Trying to convert strings like 9 weeks ago, 1 year, 6 months ago, 20 hours ago into a ruby time object like Tue, 10 Mar 2015 12:06:15 PDT -07:00.
I've been doing this:
eval("10 days ago".gsub(' ', '.'))
This works fine, but for strings like 1 year, 6 months ago blows up.
I just need to do comparisons like:
eval("10 days ago".gsub(' ', '.')) < (Time.now - 7.days)
I'm using sinatra so no fancy rails helpers.
Please never use eval in production code..
Converting from timeago notation would be quite complex and resource intensive.
However, this way seems the least error prone: It will convert a string like "5 seconds ago" to "5S" and use mapping to find what it means in time, after which it will subtract that time from the current time.
The parse string is dynamically built so it can accomodate most every timeago notation.
require('date')
mapping = {"D"=> "%d","W"=>"%U","H"=>"%T","Y"=>"%Y","M"=>"%m","S"=>"%S"}
timerel = "1 year, 6 months ago".split(",").map { |n| n.gsub(/\s+/, "").upcase()[0,2].split('')}
Date.strptime(
timerel.map {|n| n[0]}.join(" "),
timerel.map {|n| mapping[n[1]]}.join(" ")
)
date = Date.new(0) + (Date.today - Date.strptime(timerel.map {|n| n[0]}.join(" "), timerel.map {|n| mapping[n[1]]}.join(" ")))
=> #<Date: 2014-10-10 ((2456941j,0s,0n),+0s,2299161j)>
It goes without saying that is very error prone. Use at your own risk:
def parse(date:)
eval(date.gsub(/ ?(,|and) ?/, '+').tr(' ', '.').gsub(/^(.*)(\.ago)$/, '(\1)\2'))
end
parse(date: '1 year, 6 months ago') # => Wed, 10 Sep 2014 21:29:11 BST +01:00
parse(date: '1 year, 6 months, 3 weeks, 6 days, 9 hours and 12 seconds ago')
# => Thu, 14 Aug 2014 12:33:07 BST +01:00
The idea is to convert the original string to:
'(1.year+6.months).ago'
Related
I am trying to use the Date.parse method in Ruby but it is returning odd results when I try to parse early historical dates. For example:
Date.parse("16 January 27 BC")
=> #<Date: -0026-01-16 >
This returns the correct year (although it is -26 instead of -27 but that is normal I think).
But when I try:
Date.parse("19 august 1 BC")
=> #<Date: 2000-08-19 >
Here it assumes the year 2000 as a start date. Same thing for later years:
Date.parse("19 August 14")
=> #<Date: 2014-08-19 >
yearninetynine = Date.parse("19 august AD 99")
=> #<Date: 1999-08-19 >
(I have also tried to replace AD with "CE" but that makes no difference)
The year 100 is where it starts to work properly again:
Date.parse("19 august AD 100")
=> #<Date: 0100-08-19 >
To summarize: up to the year 100 it doesn't parse properly and treats it as years occurring in the 1990s and 2000s, negative years are one less than you would intuitively expect, negative years close to 0 are also not parsed properly.
I tried a few other methods like start=Date::ITALY, but these seem to make no difference. Is there a way to make it parse these years properly?
According to the docs that is exactly what the 2nd argument is for.
Date.parse("19 august 99") #=> Thu, 19 Aug 1999
Date.parse("19 august 99", false) #=> Mon, 19 Aug 0099
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've installed the Chronic gem and while
I can get Chronic.parse('Saturday', :context => :past) to return last Saturday's date but not
'5 last saturdays' returns 'nil'
I would like it to return in this format: strftime "%m%d%Y%H%M"
I also can't tag on any methods like .exists? or should ==true so I can use it to validate that those reports were run
I would just like to verify that 02/11/2017 02/04/2017 01/28/2017 01/28/2017 01/21/2017 appear as text on the page, but since they will change every week i need logic that will return last 5 Saturdays from now
Use "weeks ago" to get the Saturday of a previous week:
Chronic.parse('1 week ago Saturday')
#=> "2017-02-11 12:00:00 -0500"
Chronic.parse('2 weeks ago Saturday')
#=> "2017-02-04 12:00:00 -0500"
Chronic.parse('3 weeks ago Saturday')
#=> "2017-01-28 12:00:00 -0500"
Chronic.parse('4 weeks ago Saturday')
#=> "2017-01-21 12:00:00 -0500"
Chronic.parse('5 weeks ago Saturday')
#=> "2017-01-14 12:00:00 -0500"
You can format the value returned using #strftime:
Chronic.parse('1 week ago Saturday').strftime("%m%d%Y%H%M")
#=> "021120171200"
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"]
Is there a generic parser to parse "... ago" strings and turn them into DateTime objects?
Possible sentences are:
1 year|#count years (and 1 week|#count weeks) ago
1 week|#count weeks (and 1 day|#count days) ago
1 day|#count days (and 1 hour|#count hours) ago
1 hour|#count hours ago (and 1 min|#count min) ago
1 min|#count min ago (and 1 sec|#count sec) ago
1 sec|#count sec ago
So its either a combination of two (Foo and Bar ago) or only one (Foo ago). And it can be singular or plural.
Ruby's DateTime::parse() cannot handle this, nor can DateTime::strptime().
I am hoping for a gem or a snippet somewhere that handles this.
Else I will have to create my own parser, in which case a pointer to how-to-create own DateTime Parsers would be very welcome.
Sidenote: For Future Reference: These are the timestrings generated by Drupals Format Interval
The Chronic gem may be what you're looking for.
require 'chronic'
Chronic.parse "2 days ago"
# => 2011-06-14 14:13:59 -0700
Per #injekt (one of Chronic's maintainers), you can handle "1 year and 1 week ago" like this:
str = "one year and 1 week ago"
chunks = str.split('and')
Chronic.parse(chunks[1], :now => Chronic.parse(chunks[0] + 'ago'))
#=> 2010-06-09 14:29:37 -0700