C#取两个时间的时间差并去除非工作日

来源:互联网 发布:杭州游戏公司美工招聘 编辑:程序博客网 时间:2024/05/29 08:22
public bool isLate() 
    {
  DateTime start = Convert.ToDateTime("前一时间");
        DateTime end= Convert.ToDateTime("后一时间");
        TimeSpan span = end - start;
        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;
        }
if(lastDay>3)
return false;
    }