Question

输入：

2015.0596924

期望的输出：

2015年1月22日

Answer 1

您可以按如下方式获取格式化日期：

require 'date'

d = 2015.0596924
year = Date.new(d)
  #=> #<Date: 2015-01-01 ((2457024j,0s,0n),+0s,2299161j)> 
date = (year + (year.leap? ? 366 : 365) * (d % 1))
  #=> #<Date: 2015-01-22 ((2457045j,68059s,526396957n),+0s,2299161j)> 
date.strftime("%B %d, %Y") 
  #=> "January 22, 2015"

Answer 2

这就是我想出来的。

# @param decimalDate [String] the date in decimal form, but can be a String or a Float
# @return [DateTime] the converted date
def decimal_year_to_datetime(decimalDate)
  decimalDate = decimalDate.to_f
  year = decimalDate.to_i # Get just the integer part for the year
  daysPerYear = leap?(year) ? 366 : 365 # Set days per year based on leap year or not
  decimalYear = decimalDate - year # A decimal representing portion of the year left
  dayOfYear = (decimalYear * daysPerYear).ceil # day of Year: 1 to 355 (or 366)
  md = getMonthAndDayFromDayOfYear(dayOfYear, year)

  DateTime.new(year,md[:month],md[:day])
end

# @param dayOfYear [Integer] the date in decimal form
# @return [Object] month and day in an object
def getMonthAndDayFromDayOfYear(dayOfYear, year)
  daysInMonthArray = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
  daysInMonthArray[2] = 29 if leap?(year)

  daysLeft = dayOfYear
  month = 0
  daysInMonthArray.each do |daysInThisMonth|
    if daysLeft > daysInThisMonth
      month += 1 
      daysLeft -= daysInThisMonth
    else
      break
    end
  end
  return { month: month, day: daysLeft }
end

# @param year [Integer]
# @return [Boolean] is this year a leap year?
def leap?(year)
  return year % 4 == 0 && year % 100 != 0 || year % 400 == 0
end

如何将十进制年份值转换为Ruby中的日期？

2 个答案: