我想处理Timeblocks,这意味着一组两个DateTimes,例如代表员工的存在。是否已经有任何结构可用于在特定时间之前或之后搜索块? 我可以通过很多方式来表达这种情况,就像我说的有两个DateTimes的开始和结束,或者有一个Datetime for start和TimeSpan。但是我希望它们能够在一种集合中处理。那么有什么类似我可以使用或者我必须完全自己实现它吗?
由于
答案 0 :(得分:2)
这个图书馆很棒 - 可能会受到启发
答案 1 :(得分:1)
班级:
public class TimePeriod
{
public DateTime Oldest { get; set; }
public DateTime Newest { get; set; }
public TimePeriod(DateTime oldest, DateTime newest)
{
Oldest = oldest;
Newest = newest;
}
public bool Contains (DateTime time)
{
return Oldest.CompareTo(time) <= 0 && Newest.CompareTo(time) >= 0;
}
public bool IsAfter(DateTime time)
{
return Newest.CompareTo(time) <= 0;
}
public bool IsBefore(DateTime time)
{
return Oldest.CompareTo(time) >= 0;
}
}
测试:
class Program
{
static void Main(string[] args)
{
var period = new TimePeriod(
DateTime.Now.AddDays(-2),
DateTime.Now.AddDays(1));
var date = DateTime.Now;
var contains = period.Contains(date); // true
var isBefore = period.IsBefore(date); // false
var isAfter = period.IsAfter(date); // false
date = DateTime.Now.AddDays(-10);
contains = period.Contains(date); // false
isBefore = period.IsBefore(date); // true
isAfter = period.IsAfter(date); // false
date = DateTime.Now.AddDays(10);
contains = period.Contains(date); // false
isBefore = period.IsBefore(date); // false
isAfter = period.IsAfter(date); // true
}
}
现在,您可以使用集合和linq以及扩展方法和lambda表达式来查找时间块。
答案 2 :(得分:0)
这不是内置的。如果您想自己实现,可能需要创建 struct 。这将为您提供值类型的复制语义。此类值的行为与int
或DateTime
等内置类型相似。使用起来非常直观。
答案 3 :(得分:0)
您可以查看TimeSpan。这是一个处理“时间块”的结构
答案 4 :(得分:0)
之前我使用过DateSpan
结构。你可以随意扩展,但这会给你一个起点。
using System;
using System.Collections.Generic;
using System.Runtime.InteropServices;
namespace StackOverFlowDateSpan
{
[StructLayout(LayoutKind.Auto)]
[Serializable]
public struct DateSpan : IComparable, IComparable<DateSpan>, IEquatable<DateSpan>
{
public DateSpan(DateTime start, DateTime end)
: this()
{
Start = start;
End = end;
}
#region Properties
public TimeSpan Duration
{
get { return TimeSpan.FromTicks((End - Start).Ticks); }
}
public DateTime End { get; private set; }
public DateTime Start { get; private set; }
#endregion
public int CompareTo(DateSpan other)
{
long otherTicks = other.Duration.Ticks;
long internalTicks = Duration.Ticks;
return internalTicks > otherTicks ? 1 : (internalTicks < otherTicks ? -1 : 0);
}
public bool Equals(DateSpan other)
{
return End.Equals(other.End) && Start.Equals(other.Start);
}
public int CompareTo(object other)
{
if (other == null)
{
return 1;
}
if (!(other is DateSpan))
{
throw new ArgumentNullException("other");
}
return CompareTo((DateSpan)other);
}
public override bool Equals(object other)
{
if (ReferenceEquals(null, other))
{
return false;
}
return other is DateSpan && Equals((DateSpan)other);
}
public override int GetHashCode()
{
unchecked
{
return (End.GetHashCode() * 397) ^ Start.GetHashCode();
}
}
public static bool operator ==(DateSpan left, DateSpan right)
{
return left.Equals(right);
}
public static bool operator !=(DateSpan left, DateSpan right)
{
return !left.Equals(right);
}
private sealed class EndStartEqualityComparer : IEqualityComparer<DateSpan>
{
#region IEqualityComparer<DateSpan> Members
public bool Equals(DateSpan x, DateSpan y)
{
return x.End.Equals(y.End) && x.Start.Equals(y.Start);
}
public int GetHashCode(DateSpan obj)
{
unchecked
{
return (obj.End.GetHashCode() * 397) ^ obj.Start.GetHashCode();
}
}
#endregion
}
private static readonly IEqualityComparer<DateSpan> _endStartComparerInstance = new EndStartEqualityComparer();
public static IEqualityComparer<DateSpan> EndStartComparer
{
get { return _endStartComparerInstance; }
}
}
}
答案 5 :(得分:0)
感谢您的帮助!我将仔细研究TimePeriod库并对Linq进行一些实验。我已经有了一个实现二进制搜索的approch,所以如果有人感兴趣你可以写我;)