如何计算日期..^^

Question

用户选择一个开始日期和一个结束日期,, 并有七个复选框分别为星期一至星期日,,用户根据情况选择星期几,,如何好效计算出用户选择的天数????比如:用户选择开始日期为6月1日,结束日期为7月15日,在复选框选择了星期一至星期五,,如何计算6月1日至7月15日星期一至星期五有多少天??

Answer 1

  DateTime dt1 = DateTime.Parse("2009-06-01");
            DateTime dt2 = DateTime.Parse("2009-07-15");
            int wd1 = (int)dt1.DayOfWeek;
            int wd2 = (int)dt2.DayOfWeek;            int iDays =((TimeSpan) (dt2 - dt1)).Days;
            int WorkDay = wd1;
            iDays -= wd1;            if (iDays % 7 == 0)
            {
                WorkDay += (iDays / 7) * 5;
            }
            else
            {
                WorkDay += (iDays / 7) * 5 + wd2;
            }
WorkDay +=1;            Response.Write(WorkDay.ToString());

Answer 2

DayOfWeek直接获得吧
using System;
using System.Collections.Generic;
using System.Text;namespace WeekCount
{
    class Program
    {
        static void Main(string[] args)
        {
            int i = 0;
            for (DateTime dt = DateTime.Parse("2009-6-17"); dt < DateTime.Parse("2009-7-17"); dt=dt.AddDays(1))
            {
                string str = dt.DayOfWeek.ToString("D");
                if (str == "1" || str == "6")
                {
                    i++;
                }
            }
            Console.WriteLine(i);
            Console.Read();
        }
    }
}

Answer 3

 DateTime dt1 = Convert.ToDateTime(TextBox1.Text);
            DateTime dt2 = Convert.ToDateTime(TextBox2.Text);
            TimeSpan ts = dt2 - dt1;
            Label1.Text = ts.Days.ToString();
            int n = ts.Days;            switch (dt2.DayOfWeek.ToString())
            {
                case "Monday":
                    n = (n) / 7;
                    break;
                case "Tuesday":
                    n = (n + 1) / 7;
                    break;
                case "Wednesday":
                    n = (n + 2) / 7;
                    break;
                case "Thursday":
                    n = (n + 3) / 7;
                    break;
                case "Friday":
                    n = (n + 4) / 7;
                    break;
                case "Saturday":
                    n = (n + 5) / 7;
                    break;
                case "Sunday":
                    n = (n + 6) / 7;
                    break;
                default:
                    n = 999;
                    break;
            }
            Label1.Text += "|"+n.ToString();

Answer 4

using System;
using System.Collections.Generic;
using System.Text;
using System.Drawing;
using System.Drawing.Imaging;
using System.Windows.Forms;
using System.Reflection;
using System.Collections;
namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int i = 0;
            for (DateTime dt = DateTime.Parse("2009-6-1"); dt < DateTime.Parse("2009-7-15"); dt = dt.AddDays(1))
            {
                string str = dt.DayOfWeek.ToString("D");
                //Console.WriteLine(str);
                if (str == "1" || str == "6")
                {
                    i++;
                }
            }
            Console.WriteLine(i);
            Console.Read();        }       
    }
}

Answer 5

using System;
using System.Collections.Generic;
using System.Text;
using System.Drawing;
using System.Drawing.Imaging;
using System.Windows.Forms;
using System.Reflection;
using System.Collections;
namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int i = 0;
            for (DateTime dt = DateTime.Parse("2009-6-1"); dt < DateTime.Parse("2009-7-15"); dt = dt.AddDays(1))
            {
                string str = dt.DayOfWeek.ToString("D");
              //  Console.WriteLine(dt.DayOfWeek.ToString());
               // Console.WriteLine(str);
                if (str != "6" && str != "0")
                {
                    i++;
                }
            }
            Console.WriteLine(i);
            Console.Read();        }       
    }
}这个事对的

Answer 6

   List<DateTime> lst=new List<DateTime>();
            for (DateTime dt = DateTime.Parse("2009-6-1"); dt < DateTime.Parse("2009-7-15"); dt = dt.AddDays(1))
            {
              if(Convert.ToInt16(dt.DayOfWeek)>0 and Convert.ToInt16(dt.DayOfWeek)<6)  
              {
                   lst.Add(dt);
              }
            }
          

