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。