检查经过了多少小时/天

Question

我正在尝试创建一个功能，以每小时和每天进行奖励。

为此，我想实现一个给定 2 DateTime 检查经过多少小时/天的函数。

我已经有一个协程可以每秒调用该函数。

例如，我想每隔 00:00:00、01:00:00 ... 12:00:00 ... 23:00:00 颁奖

例子：

DateTime date0 = new DateTime(2018, 8, 9, 20, 59, 59);

DateTime date1 = new DateTime(2018, 8, 9, 21, 0, 0);

int hoursPassed = (int) date1.Subtract(date0).TotalHours;

Debug.Log("Hours "+ hoursPassed);

在这种情况下，hoursPassed 将等于 0。我想创建一个函数（可能已经存在）来检索我 1 。日子也一样。

示例 2

DateTime date0 = new DateTime(2018, 8, 9, 23, 59, 59);

    DateTime date1 = new DateTime(2018, 8, 10, 00, 00, 00);

    int DaysPassed = (int) date1.Subtract(date0).TotalDays;

    Debug.Log("Days "+ DaysPassed);

它将检索 0，我想检索 1。

我也希望它在多个小时/天过去后工作。

有谁能够帮我？

谢谢

Answer 1

如果我理解正确：private static int GetHours(DateTime date1, DateTime date2){    return Math.Max(0, Convert.ToInt32((date1 - date2).TotalMinutes / 60 + ((date1 - date2).TotalMinutes % 60 > 0 ? 1 : 0)));}用法：    DateTime date0 = new DateTime(2018, 8, 9, 20, 59, 59);    DateTime date1 = new DateTime(2018, 8, 9, 21, 0, 0);                            Console.Write(GetHours(date1, date0).ToString());返回 1跨越一天以上：    DateTime date0 = new DateTime(2018, 8, 9, 20, 59, 59);    DateTime date1 = new DateTime(2018, 8, 10, 21, 0, 0);                           Console.Write(GetHours(date1, date0).ToString());返回 25

Answer 2

在我看来，这Math.Ceiling是完成这项工作的最简单方法：例子：DateTime date0 = new DateTime(2018, 8, 9, 20, 59, 59);DateTime date1 = new DateTime(2018, 8, 9, 21, 0, 0);int hoursPassed = (int)Math.Ceiling(date1.Subtract(date0).TotalHours);Console.WriteLine("Hours " + hoursPassed);示例 2：DateTime date0 = new DateTime(2018, 8, 9, 23, 59, 59);DateTime date1 = new DateTime(2018, 8, 10, 00, 00, 00);int DaysPassed = (int)Math.Ceiling(date1.Subtract(date0).TotalDays);Console.WriteLine("Days "+ DaysPassed);

Answer 3

最简单的方法是通过在仅对小时感兴趣时去除分钟/秒以及在仅对天感兴趣时通过修剪小时/分钟/秒来创建修剪 DateTime变量。然后计算小时或天的差异。// Calculates the number of hour strikes between the two given timespublic static int HourStrikesBetween(DateTime from, DateTime to){    if(from > to)    {        throw new ArgumentException("from must not be after to");    }    // Trim to hours    DateTime fromTrimmed = new DateTime(from.Year, from.Month, from.Day, from.Hour, 0, 0);    DateTime toTrimmed = new DateTime(to.Year, to.Month, to.Day, to.Hour, 0, 0);    int hours = (int)(toTrimmed - fromTrimmed).TotalHours;    return hours;}// Calculates the number of midnights between the two given timespublic static int MidnightsBetween(DateTime from, DateTime to){    if (from > to)    {        throw new ArgumentException("from must not be after to");    }    // Trim to days    DateTime fromTrimmed = new DateTime(from.Year, from.Month, from.Day);    DateTime toTrimmed = new DateTime(to.Year, to.Month, to.Day);    int days = (toTrimmed - fromTrimmed).Days;    return days;}样本：DateTime date0 = new DateTime(2018, 8, 9, 20, 59, 59);DateTime date1 = new DateTime(2018, 8, 9, 21, 0, 0);int hourspassed = HourStrikesBetween(date0, date1);  // = 1date0 = new DateTime(2018, 8, 9, 20, 00, 00);date1 = new DateTime(2018, 8, 9, 21, 50, 0);hourspassed = HourStrikesBetween(date0, date1);  // = still 1date0 = new DateTime(2018, 8, 9, 20, 59, 59);date1 = new DateTime(2018, 8, 9, 22, 10, 0);hourspassed = HourStrikesBetween(date0, date1);  // = 2date0 = new DateTime(2018, 8, 9, 23, 59, 59);date1 = new DateTime(2018, 8, 10, 00, 00, 00);int daysPassed = MidnightsBetween(date0, date1);  // = 1请注意，如果您始终使用 UTC 日期时间而不是本地日期时间，这只会在跨越夏令时边界时起作用。