计算两个日期相差的天数

曾经面试的时候被问到写一段函数，实现不用系统已经提供的方法计算两个日期的差值。当时时间很短，脑袋很糊涂，今天细想了一下，实现如下:
首先定义了一个MyDate类：

class  Mydate
    {
         private   int  year;
         private   int  month;
         private   int  day;

         public   int  Year
        {
             get  {  return  year; }
        }
        
         public   int  Month
        {
             get  {  return  month; }
        }

         public   int  Day
        {
             get  {  return  day; }
        }

         public  Mydate( int  year,  int  month,  int  day)
        {
             this .year  =  year;
             this .month  =  month;
             this .day  =  day;
        }
    }

接下来我取较小的日期的年和一月一号组成一个参照的日期，分别计算两个日期和它的天数的差值，最后这两个值相减就是结果了，我觉得这样比较简单:

Code
/// <summary>
        /// here we assume that the date1 is less than date2, we choose the day(date1.year/01/01) as a reference,
        /// then calculate the the difference to the referred date separately.At last, get the result.
        /// </summary>
        /// <param name="date1"></param>
        /// <param name="date2"></param>
        /// <returns></returns>
        private int DiffDate(Mydate date1,Mydate date2)
        {
            int diffdays1 = 0;
            int diffdays2 = 0;

            diffdays1 = GetMonthDayDifftoJanuaryFirst(date1);

            for (int i = date1.Year; i < date2.Year; i++)
            {
                diffdays2 += isLeapYear(i) ? 366 : 365;
            }

            diffdays2 += GetMonthDayDifftoJanuaryFirst(date2);

            return diffdays2 - diffdays1;

        }

        /// <summary>
        /// Gets the difference to the January first.
        /// </summary>
        /// <param name="date"></param>
        /// <returns></returns>
        private int GetMonthDayDifftoJanuaryFirst(Mydate date)
        {
            int diffdays = 0;
            int[] leapdays = new int[] { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
            int[] normaldays = new int[] { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

            if (isLeapYear(date.Year))
            {
                for (int i = 0; i < date.Month - 1; i++)
                {
                    diffdays += leapdays[i];
                }
            }
            else
            {
                for (int i = 0; i < date.Month - 1; i++)
                {
                    diffdays += normaldays[i];
                }
            }
            diffdays += date.Day;

            return diffdays;
        }

        private bool isLeapYear(int year)
        {
            return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
        }

我们可以使用系统提供的类来验证，例如：

DateTime

dt1 = new DateTime(1999, 9, 10);

DateTime dt2 = new DateTime(2005, 2, 17);

int diff2 = ((TimeSpan)(dt1.Subtract(dt2))).Days;

 

改进一下：

View Code

class MyTest
    {
        private static readonly int[] _leapdays = new int[] { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
        private static readonly int[] _normaldays = new int[] { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

        public static int GetDiff2(DateTime d1, DateTime d2)
        {
            if (d1.Year == d2.Year && d1.Month == d2.Month)
                return Math.Abs(d1.Day - d2.Day);

            // make sure d1 < d2.
            if ((d1.Year > d2.Year) || (d1.Year == d2.Year && d1.Month > d2.Month))
            {
                var temp = d2;
                d2 = d1;
                d1 = temp;
            }

            int sum1 = 0;
            int sum2 = 0;
            for (int i = 1; i < d1.Month; i++)
                sum1 += GetDaysofMonth(d1.Year, i);
            sum1 += d1.Day;
            for (int i = 1; i < d2.Month; i++)
                sum2 += GetDaysofMonth(d2.Year, i);
            sum2 += d2.Day;

            for (int i = d1.Year; i < d2.Year; i++)
                sum2 += DateTime.IsLeapYear(i) ? 366 : 365;

            return sum2 - sum1;
        }

        /// <summary>
        /// Gets days of a specified year, month.
        /// </summary>
        /// <param name="year"></param>
        /// <param name="month">1-based, from 1 to 12</param>
        /// <returns></returns>
        public static int GetDaysofMonth(int year, int month)
        {
            if (DateTime.IsLeapYear(year))
                return _leapdays[month - 1];
            else
                return _normaldays[month - 1];
        }
    }

以下是测试代码：

View Code

        private void Test()
        {
            Random r = new Random();
            Debug.WriteLine("date1\tdate2\tdiffmytest\tdiffsystem");
            for (int i = 0; i < 10000; i++)
            {
                var y1 = r.Next(1957, 2012);
                var m1 = r.Next(1, 12);
                var d1 = r.Next(1, MyTest.GetDaysofMonth(y1, m1));

                var y2 = r.Next(1957, 2012);
                var m2 = r.Next(1, 12);
                var d2 = r.Next(1, MyTest.GetDaysofMonth(y2, m2));

                var date1 = new DateTime(y1, m1, d1);
                var date2 = new DateTime(y2, m2, d2);
                var v1 = MyTest.GetDiff2(date1, date2);
                var v2 = Math.Abs(date1.Subtract(date2).Days);
                Debug.WriteLine(date1.ToShortDateString() + "\t" + date2.ToShortDateString() + "\t" + v1 + "\t" + v2);
                if (v1 != v2)
                {
                    MessageBox.Show("failed" + date1.ToShortDateString() + "\t" + date2.ToShortDateString() + "\t" + v1 + "\t" + v2);
                }
            }
        }