Question

假设我有一些截止时间对象deadline = project_start + 34.days

现在让我们假设某种假期/推迟的业务逻辑必须让我延长这些假期的最后期限。

让我们假设我的假期是阵列或范围（或几个开始/结束日期，无论如何）并且该项目于12月15日开始

project_start = Time.parse('2017-12-15 0:00') # December 15th
deadline = project_start + 34.days            # January 18th

vacations = [
  Time.parse('2017-12-10 0:00')..Time.parse('2018-01-20 0:00'), # Starts 5 days before the project start, so only 5 out of 10 days extend the deadline
  Time.parse('2018-01-05 0:00')..Time.parse('2018-01-10 0:00'), # Starts during the project so whole vacation period must extend the deadline
  Time.parse('2018-01-20 0:00')..Time.parse('2018-01-25 0:00'), # STarts after the original deadline, but dued to previous vacations, this vacation fully extends the deadline
]

考虑到假期，我如何（轻松）计算延长的截止日期？

我尝试寻找现有的解决方案，但我发现大多数宝石涉及静态假期，但我对每个用户的内容感兴趣。

基本上我在考虑订购假期，并延长每个步骤的截止日期，但我需要考虑到在实际项目开始之前可能已经开始休假。要弄清楚正确的事情变得有点困难。我将发布第一次尝试

Answer 1

所以这是我设计的第一种方法，我正在编写规格，也许它已经很好但我有疑问

# Extends a deadline with extra time provided as "vacations"
#
# @param range: [Range<Time>]
# @param vacations: [Array<Range<Time>>]
#
# @return [Range]
def extend_time_range_with_vacations(range:, vacations:)
  extended_deadline = range.last.dup
  vacations.each do |vacation|
    cover = (range.first..extended_deadline) & vacation
    next unless cover.present?
    # Here I need to pick more than just the intersection
    #   For eg maybe the vacation is 2.weeks long
    #   but the range started only in the 2nd week of the vacation,
    #   hence the time extension is 1.week only)
    actual_cover = cover.first..vacation.last
    extended_deadline += (actual_cover.last - actual_cover.first)
  end
  extended_deadline
end

但感觉还有很多工作要做，也许更简单的事情存在？

编辑：代码已更改为使用this range intersection monkeypatch

编辑：规范

describe '#extend_time_range_with_vacations' do
  let(:start_d) { Time.parse('2018-04-20 8:42') }
  let(:end_d) { start_d + 42.days }
  let(:range) { start_d..end_d }
  let(:vacations) do
    [
      (start_d - 4.days)..(start_d + 4.days), # 4 days extension out of 8
      (start_d + 10.days)..(start_d + 14.days), # 4 days extension
      (end_d + 4.days)..(end_d + 8.days), # 4 days extensions unlocked because of previous vacations
    ]
  end

  it 'returns the expected extended deadline' do
    new_range = Utility.extend_time_range_with_vacations(range: range, vacations: vacations)
    expect(new_range).to be_within(1).of(end_d + (4 * 3).days)
  end
end

Answer 2

我已经解决了以下问题。我们将获得完成项目所需的开始日期和工作天数。我们还有一系列假期。如果项目已经开始并且在给定日期尚未完成的情况下，如果该日期属于任何休假期，则该日期不会执行任何工作。每个休假期是连续几天的范围，可以在开始日期之前开始或结束，并且可以与其他休假期重叠。¹。

目标是确定项目完成的日期。

问题的表现方式不同，但如果我的解释是准确的而不是单位（例如，我假设日期输入和日期输出是字符串而不是Time对象），它会不难修改我的答案以符合要求。

<强>代码

require 'date'
require 'set'

DATE_FMT = "%Y-%m-%d"

def end_date(start_date, duration, vacations)
  start_date = Date.strptime(start_date, DATE_FMT)
  vacation_dates = construct_vacations_dates_set(vacations)
  d = start_date
  while duration > 0
    duration -= 1 unless vacation_dates.include?(d)
    d = d.next
  end
  d.strftime(DATE_FMT)
end

def construct_vacations_dates_set(vacations)
  vacations.map do |a|
    s, e = a.map { |s| Date.strptime(s, DATE_FMT) }
    (s..e).to_a
  end.
  reduce(:|).
  to_set
end

