使用TimePeriod.NET的CalendarPeriodCollector计算"工作时间"会产生意外结果

Question

我正在尝试计算服务级别协议的截止日期,同时,我还需要在另一个方向上重新计算服务级别协议.

我一直在努力计算"工作时间"(即在一系列日子里工作的时间),并决定使用名为TimePeriodLibrary.NET的第三方库来完成任务.我需要做两件事:

• 如果有一个开始DateTime和一个TimeSpan,您应该收到DateTime一个服务级别协议日期到期(到期日期).
• 在开始DateTime和结束时DateTime,您应该收到TimeSpan服务级别协议应该花多长时间.

所有源代码(测试项目都在GitHub上).我有一个ServiceLevelManager完成所有工作的课程.它采取的列表WorkDays和HolidayPeriods,以计算出它们可被工作时间.这CalendarPeriodCollector堂课给出了意想不到的结果.在从时间跨度确定到期日期时起作用的期望,在我重新计算它们时不能正确计算.

任何人都可以看到我做错了什么,或者图书馆是否有错误？

namespace ServicePlanner
{
    using System;
    using System.Collections.Generic;
    using Itenso.TimePeriod;

    public class ServicePlannerManager
    {
        public ServicePlannerManager(IEnumerable<WorkDay> workDays, IEnumerable<HolidayPeriod> holidays)
        {
            this.WorkDays = workDays;
            this.Holidays = holidays;
        }

        public IEnumerable<WorkDay> WorkDays { get; set; }

        public IEnumerable<HolidayPeriod> Holidays { get; set; }

        public TimeSpan GetRemainingWorkingTime(DateTime start, DateTime dueDate)
        {
            var filter = new CalendarPeriodCollectorFilter();
            foreach (var dayOfWeek in this.WorkDays)
            {
                filter.CollectingDayHours.Add(new DayHourRange(dayOfWeek.DayOfWeek, new Time(dayOfWeek.StartTime), new Time(dayOfWeek.EndTime)));
            }

            foreach (var holiday in this.Holidays)
            {
                filter.ExcludePeriods.Add(new TimeBlock(holiday.StartTime, holiday.EndTime));
            }

            var range = new CalendarTimeRange(start, dueDate);
            var collector = new CalendarPeriodCollector(filter, range);
            collector.CollectHours();

            var duration = collector.Periods.GetTotalDuration(new TimeZoneDurationProvider(TimeZoneInfo.FindSystemTimeZoneById("UTC")));
            return duration;
            //var rounded = Math.Round(duration.TotalMinutes, MidpointRounding.AwayFromZero);
            //return TimeSpan.FromMinutes(rounded);
        }
    }
}

下面提取了失败的单元测试:

[TestFixture]
public class ServicePlannerManagerTest
{
        [Test, TestCaseSource("LocalSource")]
    public void GetRemainingWorkingTimeWithHolidayShouldOnlyEnumerateWorkingTime(DateTime startTime, TimeSpan workingHours, DateTime expectedDueDate, string expectation)
    {
        // Arrange
        var workDays = new List<WorkDay>
        { 
            new WorkDay(DayOfWeek.Monday, new DateTime(1, 1, 1, 9, 0, 0), new DateTime(1, 1, 1, 17, 0, 0)),
            new WorkDay(DayOfWeek.Tuesday, new DateTime(1, 1, 1, 9, 0, 0), new DateTime(1, 1, 1, 17, 0, 0)),
            new WorkDay(DayOfWeek.Wednesday, new DateTime(1, 1, 1, 9, 0, 0), new DateTime(1, 1, 1, 17, 0, 0)),
            new WorkDay(DayOfWeek.Thursday, new DateTime(1, 1, 1, 9, 0, 0), new DateTime(1, 1, 1, 17, 0, 0)),
            new WorkDay(DayOfWeek.Friday, new DateTime(1, 1, 1, 9, 0, 0), new DateTime(1, 1, 1, 17, 0, 0)),
        };
        var holidayPeriods = new List<HolidayPeriod>
        { 
            new HolidayPeriod(new DateTime(2015, 9, 15, 00, 0, 0), new DateTime(2015, 9, 16, 0, 0, 0))
        };
        var service = new ServicePlannerManager(workDays, holidayPeriods);