Answer 7

    /// <summary>
    /// 获取时间段内星期几个天数综合
    /// </summary>
    /// <param name="beginDate">开始日期</param>
    /// <param name="endDate">结束日期</param>
    /// <param name="DayOfWeek">星期几</param>
    /// <returns></returns>
    private int GetDayCount(DateTime beginDate, DateTime endDate, int DayOfWeek)
    {
        int i = 0;        for (DateTime date = beginDate; date <= endDate; date = date.AddDays(1))
        {
            int day = (int)date.DayOfWeek;
            if (day == DayOfWeek)
            {
                i++;
            }
        }        return i;
    }

Answer 8

任意两个时间之间的星期几的次数-横.sqlif exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_weekdaycount]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[f_weekdaycount]
GO/**//*--计算任意两个时间之间的星期几的次数(横向显示)    本方法直接判断 @@datefirst 做对应处理
    不受 sp_language 及 set datefirst 的影响     --邹建 2004.08(引用请保留此信息)--*//**//*--调用示例
    
    select * from f_weekdaycount('2004-9-01','2004-9-02')
--*/
create function f_weekdaycount(
@dt_begin datetime,
@dt_end datetime
)returns table
as
return(
    select 跨周数
        ,周一=case a
            when -1 then case when 1 between b and c then 1 else 0 end
            when  0 then case when b<=1 then 1 else 0 end
                    +case when c>=1 then 1 else 0 end
            else a+case when b<=1 then 1 else 0 end
                +case when c>=1 then 1 else 0 end
            end
        ,周二=case a
            when -1 then case when 2 between b and c then 1 else 0 end
            when  0 then case when b<=2 then 1 else 0 end
                    +case when c>=2 then 1 else 0 end
            else a+case when b<=2 then 1 else 0 end
                +case when c>=2 then 1 else 0 end
            end
        ,周三=case a
            when -1 then case when 3 between b and c then 1 else 0 end
            when  0 then case when b<=3 then 1 else 0 end
                    +case when c>=3 then 1 else 0 end
            else a+case when b<=3 then 1 else 0 end
                +case when c>=3 then 1 else 0 end
            end
        ,周四=case a
            when -1 then case when 4 between b and c then 1 else 0 end
            when  0 then case when b<=4 then 1 else 0 end
                    +case when c>=4 then 1 else 0 end
            else a+case when b<=4 then 1 else 0 end
                +case when c>=4 then 1 else 0 end
            end
        ,周五=case a
            when -1 then case when 5 between b and c then 1 else 0 end
            when  0 then case when b<=5 then 1 else 0 end
                    +case when c>=5 then 1 else 0 end
            else a+case when b<=5 then 1 else 0 end
                +case when c>=5 then 1 else 0 end
            end
        ,周六=case a
            when -1 then case when 6 between b and c then 1 else 0 end
            when  0 then case when b<=6 then 1 else 0 end
                    +case when c>=6 then 1 else 0 end
            else a+case when b<=6 then 1 else 0 end
                +case when c>=6 then 1 else 0 end
            end
        ,周日=case a
            when -1 then case when 0 between b and c then 1 else 0 end
            when  0 then case when b<=0 then 1 else 0 end
                    +case when c>=0 then 1 else 0 end
            else a+case when b<=0 then 1 else 0 end
                +case when c>=0 then 1 else 0 end
            end
    from(
        select 跨周数=case when @dt_begin<@dt_end
                then (datediff(day,@dt_begin,@dt_end)+7)/7
                else (datediff(day,@dt_end,@dt_begin)+7)/7 end
            ,a=case when @dt_begin<@dt_end
                then datediff(week,@dt_begin,@dt_end)-1
                else datediff(week,@dt_end,@dt_begin)-1 end
            ,b=case when @dt_begin<@dt_end
                then (@@datefirst+datepart(weekday,@dt_begin)-1)%7
                else (@@datefirst+datepart(weekday,@dt_end)-1)%7 end
            ,c=case when @dt_begin<@dt_end
                then (@@datefirst+datepart(weekday,@dt_end)-1)%7
                else (@@datefirst+datepart(weekday,@dt_begin)-1)%7 end)a
)
go

