I made new object Date.new with args (year, month). After create ruby added 01 number of day to this object by default. Is there any way to add not first day, but last day of month that i passed as arg(e.g. 28 if it will be 02month or 31 if it will be 01month) ?
use Date.civil
With Date.civil(y, m, d) or its alias .new(y, m, d), you can create a new Date object. The values for day (d) and month (m) can be negative in which case they count backwards from the end of the year and the end of the month respectively.
=> Date.civil(2010, 02, -1)
=> Sun, 28 Feb 2010
>> Date.civil(2010, -1, -5)
=> Mon, 27 Dec 2010
To get the end of the month you can also use ActiveSupport's helper end_of_month.
# Require extensions explicitly if you are not in a Rails environment
require 'active_support/core_ext'
p Time.now.utc.end_of_month # => 2013-01-31 23:59:59 UTC
p Date.today.end_of_month # => Thu, 31 Jan 2013
You can find out more on end_of_month in the Rails API Docs.
So I was searching in Google for the same thing here...
I wasn't happy with above so my solution after reading documentation
in RUBY-DOC was:
Example to get 10/31/2014
Date.new(2014,10,1).next_month.prev_day
require "date"
def find_last_day_of_month(_date)
if(_date.instance_of? String)
#end_of_the_month = Date.parse(_date.next_month.strftime("%Y-%m-01")) - 1
else if(_date.instance_of? Date)
#end_of_the_month = _date.next_month.strftime("%Y-%m-01") - 1
end
return #end_of_the_month
end
find_last_day_of_month("2018-01-01")
This is another way to find
You can do something like that:
def last_day_of_month?
(Time.zone.now.month + 1.day) > Time.zone.now.month
end
Time.zone.now.day if last_day-of_month?
This is my Time based solution. I have a personal preference to it compared to Date although the Date solutions proposed above read somehow better.
reference_time ||= Time.now
return (Time.new(reference_time.year, (reference_time.month % 12) + 1) - 1).day
btw for December you can see that year is not flipped. But this is irrelevant for the question because december always has 31 day. And for February year does not need flipping. So if you have another use case that needs year to be correct, then make sure to also change year.
Here is taking the first and third answers to find the last day of the previous month.
today_c = Date.civil(Date.today.prev_month.year, -1, -1)
p today_c
Related
I need to output to the console the calendar of the current month in Ruby. The result should be similar to ncal on UNIX-like systems. I found a solution for C ++ but can't adapt for Ruby. So far, I only realized that I need to use nested loops to output the height and width. Tell me in which direction to move?
require 'date'
days = %w[Mun Tue Wed Thu Fri Sat Sun]
puts " #{Date::MONTHNAMES[Date.today.month]} #{Date.today.year}"
i = 0
start_month = (Date.today - Date.today.mday + 1).strftime("%a")
while i < days.size
print days[i]
j = 1
while j <= 31
if days[i] == start_month
print " #{j}"
end
j += 7
end
i += 1
puts
end
I'll take your solution so far, and try to give some specific pointers for how to progress with it - but of course, there are many different ways to approach this problem in general, so this is by no means the only approach!
The first critical issue (as you're aware!) is that you're only printing things for the row starting on the 1st of the month, due to this line:
if days[i] == start_month
Sticking with the current overall design, we know we'll need to print something for every line, so clearly a conditional like this isn't going to work. Let's try removing it.
Firstly, it will be more convenient to know which day of the week the month started on as a number, not a string, so we can easily calculate offsets against another day. Let's do that with:
# e.g. for 1st July 2021 this was a Thursday, so we get `4`.
start_of_month_weekday = (Date.today - Date.today.mday + 1).cwday
Next (and this is the crucial step!), we can use the above information to find out "which day of the month is it, on this day of the week?"
Here a first version of that calculation, incorporated into your solution so far:
require 'date'
days = %w[Mon Tue Wed Thu Fri Sat Sun]
puts " #{Date::MONTHNAMES[Date.today.month]} #{Date.today.year}"
i = 0
# e.g. for 1st July 2021 this was a Thursday, so we get `4`.
start_of_month_weekday = (Date.today - Date.today.mday + 1).cwday
while i < days.size
print days[i]
day_of_month = i - start_of_month_weekday + 2 # !!!
while day_of_month <= 31
print " #{day_of_month}"
day_of_month += 7
end
i += 1
puts
end
This outputs:
July 2021
Mon -2 5 12 19 26
Tue -1 6 13 20 27
Wed 0 7 14 21 28
Thu 1 8 15 22 29
Fri 2 9 16 23 30
Sat 3 10 17 24 31
Sun 4 11 18 25
Not bad! Now we're getting somewhere!
