C#获取两个时间的时间差并去除周末（取工作日）的方法

本文实例讲述了C#获取两个时间的时间差并去除周末的方法。分享给大家供大家参考。具体分析如下：

一般来说取时间差的代码很多，但是能够只取工作日的时间差的代码很少，这段代码就来实现这一功能。

protected void Page_Load(object sender, EventArgs e)
{
DateTime start = Convert.ToDateTime("2012-12-10");
DateTime end= Convert.ToDateTime("2012-12-18");
TimeSpan span = end - start;
//int totleDay=span.Days;
//DateTime spanNu = DateTime.Now.Subtract(span);
int AllDays=Convert.ToInt32(span.TotalDays)+1;//差距的所有天数
int totleWeek = AllDays / 7;//差别多少周
int yuDay = AllDays ％ 7; //除了整个星期的天数
int lastDay = 0;
if (yuDay == 0) //正好整个周
{
 lastDay = AllDays - (totleWeek * 2);
}
else
{
 int weekDay = 0;
 int endWeekDay = 0; //多余的天数有几天是周六或者周日
 switch (start.DayOfWeek)
 {
 case DayOfWeek.Monday:
  weekDay = 1;
  break;
 case DayOfWeek.Tuesday:
  weekDay = 2;
  break;
 case DayOfWeek.Wednesday:
  weekDay = 3;
  break;
 case DayOfWeek.Thursday:
  weekDay = 4;
  break;
 case DayOfWeek.Friday:
  weekDay = 5;
  break;
 case DayOfWeek.Saturday:
  weekDay = 6;
  break;
 case DayOfWeek.Sunday:
  weekDay = 7;
  break;
 }
 if ((weekDay == 6 && yuDay >= 2) || (weekDay == 7 && yuDay >= 1) || (weekDay == 5 && yuDay >= 3) || (weekDay == 4 && yuDay >= 4) || (weekDay == 3 && yuDay >= 5) || (weekDay == 2 && yuDay >= 6) || (weekDay == 1 && yuDay >=7))
 {
 endWeekDay =2;
 }
 if ((weekDay == 6 && yuDay < 1) || (weekDay == 7 && yuDay <5) || (weekDay == 5 && yuDay < 2) || (weekDay == 4 && yuDay < 3) || (weekDay == 3 && yuDay < 4) || (weekDay == 2 && yuDay < 5) || (weekDay == 1 && yuDay < 6))  {
 endWeekDay = 1;
 }
 lastDay = AllDays - (totleWeek * 2) - endWeekDay;
}
lblTime.Text = lastDay.ToString();
}

希望本文所述对大家的C#程序设计有所帮助。