[英]Ruby on Rails parsing time from datetime, not as string
I'm trying to extract the time component from a DateTime object (which is represented as "at" in my example). 我正在尝试从DateTime对象(在我的示例中表示为“ at”)中提取时间分量。 How do I do this, I am absolutely stumped?
我该怎么办,我很沮丧吗? (I don't want to parse it to a string with
strftime
as i did here): (我不想像我在这里那样用
strftime
将其解析为字符串):
@session_date.at.strftime("%H:%M")
I would really like to return the hours and minutes as a Time object. 我真的很想将小时和分钟作为Time对象返回。
If you have a DateTime object: 如果您有一个DateTime对象:
date_time = DateTime.now
date_time.hour
# => 16
date_time.minute
# => 1
Is there a specific reason you want a Time object? 您是否需要时间对象的特定原因?
Just so we're clear, the Time class in Ruby isn't just "DateTime without the date." 就像我们清楚地知道的那样,Ruby中的Time类不仅仅是“没有日期的DateTime”。 As " What's the difference between DateTime and Time in Ruby? " explains, "Time is a wrapper around POSIX-standard time_t, or seconds since January 1, 1970."
正如“ Ruby中的DateTime和Time有什么区别? ”解释说,“ Time是POSIX标准time_t的包装,或者是自1970年1月1日以来的秒数。” Like DateTime, a Time object still has year, month, and day, so you don't really gain anything by using Time instead.
与DateTime一样,Time对象仍然具有年,月和日,因此使用Time并不会带来任何好处。 There's not really a way to represent just hour and minute using either Time or DateTime.
确实没有一种方法可以使用Time或DateTime仅表示小时和分钟。
The best you could do with Time, I think, would be this: 我认为,使用Time可以做的最好的事情是:
date_time = DateTime.now
seconds = date_time.hour * 60 * 60 + date_time.minute * 60
time = Time.at(seconds)
# => 1970-01-01 09:58
...but then you still have to call time.hour
and time.min
to get at the hour and minute. ...但是您仍然必须致电
time.hour
和time.min
来获取小时和分钟。
If you're just looking for a lightweight data structure to represent an hour and minute pair, though, you might as well just roll your own: 但是,如果您只是在寻找一个轻量级的数据结构来表示小时和分钟,那么您也可以自己滚动:
HourAndMinute = Struct.new(:hour, :minute) do
def self.from_datetime(date_time)
new(date_time.hour, date_time.minute)
end
end
hm = HourAndMinute.from_datetime(DateTime.now)
# => #<struct HourAndMinute hour=15, minute=58>
hm.to_h
# => { :hour => 15, :minute => 58 }
hm.to_a
# => [ 15, 58 ]
I have a variable that stores an appointment -- this variable is a DateTime object.
我有一个存储约会的变量-该变量是DateTime对象。 I have two table fields that store the start and end times of a location.
我有两个表字段,用于存储位置的开始时间和结束时间。 I need to check if the time scheduled for that appointment lies between the start and end times.
我需要检查为该约会安排的时间是否在开始时间和结束时间之间。
Ah, it seems you had a bit of a XY problem . 嗯,看来您有点XY问题 。 This makes a lot more sense now.
现在,这变得更加有意义。
Absent any more information, I'm going to assume your "fields that store the start and end times of a location" are MySQL TIME
columns called start_time
and end_time
. 缺少更多信息,我将假设您的“存储位置开始和结束时间的字段”是MySQL
TIME
列,称为start_time
和end_time
。 Given MySQL TIME columns, Rails casts the values to Time objects with the date component set to 1/1/2000 . 给定MySQL TIME列,Rails将日期分量设置为1/1/2000的值转换为Time对象。 So if your database has the values
start_time = '09:00'
and end_time = '17:00'
, Rails will give you Time objects like this: 因此,如果您的数据库具有值
start_time = '09:00'
和end_time = '17:00'
,Rails将为您提供Time对象,如下所示:
start_time = Time.new(2000, 1, 1, 9, 0) # => 2000-01-01 09:00:00 ...
end_time = Time.new(2000, 1, 1, 17, 0) # => 2000-01-01 17:00:00 ...
Now you say your appointment time is a DateTime, so let's call it appointment_datetime
and suppose it's at 10:30am tomorrow: 现在你说你的预约时间是一个DateTime,让我们把它叫做
appointment_datetime
并假设它在上午10:30明天:
appointment_datetime = DateTime.new(2014, 11, 18, 10, 30) # => 2014-11-18 10:30:00 ...
So now to rephrase your question: How do we tell if the time part of appointment_datetime
is between the time part of start_time
and end_time
. 所以,现在要改一下你的问题:我们如何告诉我们,如果在部分时间
appointment_datetime
是部分时间之间start_time
和end_time
。 The answer is, we need to either change the date part of start_time
and end_time
to match the date part of appointment_datetime
, or the other way around. 答案是,我们需要更改的日期部分
start_time
和end_time
匹配的日期部分appointment_datetime
,或者反过来。 Since it's easier to change one thing than two, let's do it the other way around and change appointment_datetime
to match start_time
and end_time
(and, since those two are Time objects, we'll create a Time object): 由于更改一件事情要比更改一件事情容易,让我们以另一种方式来做,并更改
appointment_datetime
以匹配start_time
和end_time
(并且,由于这两个是Time对象,因此我们将创建一个Time对象):
appointment_time = DateTime.new(2000, 1, 1, appointment_datetime.hour, appointment_datetime.minute)
# => 2000-01-01 10:30:00 ...
Now we can compare them directly: 现在我们可以直接比较它们:
if appointment_time >= start_time && appointment_time <= end_time
puts "Appointment time is good!"
end
# Or, more succinctly:
if (start_time..end_time).cover?(appointment_time)
puts "Appointment time is good!"
end
You would, of course, want to wrap all of this up in a method, perhaps in your Location model (which, again, I'm assuming has start_time
and end_time
attributes): 当然,您可能希望将所有这些都包装在一个方法中,也许是在您的Location模型中(我再次假设它具有
start_time
和end_time
属性):
class Location < ActiveRecord::Base
# ...
def appointment_time_good?(appointment_datetime)
appointment_time = DateTime.new(2000, 1, 1,
appointment_datetime.hour, appointment_datetime.minute)
(start_time..end_time).cover?(appointment_time)
end
end
location = Location.find(12) # => #<Location id: 12, ...>
location.appointment_time_good?(appointment_time) # => true
I hope that's helpful! 希望对您有所帮助!
PS Another way to implement this would be to ditch the date/time objects entirely and do a straight numeric comparison: PS实现此目的的另一种方法是完全放弃日期/时间对象并进行直接的数值比较:
def appointment_time_good?(appointment_datetime)
appointment_hour_min = [ appointment_datetime.hour, appointment_datetime.minute ]
appointment_hour_min >= [ start_time.hour, start_time.min ]
&& appointment_hour_min <= [ end_time.hour, end_time.min ]
end
如果您正在寻找从现在开始的时间(在Rails应用程序中很常见),那么这可能是一本好书: http : //api.rubyonrails.org/classes/ActionView/Helpers/DateHelper.html#method- i-time_ago_in_words
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