示例

start_date = '2017-12-15' vacations = [['2017-11-20', '2017-12-03'], ['2017-12-13', '2017-12-19'], ['2017-12-23', '2018-01-02'], ['2017-12-29', '2018-01-08'], ['2018-01-14', '2018-02-28'], ['2018-05-18', '2018-06-01']] (1..20).each { |n| puts "for duration %2d days: %s" % [n, end_date(start_date, n, vacations)] }

打印

for duration 1 days: 2017-12-21 for duration 2 days: 2017-12-22 for duration 3 days: 2017-12-23 for duration 4 days: 2018-01-10 for duration 5 days: 2018-01-11 for duration 6 days: 2018-01-12 for duration 7 days: 2018-01-13 for duration 8 days: 2018-01-14 for duration 9 days: 2018-03-02 for duration 10 days: 2018-03-03 for duration 11 days: 2018-03-04 for duration 12 days: 2018-03-05 for duration 13 days: 2018-03-06 for duration 14 days: 2018-03-07 for duration 15 days: 2018-03-08 for duration 16 days: 2018-03-09 for duration 17 days: 2018-03-10 for duration 18 days: 2018-03-11 for duration 19 days: 2018-03-12 for duration 20 days: 2018-03-13

<强>解释

见Date#strftime， Date::strptime，Date#next（又名next_day）和Enumerable#reduce（又名inject）。 reduce采用＆＃34; set union＆＃34;通过应用方法Array#|来度假数组（连续日期）。

我将休假日期数组转换为快速查找集。对于给出的示例，如果将vacation_dates的元素转换为非重叠范围的数组，然后将范围端点转换为日期字符串，则该数组将如下所示。

["2017-11-20".."2017-12-03", "2017-12-13".."2017-12-19", "2017-12-23".."2018-01-08", "2018-01-14".."2018-02-28", "2018-05-18".."2018-06-01"]

第一个范围完全在start_date之前，但没有必要删除它。更一般地说，没有必要在start_date之前删除休假日期。

通过添加一些puts语句可以看到计算假期日期集的步骤。为了提高可读性，我删除了返回值和打印行范围内的Date个对象，用,...,替换它们的组。我还缩写了Date个对象。例如，我将#<Date: 2017-12-13 ((2458101j,0s,0n),+0s,2299161j)>缩短为#<Date: 2017-12-13...>。

def construct_vacations_dates_set(vacations) vacations.map do |a| puts "a = #{a}" s, e = a.map { |s| Date.strptime(s, DATE_FMT) } puts " s = #{s}, e = #{e}" (s..e).to_a. tap { |o| puts " (s..e).to_a = #{o}" } end. reduce(:|). tap { |o| puts "after reduce(:|) = #{o}" }. to_set end construct_vacations_dates_set([['2017-12-13', '2017-12-19'], ['2017-12-23', '2018-01-02'], ['2017-12-29', '2018-01-08']]) #=> #<Set: {#<Date: 2017-12-13...>,..., #<Date: 2017-12-19...>, # #<Date: 2017-12-23...>,..., #<Date: 2018-01-08...>}>

打印（范围未显示的元素）

a = ["2017-12-13", "2017-12-19"] s = 2017-12-13, e = 2017-12-19 (s..e).to_a = [#<Date: 2017-12-13...>,.., #<Date: 2017-12-19..>] a = ["2017-12-23", "2018-01-02"] s = 2017-12-23, e = 2018-01-02 (s..e).to_a = [#<Date: 2017-12-23...>,..., #<Date: 2018-01-02...>] a = ["2017-12-29", "2018-01-08"] s = 2017-12-29, e = 2018-01-08 (s..e).to_a = [#<Date: 2017-12-29...>,..., #<Date: 2018-01-08...>] after reduce(:|) = [#<Date: 2017-12-13...>,..., #<Date: 2017-12-19...>, #<Date: 2017-12-23...>,..., #<Date: 2018-01-08...>]

^{1如果保证度假范围不重叠，我的解决方案将是相同的。}

在休假期限内延长截止日期

2 个答案: