C#获取两个时间的时间差并去除周末(取工作日)的方法
作者:niuniu
这篇文章主要介绍了C#获取两个时间的时间差并去除周末(取工作日)的方法,可有效的实现获取工作日的功能,涉及C#时间操作的相关技巧,需要的朋友可以参考下
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。