C#獲得兩個時光的時光差並去除周末（取任務日）的辦法

C#獲得兩個時光的時光差並去除周末（取任務日）的辦法。本站提示廣大學習愛好者：（C#獲得兩個時光的時光差並去除周末（取任務日）的辦法）文章只能為提供參考，不一定能成為您想要的結果。以下是C#獲得兩個時光的時光差並去除周末（取任務日）的辦法正文

本文實例講述了C#獲得兩個時光的時光差並去除周末的辦法。分享給年夜家供年夜家參考。詳細剖析以下：

普通來講取時光差的代碼許多，然則可以或許只取任務日的時光差的代碼很少，這段代碼就來完成這一功效。

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;//差距的一切天數
 int totleWeek = AllDays / 7;//差異若干周
 int yuDay = AllDays % 7; //除全部禮拜的天數
 int lastDay = 0;
 if (yuDay == 0) //正好全部周
 {
  lastDay = AllDays - (totleWeek * 2);
 }
 else
 {
  int weekDay = 0;
  int endWeekDay = 0; //過剩的天數有幾天是周六或許周日
  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();
}

願望本文所述對年夜家的C#法式設計有所贊助。