Answer 9

上面的有点错误,最后不用+1DateTime dt1 = DateTime.Parse("2009-06-01"); 
            DateTime dt2 = DateTime.Parse("2009-07-15"); 
            int wd1 = (int)dt1.DayOfWeek; 
            int wd2 = (int)dt2.DayOfWeek;             int iDays =((TimeSpan) (dt2 - dt1)).Days; 
            int WorkDay = wd1; 
            iDays -= wd1;             if (iDays % 7 == 0) 
            { 
                WorkDay += (iDays / 7) * 5; 
            } 
            else 
            { 
                WorkDay += (iDays / 7) * 5 + wd2; 
            } 
             Response.Write(WorkDay.ToString());

Answer 10

DateTime dt1 = DateTime.Parse("2009-06-01"); 
            DateTime dt2 = DateTime.Parse("2009-07-15"); 
            int wd1 = (int)dt1.DayOfWeek; 
            int wd2 = (int)dt2.DayOfWeek;             int iDays =((TimeSpan) (dt2 - dt1)).Days; 
            int WorkDay = wd1; 
            iDays -= wd1;             if (iDays % 7 == 0) 
            { 
                WorkDay += (iDays / 7) * 5; 
            } 
            else 
            { 
                WorkDay += (iDays / 7) * 5 + wd2; 
            } 
WorkDay +=1;             Response.Write(WorkDay.ToString());

Answer 11

肯定不是最简单的，但是思路却是非常清晰的。public virtual class DateTimeHelper
    {
        /// <summary>
        /// 功能：用于统计从开始到结束时间段内的特定的DayOfWeek的天数
        /// </summary>
        /// <param name="paramStart">开始时间</param>
        /// <param name="paramEnd">结束时间</param>
        /// <param name="paramDayOfWeek">要统计的DayOfWeek</param>
        /// <returns>返回从开始到结束时间段内的特定的DayOfWeek的天数</returns>
        public static Int32 CalcualteDays(DateTime paramStart, DateTime paramEnd, params DayOfWeek[] paramDayOfWeek)
        {
            TimeSpan tempStartEndTimeSpan = new TimeSpan();
            Int32 tempEffectiveDaysCount = 0;
            Int32 tempTotalDayCount = 0;
            int tempLeftDays = 0;
            switch (paramDayOfWeek.Length)
            { 
                case 0:
                    tempTotalDayCount = 0;
                    tempEffectiveDaysCount = 0;
                    break;
                default:                    
                    if (paramEnd < paramStart)
                    {
                        DateTime? tempExchange = null;
                        tempExchange = paramEnd;
                        paramEnd = paramStart;
                        paramStart = (DateTime)tempExchange;
                    }
                    tempStartEndTimeSpan = paramEnd - paramStart;
                    tempTotalDayCount = tempStartEndTimeSpan.Days + 1;
                    if (tempTotalDayCount >= 7)
                    {
                        tempEffectiveDaysCount = (tempTotalDayCount / 7) * paramDayOfWeek.Length;
                        tempLeftDays = tempTotalDayCount % 7;
                        if (tempLeftDays > 0)
                        {
                            CalcualteDays(tempLeftDays, ref tempEffectiveDaysCount, paramStart, paramDayOfWeek);
                        }                    }
                    else
                    {
                        tempLeftDays = tempTotalDayCount;
                        CalcualteDays(tempLeftDays, ref tempEffectiveDaysCount, paramStart, paramDayOfWeek);
                    }
                    break;
            }            return tempEffectiveDaysCount;
        }
        /// <summary>
        /// 用于统计剩余天数少于7天的特定的DayOfWeek的天数
        /// </summary>
        /// <param name="paramLeftDays">剩余的天数（少于7天）</param>
        /// <param name="paramEffectiveDaysCount">用于累加的符合特定DayOfWeek的累加器</param>
        /// <param name="paramStart">用于统计的开始日期</param>
        /// <param name="paramDayOfWeek">特定的DayOfWeek</param>
        private static void CalcualteDays(Int32 paramLeftDays, ref Int32 paramEffectiveDaysCount, DateTime paramStart, DayOfWeek[] paramDayOfWeek)
        {
            for (int i = 0; i < paramLeftDays; i++)
            {
                paramStart = paramStart.AddDays(i);
                for (int j = 0; j < paramDayOfWeek.Length; j++)
                {
                    if (paramDayOfWeek[j] == paramStart.DayOfWeek)
                    {
                        paramEffectiveDaysCount++;
                    }
                }
            }
        }
    }