Rails - Get the Time Difference in Hours, Minutes and Seconds

Rails - Get the time difference in hours, minutes and seconds

You can try with this:

def time_diff(start_time, end_time)
  seconds_diff = (start_time - end_time).to_i.abs

  hours = seconds_diff / 3600
  seconds_diff -= hours * 3600

  minutes = seconds_diff / 60
  seconds_diff -= minutes * 60

  seconds = seconds_diff

  "#{hours.to_s.rjust(2, '0')}:#{minutes.to_s.rjust(2, '0')}:#{seconds.to_s.rjust(2, '0')}"
  # or, as hagello suggested in the comments:
  # '%02d:%02d:%02d' % [hours, minutes, seconds]
end

And use it:

time_diff(Time.now, Time.now-2.days-3.hours-4.minutes-5.seconds)
# => "51:04:04"

how to get time difference in minutes in rails for two date time fields?

As created_at and updated_at are of type DateTime, this should work.

created = article.created_at
updated = article.updated_at
minutes = ((updated - created) * 24 * 60).to_i

Explanation:

Subtracting two DateTimes returns the elapsed time in days. In the below example e-d will return (28807336643183/28800000000000). So to convert this into minutes, we need to multiply it by 24*60(As the day has 24 hours and each hour has 60 minutes)

Example(tested):

d = DateTime.now
 => Tue, 13 Jun 2017 10:21:59 +0530 
2.3.3 :003 > e = DateTime.now + 1.day
 => Wed, 14 Jun 2017 10:22:21 +0530
g = ((e-d) * 24 * 60).to_i
 => 1440

Ruby: Calculate time difference between 2 times

I don't understand why you think it should return 4 hours or why it does return 22 hours. 20 hours would be correct for your example:

require 'time'

a = Time.parse('2015-12-31 22:00:00 +0100')
b = Time.parse('2015-12-31 02:00:00 +0100')

a - b
#=> 72000.0 # difference in seconds

(a - b) / 3600
#=> 20.0  # difference in hours

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Update: It seems like you are dealing only with the time portion and not with the actual date. And I assume the maximum difference you will have to deal with is 24 hours:

def time_difference(time_a, time_b)
  difference = time_b - time_a

  if difference > 0
    difference
  else
    24 * 3600 + difference 
  end
end

a = Time.parse('2015-12-31 22:00:00 +0100')
b = Time.parse('2015-12-31 02:00:00 +0100')
time_difference(a, b) / 3600
# => 4 # hours

a = Time.parse('2015-12-31 02:00:00 +0100')
b = Time.parse('2015-12-31 22:00:00 +0100')
time_difference(a, b) / 3600
# => 20 # hours

Get time difference in minutes in rails the most effective way

This is not a problem, Ruby handles timezones internally.

time1 = DateTime.parse('2000-01-01 10:00:00 UTC')
time2 = DateTime.parse('2000-01-01 10:00:00 -0100')
time2 - time1
# => (1/24)

I.e. Ruby already knows that the difference between the same time in UTC and in -0100 is one hour. The time difference operator is already doing the right thing, if you let it :)

However, DateTime is Ruby, Time is C - the former has a lot nicer API and works awesomely with databases, but the latter is faster.

EDIT: Here's an example using Time, still same:

time1 = Time.new(2000, 1, 1, 10, 0, 0, "+00:00")
time2 = Time.new(2000, 1, 1, 10, 0, 0, "-01:00")
time2 - time1
# => 3600.0

EDIT: Year 2999 2000

calculate difference in time in days, hours and minutes

I've used time_diff to achieve this sort of thing easily before, you may want to check it out.

rails How to calculate time difference with integer between 3 dates

I have also faced the same scenario in one of my projects and there i used the Time Difference Gem . It is very simple and easy to integrate and provides you additional helper methods to get the difference in year, month, week, day, hour, minute, and seconds.

Set the start and end date

start_time = Time.new(2013,1)
end_time = Time.new(2014,1)

Now call the calculate the difference as

TimeDifference.between(start_time, end_time).in_days
=> 365.0

Additionally it provides you liberty to fetch all differences at the same time

TimeDifference.between(start_time, end_time).in_each_component
=> {:years=>1.0, :months=>12.0, :weeks=>52.14, :days=>365.0, :hours=>8760.0, :minutes=>525600.0, :seconds=>31536000.0}

Time difference in hours

((date_2 - date_1) / 3600).round

or

((date_2 - date_1) / 1.hour).round

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