I'll leave you to figure out the rest 😉 .... But here are some clues, for what I'd tackle next:
This code, print " #{day_of_month}", needs to print a "blank space" if the day number is less than 1. This could be done with a simple if statement.
Similarly, since you want this calendar to line up neatly in a grid, you need this code to always print a something two characters wide. sprintf is your friend here! Check out the "Examples of width", about halfway down the page.
You've hardcoded 31 for the number of days in the month. This should be fixed, of course. (Use the Date library!)
It's funny how you used strftime("%a") in one place, yet constructed the calendar title awkwardly in the line above! 😄 Take a look at the documentation for formatting dates; it's extremely flexible. I think you can use: Date.today.strftime("%B %Y").
If you'd like to add some colour (or background colour?) to the current day of the month, consider doing something like this, or use a library to assist.
Using while loops works OK, but is quite un-rubyish. In 99% of cases, ruby has even better tools for the job; it's a very expressive language - iterators are king! (I'm guessing you first learned another language, before ruby? Seeing while loops, and/or for loops, is a dead giveaway that you're more familiar with a different language.) Instead of the outer while loop (while i < days.size), you could use days.each_with_index. And instead of the inner while loop (while j < 31), you could use day_of_month.step(31, 7) (how cool is that!!).
This is one way:
Construct a one-dimensional array, beginning with the daynames (Mon Tue ...).
Figure out a way to determine with how many "blanks" the month starts (these are days from the previous month. wday might help). Attach that amount of empty strings to the array.
Determine how many days the month has (hint Date.new(2021,7,-1), and attach all these daynumbers to the array.
Attach empty strings to the array until the size of the array is divisible by 7 (or better, calculate). Skip this if you're skipping the last bullet.
Convert all elements of this array to right-adjusted strings of size 3 or some-such.
Use each_slice(7) to slice the array into weeks.
If desired, transpose this array of week-slices to mimic the ncal output.
Thank you for your help, literally 10 hours and I figured it out thanks to you. I apologize once again for the initially incorrectly posed question.
With the help of hints, I assembled such a solution.
require 'date'
days = %w[Mon Tue Wed Thu Fri Sat Sun]
p days
blanks = Date.new(2021,7,1).wday - 1
blanks.times do
days.push(' ')
end
days_in_month = Date.new(2021, 7, -1).day
days_in_month
day = 1
while day <= days_in_month
days.push(day)
day += 1
end
unless (days.size % 7) == 0
days.push(' ')
end
days.join(', ')
new_arr = days.each_slice(7).to_a
puts"Массив дней: #{new_arr}"
for i in 0...7
for j in 0...new_arr.size
print " #{new_arr[j][i]}"
end
puts
end
require 'date'
# init
DAYS_ORDER = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun']
today = Date.today
month = today.month
year = today.year
first_day = Date.new(year, month, 1)
last_day = Date.new(year, month, -1)
hash_days = {}
# get all current months days and add to hash_days
first_day.upto(last_day) { |day| hash_days[day.day] = day.strftime('%a') }
# group by wday
grouped_hash = hash_days.group_by { |day| day.pop }.transform_values { |days| days.flatten }
# sort by wday from DAYS_ORDER
sorted_arr = grouped_hash.sort_by { |k, v| DAYS_ORDER.index(k) }
# rendering current month's calendar with mark current day
## title
print "\x1b[4m#{today.strftime("%B %Y")}\x1b[0m\n"
## calendar
indent = true
sorted_arr.each do |wday, days|
print wday
if days[0] != 1 && indent == true
print " "
else
indent = false
end
days.each do |value|
spaces = " " * (value > 9 ? 1 : 2)
str_day = spaces + value.to_s
current_day = "\x1b[1;31m#{str_day}\x1b[0m"
print value == today.day ? current_day : str_day
end
puts
end
view
I want to check if a day is the last day of the month and if it is, for a function to return true, otherwise return false.
For example, if I pass in an argument of "Sun, 30 Jun 2013", the function is to return true, because it is the last day of the month, however if I pass in the argument "Mon, 03 Jun 2013" the function is to return false.
How can this be accomplished using Ruby.
If you're using Rails, you can always do this as well:
date == date.end_of_month
or to check the end of this month:
date == Date.today.end_of_month
I would do something like this
def is_last_day(mydate)
mydate.month != mydate.next_day.month
end
Parse the date with DateTime.parse. DateTime.parse has built-in support for many date formats (including those in your example), but you can always use DateTime.strptime for more complex formats.
