您的位置：首页 > 编程语言 > C#

C#获取两个时间的时间差并去除周末（取工作日）的方法

2015-05-07 10:42 1211 查看

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。

您可能感兴趣的文章:

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签： C# 时间差

相关文章推荐

新的分享

章节导航