在DateTime中存储超过24小时

Question

我在一个奇怪且非理性的行业工作,我们需要能够将时间表示为06:00:00到30:00:00而不是0:00:00到24:00:00.有没有办法使用DateTime类型？如果我尝试构造一个小于24的小时值的日期时间,则会抛出异常.

Answer 1

我认为这应该只是一个演示问题.

允许您的用户以这种奇怪的格式输入数据,并立即将其转换为UTC.对UTC时间进行所有计算.然后创建一个ToString方法将结果转换回您的奇怪格式.您可能还需要一些其他实用程序方法和属性,例如实现WeirdDateTime.Day.

您可以围绕DateTime编写一个包装类,并拥有该类所需的所有转换和实用程序方法.我已经开始了 - 通过以奇怪的格式从字符串中实现解析.这不是任何准备好的生产代码,但也许它可以为您提供一些如何处理此问题的想法:

class WeirdDateTime
{
    public DateTime DateTime { get; set; }

    public WeirdDateTime(int year, int month, int day, int hour, int minute, int second, DateTimeKind kind)
    {
        if (hour < 6 || hour >= 30)
            throw new ArgumentException("Not a valid WeirdDateTime", "hour");

        bool addDay;
        if (hour >= 24)
        {
            addDay = true;
            hour -= 24;
        }
        else
        {
            addDay = false;
        }

        DateTime dateTime = new DateTime(year, month, day, hour, minute, second, kind);
        if (addDay)
            dateTime = dateTime.AddDays(1);

        DateTime = dateTime;
    }

    public static WeirdDateTime Parse(string s)
    {
        Regex regex = new Regex(@"(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2})");
        Match match = regex.Match(s);
        if (!match.Success)
            throw new FormatException("Not a valid WeirdDateTime");

        int[] parts = match.Groups.Cast<Group>()
            .Skip(1)
            .Select(x => int.Parse(x.Value))
            .ToArray();

        int year = parts[0];
        int month = parts[1];
        int day = parts[2];
        int hour = parts[3];
        int minute = parts[4];
        int second = parts[5];

        return new WeirdDateTime(year, month, day, hour, minute, second, DateTimeKind.Unspecified);
    }

    public override string ToString()
    {
        throw new NotImplementedException("Write this!");
    }
}

class Program
{
    public static void Main()
    {
        WeirdDateTime weirdDateTime = WeirdDateTime.Parse("2010-01-19 27:00:00");
        DateTime dateTimeUtc = weirdDateTime.DateTime.ToUniversalTime();
        Console.WriteLine(dateTimeUtc);
    }
}

Answer 2

使用TimeSpan来代替怎么样？

DateTime departure = new DateTime(2010, 6, 12, 18, 32, 0);
DateTime arrival = new DateTime(2010, 6, 13, 22, 47, 0);
TimeSpan travelTime = arrival - departure;  
Console.WriteLine("{0} - {1} = {2}", arrival, departure, travelTime);  

然后使用TotalHoursTimeSpan 对象的属性