C#将天数添加到给定日期

时间:2016-03-05 17:41:11

标签: c# datetime console console-application leap-year

我在第一次看的时候做了相当简单的练习,但结果却相当困难。我需要将日期作为输入,并且还要添加日期,也可以从用户输入中获取。我已经做了一些函数和一些简单的计算,现在我已经从日期(01,01,0001为零)获取所有日期,例如:

第一年1月第二年(01.01.0002)+ 0天等于365天它也正确计算如果我添加一些天:01.01.0002 + 12天= 387 ..它也计算闰年。现在我有总天数,我只需要将其转换为正常的日/月/年格式..

我不允许使用DATETIME

这是我的代码:

private static int[] daysPerMonth = new int[12];
    private static int days;
    private static int months;
    private static int years;
    private static int add;

    private static void Main()
    {
        Console.Write("Enter day : ");
        int.TryParse(Console.ReadLine(), out days);
        Console.Write("Enter Month : ");
        int.TryParse(Console.ReadLine(), out months);
        Console.Write("Enter Year : ");
        int.TryParse(Console.ReadLine(), out years);
        Console.Write("Enter days to add : ");
        int.TryParse(Console.ReadLine(), out add);
        int totalDays = GetTotalDays(new[] {days, months, years});
        totalDays += add;
        TransformIntoDate(totalDays);

        Console.ReadKey();
    }

    private static void TransformIntoDate(int inputDays)
    {

    }
    private static int GetTotalDays(IReadOnlyList<int> date)
    {
        int totalDays = 0;
        for (int i = date[2]; i > 1; i--)
        {
            if (IsLeap(i))
            {
                daysPerMonth = new[] {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
                totalDays += 366;
            }
            else
            {
                daysPerMonth = new[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
                totalDays += 365;
            }
        }
        for (int i = 1; i <= date[1]; i++)
        {
            if (i == date[1])
            {
                totalDays += date[0] - 1;
            }
            else
            {
                totalDays += daysPerMonth[i];
            }
        }
        return totalDays;
    }

    private static bool IsLeap(int year)
    {
        if (year%400 == 0) return true;
        return (year%4 == 0) && (year%100 != 0);
    }

1 个答案:

答案 0 :(得分:1)

有两种方法可以“保存”日期:分别保存年,月,日或保存总天数(或小时或分钟或秒或毫秒...此处选择一个度量单位) 0分“。例如,.NET的DateTime使用100纳秒作为Tick,将1月1日作为“0点”使用。 Unix从1970年1月1日开始使用秒数。显然,.NET和Unix的方式在内存中更紧凑(单个值可以保存),如果你想加/减数量(简单地加/减),这非常有用。问题是,将此内部数字转换为年/月/日或将年/月/日转换为此数字会更复杂。

关于年内/月/日内部数字如何完成的简单示例:

public class MyDate
{
    public int TotalDaysFrom00010101 { get; private set; }

    private const int DaysIn400YearCycle = 365 * 400 + 97;
    private const int DaysIn100YearCycleNotDivisibleBy400 = 365 * 100 + 24;
    private const int DaysIn4YearCycle = 365 * 4 + 1;

    private static readonly int[] DaysPerMonthNonLeap = new[] 
    {
        31, 
        31 + 28, 
        31 + 28 + 31, 
        31 + 28 + 31 + 30, 
        31 + 28 + 31 + 30 + 31, 
        31 + 28 + 31 + 30 + 31 + 30, 
        31 + 28 + 31 + 30 + 31 + 30 + 31, 
        31 + 28 + 31 + 30 + 31 + 30 + 31 + 31, 
        31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30, 
        31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31, 
        31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30, 
        31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 // Useless
    };

    private static readonly int[] DaysPerMonthLeap = new[] 
    {
        31, 
        31 + 29, 
        31 + 29 + 31, 
        31 + 29 + 31 + 30, 
        31 + 29 + 31 + 30 + 31, 
        31 + 29 + 31 + 30 + 31 + 30, 
        31 + 29 + 31 + 30 + 31 + 30 + 31, 
        31 + 29 + 31 + 30 + 31 + 30 + 31 + 31, 
        31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30, 
        31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31, 
        31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30, 
        31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 // Useless
    };

    public static bool IsLeap(int year)
    {
        if (year % 400 == 0) return true;
        return (year % 4 == 0) && (year % 100 != 0);
    }

    public void SetDate(int year, int month, int day)
    {
        TotalDaysFrom00010101 = 0;

        {
            int year2 = year - 1;

            // Full 400 year cycles
            TotalDaysFrom00010101 += (year2 / 400) * DaysIn400YearCycle;
            year2 %= 400;

            // Remaining 100 year cycles (0...3)
            if (year2 >= 100)
            {
                year2 -= 100;
                TotalDaysFrom00010101 += DaysIn100YearCycleNotDivisibleBy400;

                if (year2 >= 100)
                {
                    year2 -= 100;
                    TotalDaysFrom00010101 += DaysIn100YearCycleNotDivisibleBy400;

                    if (year2 >= 100)
                    {
                        year2 -= 100;
                        TotalDaysFrom00010101 += DaysIn100YearCycleNotDivisibleBy400;
                    }
                }
            }

