[英]How can I make this conditional riddled instance method more Ruby idiomatic?
def short_remaining_time
difference = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
# To display the short remaining time in an auction listing.
if difference[:day] == 0 and difference[:hour] >= 1
"#{difference[:minute]} minutos"
elsif difference[:day] == 0 and difference[:hour] >= 23
"#{difference[:hour]} horas"
else
if difference[:day] != 1
"#{difference[:day]} dias"
else
"#{difference[:day]} dia"
end
end
end
該方法在我的Rails應用程序的auction.rb
模型中。
在我的一種觀點中,我列出了系統中的所有拍賣,並且還顯示了拍賣結束之前還剩下多少時間。
根據時間的量,我要么顯示days
hours
或minutes
。
該代碼運行良好,外觀看上去很笨拙。 有什么辦法可以解決這個問題嗎?
您可以如下簡化。 請注意,您的代碼是多余的。 如果difference[:hour] >= 23
,那么這就意味着difference[:hour] >= 1
,並且將被后者捕獲,因此前者的條件永遠不會被評估為真。 這樣就可以刪除該部分。
def short_remaining_time
difference = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
case day = difference[:day]
when 0
if difference[:hour] >= 1 then "#{difference[:minute]} minutos"
else "#{day} dias"
end
when 1 then "#{day} dia"
else "#{day} dias"
end
end
我假設您無意間遇到了不平等現象(您需要<=
不是>=
)。 另外,如果您認為時差始終不超過23
,則不需要進行檢查(即,我們假設時差已“標准化”)。 因此,我將以這種方式修改它,以保持您的原始意圖:
def short_remaining_time
difference = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
# To display the short remaining time in an auction listing.
if difference[:day] == 0
if difference[:hour] <= 1
"#{difference[:minute]} minutos"
else
"#{difference[:hour]} horas"
end
else
"#{difference[:day]} dia" + ((difference[:day] == 1) ? "" : "s")
end
end
關於什么
def short_remaining_time
difference = Time.diff(Time.now, created_at + 7.days, '%d - %H - %N')
diff_in_minutes = difference[:day] * 3600 + difference[:hour] * 60
case diff_in_minutes
when 0..60 then "#{difference[:minute]} minutos"
when 61..3600 then "#{difference[:hour] } horas"
when 3600..7200 then "#{difference[:day] } dia"
else "#{difference[:day] } dias"
end
end
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