C#获取两个时间的时间差并去除周末（取工作日）的方法

概述本文实例讲述了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：

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。

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