        // Act
        var result = service.GetRemainingWorkingTime(startTime, expectedDueDate);

        // Assert - 
        Assert.AreEqual(workingHours.TotalHours, result.TotalHours, expectation);
    }

    protected IEnumerable LocalSource()
    {
        yield return
            new TestCaseData(
                new DateTime(2015, 9, 14, 9, 0, 0),
                new TimeSpan(23, 0, 0),
                new DateTime(2015, 9, 17, 16, 0, 0),
                    "5. Expected 23 hours of working time to end on the 17/09/2015 16:00. Monday to Thursday evening. Just short of 3 full working days by one hour. Tuesday is holiday.");
    }
}

该测试的输出是

5. Expected 23 hours of working time to end on the 17/09/2015 16:00. Monday to Thursday evening. Just short of 3 full working days by one hour. Tuesday is holiday.

Expected: 23.0d
But was:  15.999999999944444d

我想知道我是否错误地使用了收集器,或者收集器是否有错误.

Answer 1

这看起来像是一个很好的解决熟悉问题的库.

最好的办法是输出句点集合中的句点,以帮助您调试问题.

我已经重写了你的测试,以便在他们的文档中使用示例中的基类型:

        [Test, TestCaseSource("LocalSource")]
    public void SO_GetRemainingWorkingTimeWithHolidayShouldOnlyEnumerateWorkingTime(DateTime startTime,
        TimeSpan workingHours, DateTime expectedDueDate, string expectation)
    {
        CalendarPeriodCollectorFilter filter = new CalendarPeriodCollectorFilter();
        filter.Months.Add(YearMonth.September); // only Januaries
        filter.WeekDays.Add(DayOfWeek.Monday); // 
        filter.WeekDays.Add(DayOfWeek.Tuesday); // 
        filter.WeekDays.Add(DayOfWeek.Wednesday); // 
        filter.WeekDays.Add(DayOfWeek.Thursday); // 
        filter.WeekDays.Add(DayOfWeek.Friday); // 
        filter.CollectingHours.Add(new HourRange(9, 17)); // working hours

        CalendarTimeRange testPeriod = new CalendarTimeRange(startTime, expectedDueDate);//new DateTime(2015, 9, 14, 9, 0, 0), new DateTime(2015, 9, 17, 18, 0, 0));
        Console.WriteLine("Calendar period collector of period: " + testPeriod);

        filter.ExcludePeriods.Add(new TimeBlock(new DateTime(2015, 9, 15, 00, 0, 0), new DateTime(2015, 9, 16, 0, 0, 0)));

        CalendarPeriodCollector collector = new CalendarPeriodCollector(filter, testPeriod);
        collector.CollectHours();

        foreach (ITimePeriod period in collector.Periods)
        {
            Console.WriteLine("Period: " + period); // THIS WILL HELP A LOT!
        }
        var result = collector.Periods.GetTotalDuration(new TimeZoneDurationProvider(TimeZoneInfo.FindSystemTimeZoneById("UTC")));

        Console.WriteLine(result);
            //
    }

这导致:

Calendar period collector of period: 14/09/2015 09:00:00 - 17/09/2015 15:59:59 | 3.06:59
Period: 14/09/2015 09:00:00 - 14/09/2015 16:59:59 | 0.07:59
Period: 16/09/2015 09:00:00 - 16/09/2015 16:59:59 | 0.07:59
15:59:59.9999998

所以我注意到最后一段时间不见了.

如果您将期间的结束时间从下午4点更改为下午6点(因此预计额外的小时数= 24),它将会通过.(你还需要对结果进行舍入)

所以看起来期间需要完全覆盖总持续时间,不计算部分覆盖.您可以更改库的选项,或者您也可以将每个工作日的小时添加为单独的CollectingHours(hacky)

希望能让您更接近您需要的答案!