我使用ice_cube
創建計劃。從31日開始的每月計劃會在所有月份中缺少31天。我想安排那幾個月的最後一天。如果我的日程從30日開始,我想每個月的30日和2月的最後一天。閏年進一步使問題複雜化。每月計劃月份結束時間
創建可以處理29日,30日或31日的日程安排的好方法是什麼?
我使用ice_cube
創建計劃。從31日開始的每月計劃會在所有月份中缺少31天。我想安排那幾個月的最後一天。如果我的日程從30日開始,我想每個月的30日和2月的最後一天。閏年進一步使問題複雜化。每月計劃月份結束時間
創建可以處理29日,30日或31日的日程安排的好方法是什麼?
這通過了我的所有規格,但是非常可怕,並且可能會因爲時間超過一年而中斷(我現在不在意)。
class LeasePaymentSchedule
def self.monthly(a bunch of args)
case start_day
when 31
schedule = IceCube::Schedule.new(start, scheduler_options) do |s|
s.add_recurrence_rule IceCube::Rule.monthly.day_of_month(-1).until(end_time)
end
when 30,29
schedule = IceCube::Schedule.new(start, scheduler_options) do |s|
s.add_recurrence_rule IceCube::Rule.monthly.day_of_month(start_day).until(end_time)
end
schedule.all_occurrences.each do |o|
next unless [1,3,6,8,10].include? o.month
missed = (o + 1.month).yday
# Probably breaks for durations longer than 1 year
schedule.add_recurrence_rule IceCube::Rule.yearly.day_of_year(missed).count(1)
end
else
schedule = IceCube::Schedule.new(start, scheduler_options) do |s|
s.add_recurrence_rule IceCube::Rule.monthly.day_of_month(start_day).until(end_time)
end
end
schedule
end
end
這麼多的規格:
Finished in 4.17 seconds
390 examples, 0 failures
-
shared_examples_for :a_schedule do
it 'returns an IceCube Schedule' do
schedule.should be_a IceCube::Schedule
end
it 'should start on the correct day' do
schedule.start_time.should eq expected_start
end
it 'has the right number of occurrences' do
schedule.all_occurrences.size.should eq expected_occurrences
end
end
describe :monthly do
let(:expected_occurrences) { 12 }
let(:expected_start) { date.next_month.beginning_of_day }
let(:schedule) { LeasePaymentSchedule.monthly }
before do
Date.stub(:today).and_return(date)
end
shared_examples_for :on_the_28th do
let(:date) { Time.parse "#{year}-#{month}-28" }
it_behaves_like :a_schedule
end
shared_examples_for :on_the_29th do
let(:date) { Time.parse "#{year}-#{month}-29" }
it_behaves_like :on_the_28th
it_behaves_like :a_schedule
end
shared_examples_for :on_the_30th do
let(:date) { Time.parse "#{year}-#{month}-30" }
it_behaves_like :on_the_29th
it_behaves_like :a_schedule
end
shared_examples_for :on_the_31st do
let(:date) { Time.parse "#{year}-#{month}-31" }
it_behaves_like :on_the_30th
it_behaves_like :a_schedule
end
shared_examples_for :the_whole_year do
context :february do
let(:month) { 2 }
it_behaves_like :on_the_28th
end
[ 4, 7, 9, 11 ].each do |month_num|
let(:month) { month_num }
it_behaves_like :on_the_30th
end
[ 1, 3, 5, 6, 8, 10, 12].each do |month_num|
let(:month) { month_num }
it_behaves_like :on_the_31st
end
end
context :a_leap_year do
let(:year) { 2012 }
context :february_29th do
let(:month) { 2 }
it_behaves_like :on_the_29th
end
it_behaves_like :the_whole_year
end
context :before_a_leap_year do
let(:year) { 2011 }
it_behaves_like :the_whole_year
end
context :nowhere_near_a_leap_year do
let(:year) { 2010 }
it_behaves_like :the_whole_year
end
end
您可以在本月的最後一天使用day_of_month(-1)
。
如果我的復發從30日開始,這是不正確的。 –
啊,我現在明白你的問題了。我不確定你可以直接用'ice_cube'做這個。 – micahbf
您可以像平時那樣使用29日,30日和31日的day_of_month設置日程安排。然後使用函數確定哪些月份被跳過,並使用單個事件來設置它們。無可否認,這只是一個黑客攻擊,但是可以解決ice_cube的限制。
下面是一個簡單的腳本,用於確定在任何給定年份哪些月份將被跳過。
require 'date'
def months_less_than_days(days, year = Date.today.year)
months = []
(1..11).each do |m|
eom = (Date.new(year, m+1, 1) - 1)
months << eom if eom.day < days
end
eom = (Date.new(year+1, 1, 1) - 1)
months << eom if eom.day < days
months
end
puts months_less_than_days(31) # => [2013-02-28, 2013-04-30, 2013-06-30, 2013-09-30, 2013-11-30]
添加單個事件的語法是什麼?從文檔中不明顯。 –
看看這個答案http://stackoverflow.com/questions/13788926/single-occurrence-event-with-ice-cube-gem-using-start-time-and-end-time –
這已固定在最新的冰塊:
IceCube::Schedule.new(Time.parse('Oct 31, 2013 9:00am PDT')) do |s|
s.add_recurrence_rule(IceCube::Rule.monthly(1))
end.first(5)
[2013-10-31 09:00:00 -0700,
2013-11-30 09:00:00 -0800,
2013-12-31 09:00:00 -0800,
2014-01-31 09:00:00 -0800,
2014-02-28 09:00:00 -0800]
https://github.com/seejohnrun/ice_cube#每月每月 –
@ muistooshort #day_of_ the_month跳過的月份太短(請參閱doc)。因此,如果我的復發從第31次開始,我會在一年中錯過5個月.x –
您是否閱讀過「每個月的第一天和最後一天」示例? –