How do I count the number of days between these two dates?
start_date = Date.parse "2012-03-02 14:46:21 +0100"
end_date = Date.parse "2012-04-02 14:46:21 +0200"
This question is related to
ruby
to get the number of days in a time range (just a count of all days)
(start_date..end_date).count
(start_date..end_date).to_a.size
#=> 32
to get the number of days between 2 dates
(start_date...end_date).count
(start_date...end_date).to_a.size
#=> 31
I kept getting results in seconds, so this worked for me:
(Time.now - self.created_at) / 86400
To get the number of days difference by two dates:
(start.to_date...end.to_date).count - 1
or
(end.to_date - start.to_date).to_i
None of the previous answers (to this date) gives the correct difference in days between two dates.
The one that comes closest is by thatdankent. A full answer would convert to_i
and then divide:
(Time.now.to_i - 23.hours.ago.to_i) / 86400
>> 0
(Time.now.to_i - 25.hours.ago.to_i) / 86400
>> 1
(Time.now.to_i - 1.day.ago.to_i) / 86400
>> 1
In the question's specific example, one should not parse to Date
if the time passed is relevant. Use Time.parse
instead.
def business_days_between(date1, date2)
business_days = 0
date = date2
while date > date1
business_days = business_days + 1 unless date.saturday? or date.sunday?
date = date - 1.day
end
business_days
end
Assuming that end_date
and start_date
are both of class ActiveSupport::TimeWithZone
in Rails, then you can use:
(end_date.to_date - start_date.to_date).to_i
(end_date - start_date)/1000/60/60/24
any one have best practice please comment below
Rails has some built in helpers that might solve this for you. One thing to keep in mind is that this is part of the Actionview Helpers, so they wont be available directly from the console.
Try this
<% start_time = "2012-03-02 14:46:21 +0100" %>
<% end_time = "2012-04-02 14:46:21 +0200" %>
<%= distance_of_time_in_words(start_time, end_time) %>
"about 1 month"
To have the number of whole days between two dates (DateTime
objects):
((end_at - start_at).to_f / 1.day).floor
Source: Stackoverflow.com