长生不老药/凤凰从工作日+周数开始

Question

是否可以通过提供周数和工作日来获取日期(日，月，年(？基本上我需要查询本周的星期一。

我已经检查了 Erlang，有:erlang.date |> :calendar.day_of_the_week返回工作日。但是，有没有办法反转此功能？提供工作日、周数和年份以获得确切日期？

谢谢

Answer 1

基本上我需要查询本周的星期一。

您可以将日期的day_of_the_week与星期一 (1( 的day_of_the_week进行比较，并从日期中减去该天数：

def monday_in_the_week_of(date) do
  from_monday = :calendar.day_of_the_week(date) - 1
  date
  |> :calendar.date_to_gregorian_days
  |> Kernel.-(from_monday)
  |> :calendar.gregorian_days_to_date
end

演示：

defmodule A do
  def monday_in_the_week_of(date) do
    from_monday = :calendar.day_of_the_week(date) - 1
    date
    |> :calendar.date_to_gregorian_days
    |> Kernel.-(from_monday)
    |> :calendar.gregorian_days_to_date
  end
end
for d <- 1..15 do
  date = {2017, 2, d}
  IO.inspect {date, A.monday_in_the_week_of(date)}
end

输出：

{{2017, 2, 1}, {2017, 1, 30}}
{{2017, 2, 2}, {2017, 1, 30}}
{{2017, 2, 3}, {2017, 1, 30}}
{{2017, 2, 4}, {2017, 1, 30}}
{{2017, 2, 5}, {2017, 1, 30}}
{{2017, 2, 6}, {2017, 2, 6}}
{{2017, 2, 7}, {2017, 2, 6}}
{{2017, 2, 8}, {2017, 2, 6}}
{{2017, 2, 9}, {2017, 2, 6}}
{{2017, 2, 10}, {2017, 2, 6}}
{{2017, 2, 11}, {2017, 2, 6}}
{{2017, 2, 12}, {2017, 2, 6}}
{{2017, 2, 13}, {2017, 2, 13}}
{{2017, 2, 14}, {2017, 2, 13}}
{{2017, 2, 15}, {2017, 2, 13}}

Answer 2

是否可以通过提供周数和工作日来获取日期(日、月、年(？

也许这样的事情会起作用：

@doc """
iex>get_date(1, 1, 2018)
{:ok, ~D[2018-01-01]}
"""
def get_date(weekday, _week_number, _year)
    when weekday < 1 or weekday > 7,
    do: {:error, "invalid weekday: #{inspect(weekday)}"}
def get_date(_weekday, week_number, _year)
    when week_number < 0 or week_number > 51,
    do: {:error, "invalid week_number: #{inspect(week_number)}"}
def get_date(weekday, week_number, year) do
  case Date.new(year, 1, 1) do
    {:ok, first_day_of_year} ->
      Date.add(first_day_of_year, week_number * 7 + weekday - 1)
    {:error, error} ->
      {:error, error}
  end
end