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我正在尝试编写一个获取日期的天数的函数，例如，今天（3 月 29 日）是一年中的第 88 天。然后它返回一个包含月份和月份中的日期的哈希：

{"month" => "March, "day" => 29}

我不太清楚这段代码有什么问题，但它总是返回nil. 有什么想法吗？我正在使用 Ruby 1.8.7 p358。

def number_to_date(days)
  date = case days
    when days <= 31  then {"month" => "January",   "day" => days}
    when days <= 59  then {"month" => "February",  "day" => (days - 31)}
    when days <= 90  then {"month" => "March",     "day" => (days - 59)}
    when days <= 120 then {"month" => "April",     "day" => (days - 90)}
    when days <= 151 then {"month" => "May",       "day" => (days - 120)}
    when days <= 181 then {"month" => "June",      "day" => (days - 151)}
    when days <= 212 then {"month" => "July",      "day" => (days - 181)}
    when days <= 243 then {"month" => "August",    "day" => (days - 212)}
    when days <= 273 then {"month" => "September", "day" => (days - 243)}
    when days <= 304 then {"month" => "October",   "day" => (days - 273)}
    when days <= 334 then {"month" => "November",  "day" => (days - 304)}
    when days <= 365 then {"month" => "December",  "day" => (days - 334)}
  end
  return date
end
4

2 回答 2

5

如果要在每个子句case中使用表达式，则需要使用裸语句。when否则，Ruby 将调用(days <= 31) === days，这永远不会是真的。

def number_to_date(days)
  date = case
    when days <= 31  then {"month" => "January",   "day" => days}
    when days <= 59  then {"month" => "February",  "day" => (days - 31)}
    # ...
  end
  return date
end

This implementation however ignores leap days and it seems simpler and more correct to just do this:

def number_to_date(days)
  date = Date.ordinal(Date.today.year, days)
  {"month" => Date::MONTHNAMES[date.month], "day" => date.day}
end
于 2013-03-29T19:01:43.420 回答
2

你只需要一个小的语法调整。days从date = case days声明中删除。否则，您的条件语句将与days变量进行比较。

于 2013-03-29T19:00:04.920 回答