两次之间的差异

Question

如何计算两次红宝石之间的差异？

例如：

我有这个时间作为字符串

time1 = '08:00' 
time2 = '12:00' 

让我们考虑时间1为正，时间2负（这样想工作一小时，我一天上班8小时，另一天我错过12小时），我愿做这样的事情：

diff = Time.parse(time1) - Time.parse(time2) 
Time.at(diff).gmtime.strftime('%H:%M') 
# expect to get something like "-04:00" 
# but the result is "20:00" 

Answer 1

要了解什么是你的代码错误，让我们通过一个执行它一个：

▶ Time.parse('08:00') - Time.parse('12:00') 
# => -14400.0 

秒，呵呵。精细。

▶ Time.at(Time.parse('08:00') - Time.parse('12:00')) 
=> 1969-12-31 21:00:00 +0100 

哇。所以，Time.at不是我们梦想的方法。此外，我不知道标准库中的方式是从秒数中获得格式为%H:%M的小时。但是：

▶ sprintf "%+03i:%02i", -14400/3600, -14400/60 % 60 # hours, minutes 
# => "-04:00" 

唯一的问题仍然是一个标志。总结：

▶ time1 = '06:28'; time2 = '12:00' 
▶ diff = Time.parse(time1) - Time.parse(time2) 
▶ sprintf "%+03i:%02i", (diff/diff.abs) * (diff.abs/3600), diff.abs/60 % 60 
# => "-05:32" 

希望它有帮助。

UPD：对于那些，谁喜欢的monkeypatching：

class Time 
    def diff_hours other 
    return nil unless Time === other 
    diff = self - other 
    sprintf "%+03i:%02i", (diff/diff.abs) * (diff.abs/3600), diff.abs/60 % 60 
    end 
end 
# => :diff_hours 
▶ Time.parse('06:32').diff_hours Time.parse('12:00') 
# => "-05:28"