Convert.ToDouble(TS.TotalMinutes);
break;
case EnumDateCompare.second:
diff = Convert.ToDouble(TS.TotalSeconds);
break;
}
return diff;
}
Top
5 楼yufei1314(落墨长裙)回复于 2006-03-05 00:40:22 得分 0
不知道如何给分好,所以每人给了5分.大家别生气.
谢谢大家回答.祝福好运...Top