See if the next day is 1 (first day of next month) by using Date#+.
require 'date'
def last_day?(date_string)
date = DateTime.parse(date_string)
(date + 1).day == 1
end
puts last_day?('Sun, 30 Jun 2013') # true
puts last_day?('Mon, 03 Jun 2013') # false
I would like to know how to get the current week number from Rails and how do I manipulate it:
Translate the week number into date.
Make an interval based on week number.
Thanks.
Use strftime:
%U - Week number of the year. The week starts with Sunday. (00..53)
%W - Week number of the year. The week starts with Monday. (00..53)
Time.now.strftime("%U").to_i # 43
# Or...
Date.today.strftime("%U").to_i # 43
If you want to add 43 weeks (or days,years,minutes, etc...) to a date, you can use 43.weeks, provided by ActiveSupport:
irb(main):001:0> 43.weeks
=> 301 days
irb(main):002:0> Date.today + 43.weeks
=> Thu, 22 Aug 2013
irb(main):003:0> Date.today + 10.days
=> Sun, 04 Nov 2012
irb(main):004:0> Date.today + 1.years # or 1.year
=> Fri, 25 Oct 2013
irb(main):005:0> Date.today + 5.months
=> Mon, 25 Mar 2013
You are going to want to stay away from strftime("%U") and "%W".
Instead, use Date.cweek.
The problem is, if you ever want to take a week number and convert it to a date, strftime won't give you a value that you can pass back to Date.commercial.
Date.commercial expects a range of values that are 1 based.
Date.strftime("%U|%W") returns a value that is 0 based. You would think you could just +1 it and it would be fine. The problem will hit you at the end of a year when there are 53 weeks. (Like what just happened...)
For example, let's look at the end of Dec 2015 and the results from your two options for getting a week number:
Date.parse("2015-12-31").strftime("%W") = 52
Date.parse("2015-12-31").cweek = 53
Now, let's look at converting that week number to a date...
Date.commercial(2015, 52, 1) = Mon, 21 Dec 2015
Date.commercial(2015, 53, 1) = Mon, 28 Dec 2015
If you blindly just +1 the value you pass to Date.commercial, you'll end up with an invalid date in other situations:
For example, December 2014:
Date.commercial(2014, 53, 1) = ArgumentError: invalid date
If you ever have to convert that week number back to a date, the only surefire way is to use Date.cweek.
date.commercial([cwyear=-4712[, cweek=1[, cwday=1[, start=Date::ITALY]]]]) → date
Creates a date object denoting the given week date.
The week and the day of week should be a negative
or a positive number (as a relative week/day from the end of year/week when negative).
They should not be zero.
For the interval
require 'date'
def week_dates( week_num )
year = Time.now.year
week_start = Date.commercial( year, week_num, 1 )
week_end = Date.commercial( year, week_num, 7 )
week_start.strftime( "%m/%d/%y" ) + ' - ' + week_end.strftime("%m/%d/%y" )
end
puts week_dates(22)
EG: Input (Week Number): 22
Output: 06/12/08 - 06/19/08
credit: Siep Korteling http://www.ruby-forum.com/topic/125140
Date#cweek seems to get the ISO-8601 week number (a Monday-based week) like %V in strftime (mentioned by #Robban in a comment).
For example, the Monday and the Sunday of the week I'm writing this:
[ Date.new(2015, 7, 13), Date.new(2015, 7, 19) ].map { |date|
date.strftime("U: %U - W: %W - V: %V - cweek: #{date.cweek}")
}
# => ["U: 28 - W: 28 - V: 29 - cweek: 29", "U: 29 - W: 28 - V: 29 - cweek: 29"]
I've posted this question for C# but I may be working in Ruby instead. So I'm asking the same question about Ruby:
I'm looking for a Ruby class/library/module that works similarly to the Perl module Date::Manip as far as business/holiday dates. Using that module in Perl, I can pass it a date and find out whether it's a business day (ie, Mon-Fri) or a holiday. Holidays are very simple to define in a config file (see Date::Manip::Holidays). You can enter a 'fixed' date that applies to every year like:
12/25 = Christmas
or 'dynamic' dates for every year like:
last Monday in May = Memorial Day
or 'fixed' dates for a given year like:
5/22/2010 = Bob's Wedding
You can also pass in a date and get back the next/previous business day (which is any day that's not a weekend and not a holiday).