            // Full 4 year cycles
            TotalDaysFrom00010101 += (year2 / 4) * DaysIn4YearCycle;
            year2 %= 4;

            // Remaining non-leap years
            TotalDaysFrom00010101 += year2 * 365;
        }

        // Days from the previous month
        if (month > 1)
        {
            // -2 is because -1 is for the 1 January == 0 index, plus -1 
            // because we must add only the previous full months here. 
            // So if the date is 1 March 2016, we must add the days of 
            // January + February, so month 3 becomes index 1.
            TotalDaysFrom00010101 += DaysPerMonthNonLeap[month - 2];

            if (month > 2 && IsLeap(year))
            {
                TotalDaysFrom00010101 += 1;
            }
        }

        // Days (note that the "instant 0" in this class is day 1, so
        // we must add day - 1)
        TotalDaysFrom00010101 += day - 1;
    }

    public void GetDate(out int year, out int month, out int day)
    {
        int days = TotalDaysFrom00010101;

        // year
        {
            year = days / DaysIn400YearCycle * 400;
            days %= DaysIn400YearCycle;

            if (days >= DaysIn100YearCycleNotDivisibleBy400)
            {
                year += 100;
                days -= DaysIn100YearCycleNotDivisibleBy400;

                if (days >= DaysIn100YearCycleNotDivisibleBy400)
                {
                    year += 100;
                    days -= DaysIn100YearCycleNotDivisibleBy400;

                    if (days >= DaysIn100YearCycleNotDivisibleBy400)
                    {
                        year += 100;
                        days -= DaysIn100YearCycleNotDivisibleBy400;
                    }
                }
            }

            year += days / DaysIn4YearCycle * 4;
            days %= DaysIn4YearCycle;

            // Special case: 31 dec of a leap year
            if (days != 1460)
            {
                year += days / 365;
                days %= 365;
            }
            else
            {
                year += 3;
                days = 365;
            }

            year++;
        }

        // month
        {
            bool isLeap = IsLeap(year);

            int[] daysPerMonth = isLeap ? DaysPerMonthLeap : DaysPerMonthNonLeap;

            for (month = 0; month < daysPerMonth.Length; month++)
            {
                if (daysPerMonth[month] > days)
                {
                    if (month > 0)
                    {
                        days -= daysPerMonth[month - 1];
                    }

                    break;
                }
            }

            month++;
        }

        // day
        {
            day = days;
            day++;
        }
    }

    public void AddDays(int days)
    {
        TotalDaysFrom00010101 += days;
    }
}

这里的要点是我们知道有400年的“时期”,这些“周期”中的每一个都有365 * 400 + 97天。在减去这些“周期”之后,存在100年的较小“周期”,每个周期为365 * 100 + 24天。然后我们有4年的“周期”,每个周期为365 * 4 + 3天,加上剩余的年份(0 ... 3),每个周期为365天。

在添加了所有“前几年”的日期之后,我们可以添加“前几个月”的日期。在这里,我们必须考虑今年是闰年的可能性。

最后我们添加了选定的日期。

如何编写GetDate()作为练习。

现在......我如何检查结果是否正确?我们可以根据DateTime实现编写一些单元测试......类似于:

var date = default(DateTime);
var endDate = new DateTime(2017, 1, 1);

while (date < endDate)
{
    int year = date.Year;
    int month = date.Month;
    int day = date.Day;

    int totalDays = (int)(date - default(DateTime)).TotalDays;

    var md = new MyDate();
    md.SetDate(year, month, day);

    if (totalDays != md.TotalDaysFrom00010101)
    {
        Console.WriteLine("{0:d}: {1} vs {2}", date, totalDays, md.TotalDaysFrom00010101);
    }

    int year2, month2, day2;
    md.GetDate(out year2, out month2, out day2);

    if (year != year2 || month != month2 || day != day2)
    {
        Console.WriteLine("{0:d}: {1:D4}-{2:D2}-{3:D2} vs {4:D4}-{5:D2}-{6:D2}", date, year, month, day, year2, month2, day2);
    }

    date = date.AddDays(1);
}

(我知道这不是单元测试...但是展示了如何使用DateTime进行比较)

请注意,这是如何实现它的示例。我不会这样实现它。我会使用构造函数而不是struct创建一个不可变的DateTime,如SetDate()。但你的看起来似乎是一个练习,我认为应该通过一些小步骤:首先正确构建一些东西,然后使其“正式正确”。因此,第一步是构建正确的GetDate() / SetDate(),然后您可以正确地封装在更正式的正确数据结构中。这个小样本甚至缺少一些参数验证:你可以SetDate(-1, 13, 32): - )

构建GetDate()非常复杂。比我想象的要多得多。最后我能理解如何写它。请注意,最终它与Microsoft的实现非常相似(请参阅GetDatePart(int part)