Does anyone know of anything like that in the Ruby world?
You may use the holidays-gem.
http://rubygems.org/gems/holidays
Some national (and regional) holidays are already predefined, you may define your own holiday definitions.
The business_time gem should do what you need.
The example at bottom of the README doc is a good starting example:
require 'rubygems'
require 'active_support'
require 'business_time'
# We can adjust the start and end time of our business hours
BusinessTime::Config.beginning_of_workday = "8:30 am"
BusinessTime::Config.end_of_workday = "5:30 pm"
# and we can add holidays that don't count as business days
# July 5 in 2010 is a monday that the U.S. takes off because
# our independence day falls on that Sunday.
three_day_weekend = Date.parse("July 5th, 2010")
BusinessTime::Config.holidays << three_day_weekend
friday_afternoon = Time.parse("July 2nd, 2010, 4:50 pm")
tuesday_morning = 1.business_hour.after(friday_afternoon)
You probably going to need the chronic gem to help you build the holiday dates from your config file. However YMMV because your example last monday in may doesn't work in chronic. Hackaround is do something like this:
# last monday in May (2010)
Chronic.parse('last monday', :now => Time.parse('2010-06-01'))
And look at the tickle gem which works on top of chronic for a way to add recurring events.
/I3az/
You could take a look at my Workpattern gem. It allows you to specify working and resting times. It was aimed at producing a "Calendar" like is used in planning tools such as Microsoft Project and Primavera P6, so you can specify right down to the minute.
Here is a simple example:
Create a new Workpattern mywp=Workpattern.new('My Workpattern',2011,10) This is for 10 years from 2011 but you can make it longer or shorter.
Tell it you want the Weekends to be resting and that you also want to rest during the week so you work between 9 and 12 in the morning and 1 and 6 in the afternoon.
mywp.resting(:days => :weekend)
mywp.resting(:days =>:weekday, :from_time=>Workpattern.clock(0,0),:to_time=>Workpattern.clock(8,59))
mywp.resting(:days =>:weekday, :from_time=>Workpattern.clock(12,0),:to_time=>Workpattern.clock(12,59))
mywp.resting(:days =>:weekday, :from_time=>Workpattern.clock(18,0),:to_time=>Workpattern.clock(23,59))
Now just calculate using minutes
mydate=DateTime.civil(2011,9,1,9,0)
result_date = mywp.calc(mydate,1920) # => 6/9/11#18:00
1920 is 4 days * 8 hours a day * 60 minutes and hour.
I wrote the gem to learn Ruby - only scratched the surface.
Check out the biz gem.
Here's an example configuration:
require 'biz'
Biz.configure do |config|
config.hours = {
mon: {'09:00' => '17:00'},
tue: {'00:00' => '24:00'},
wed: {'09:00' => '17:00'},
thu: {'09:00' => '12:00', '13:00' => '17:00'},
sat: {'10:00' => '14:00'}
}
config.holidays = [Date.new(2014, 1, 1), Date.new(2014, 12, 25)]
config.time_zone = 'America/Los_Angeles'
end
When you use the optional core extensions, it's as easy as the following to find out if a date is a business day:
require 'biz/core_ext'
Date.new(2014, 12, 25).business_day? # => false
I'm trying to figure out how to extract dates from unstructured text using Ruby.
For example, I'd like to parse the date out of this string "Applications started after 12:00 A.M. Midnight (EST) February 1, 2010 will not be considered."
Any suggestions?
Try Chronic (http://chronic.rubyforge.org/) it might be able to parse that otherwise you're going to have to use Date.strptime.
Assuming you just want dates and not datetimes:
require 'date'
string = "Applications started after 12:00 A.M. Midnight (EST) February 1, 2010 will not be considered."
r = /(January|February|March|April|May|June|July|August|September|October|November|December) (\d+{1,2}), (\d{4})/
if string[r]
date =Date.parse(string[r])
puts date
end
Also you can try a gem that can help find date in string.
Exapmle:
input = 'circa 1960 and full date 07 Jun 1941'
dates_from_string = DatesFromString.new
dates_from_string.get_structure(input)
#=> return
# [{:type=>:year, :value=>"1960", :distance=>4, :key_words=>[]},
# {:type=>:day, :value=>"07", :distance=>1, :key_words=>[]},
# {:type=>:month, :value=>"06", :distance=>1, :key_words=>[]},
# {:type=>:year, :value=>"1941", :distance=>0, :key_words=>[]}]