I have time ranges, for example
normal time : 03:15 - 17:00
discounttime : 17:00 - 23:10
night time : 23:10 - 03:15
And I have to find to which range current time belongs_to.
Before I had such code:
when (3*60+15)..(17*60+00)
puts "normal_time"
when (17*60+00)..(23*60+10)
puts "disc_time"
else
puts "night_time"
end
But this code is not working if I have
normal time : 07:15 - 19:00
discounttime : 19:00 - 01:10
night time : 01:10 - 07:15
Which code should I use to make both configs working?
Should I use 2 days duration?
Dirty Solution
I know that the code is not optimized, but it works. I'll refactor it
time = Time.now
a = [:normal_time, :disc_time, :night_time]
b = a + [a[0]]
pp = Priceplan.last # has times sets for :normal_time, :disc_time, :night_time
check = time.hour*60+time.min
a.each_with_index do |e, i|
starts_at = pp.send(e).hour * 60 + pp.send(e).min
ends_at = pp.send(b[i+1]).hour * 60 + pp.send(b[i+1]).min
if starts_at < ends_at
if (starts_at..ends_at).include?(check)
p "ok!!!!!!!! at #{e}"
break
end
else
if (starts_at..(24*60)).include?(check)
p "ok!!!!!!!! at #{e} (divided first)"
break
elsif (0..ends_at).include?(check)
p "ok!!!!!!!! at #{e} (divided second)"
break
end
end
end
Ranges always run from low numbers to higher numbers. When you have a range like: 19:00-01:10 you're creating an invalid range. To fix it just divide up the ranges so that you have 19:00-23:59 and then a new range 00:00-01:10 to cover the rest of the timespan.
You can try using today midnight as a base point
def today_midnight
Time.now.utc.midnight
end
normal time : where(:belongs_to => (today_midnight + 7.hours + 15.minutes) .. (today_midnight + 19.hours))
discount time : where(:belongs_to => (today_midnight + 1.hours + 10.minutes) .. (today_midnight + 19.hours))
night time: where(:belongs_to => (today_midnight + 1.hour + 10.minutes) .. (today_midnight + 7.hours + 15.minutes))
Though long, I believe writing in such a manner makes the code a lot more readable. i.e. a year after, if you'll be able to understand whats going on by quickly looking at the code.
Related
I’m using Rails 4.2.7. I want to calculate the number of milliseconds given a duration, which could include hours, minutes, and seconds. So I wrote these two functions:
def convert_to_hrs(string)
if !string.nil?
string.strip!
case string.count(':')
when 0
'00:00:' + string.rjust(2, '0')
when 1
'00:' + string
else
string
end
else
"00:00:00"
end
end
def duration_in_milliseconds(input)
input = convert_to_hrs(input)
if input.match(/\d+:\d\d:\d\d\.?\d*/)
h, m, s = input.split(':').map(&:to_i)
(h.hours + m.minutes + s.seconds) * 1000
else
0
end
end
Unfortunately, when I call duration_in_milliseconds(input) with a number like 8:49, the result is zero. The result should be interpreted as 8 minutes and 49 seconds, which in milliseconds would be 529000. How do I adjust the above to account for this?
I'll just leave this one here:
def convert_to_ms(string)
string.split(':').map(&:to_i).inject(0) { |a, b| a * 60 + b } * 1000
end
convert_to_ms('8:49')
#=> 529000
Just another implementation.. Cause it's fun :)
def duration_in_milliseconds(string)
string.split(':')
.map(&:to_i)
.reverse
.zip([1,60,3600])
.map{ |segment, multiplier| segment*multiplier }
.inject(:+) * 1000
end
def some_method(x)
if x == 1
date = Date.today
elsif x == 5
date = Date.today + 2
else
date = Date.today - 2
end
date + 20
end
For visual clarity, is it possible somehow to omit date = for each statement and catch whatever the returned value is from the conditional and add 20 to it?
(The code is for example purpose, my own code has 10 if-statements.)
def some_method(x)
date = if x == 1
Date.today
elsif x == 5
Date.today + 2
else
Date.today - 2
end
date + 20
end
If you have 10 if statements it is probably better to refactor code using case-when like this:
def some_method(x)
date = case x
when 1; Date.today
when 5; Date.today + 2
else; Date.today - 2
end
date + 20
end
I have this:
<(Time.now).to_i>
it returns a integer thing
and want to convert that integer value into (mins and seconds)
only as per my requirement
Try this
Time.now.strftime("%M:%S")
Hope you are trying to find this
You can get the minutes and seconds by this
delta = Time.now.to_i
%w[ weeks days hours minutes seconds].collect do |step|
seconds = 1.send(step)
(delta / seconds).to_i.tap do
delta %= seconds
end
end
It will return an array with having 5 elements . You can get the mins and secs from
arr[3] and arr[4]
Also you can take the whatever the time in integer format as delta .
If you want to show only mins and secs when hr = 0
#hr = timedef[0]
#mn = timedef[1]
#sec = timedef[2]
time_remains = ''
unless self.timedef.blank?
if #hr > 0
time_remains = time_remains + "#{#hr} #{'hour'.pluralize(#hr)} "
elsif #mn > 0
time_remains = time_remains + "#{#mn} #{'minute'.pluralize(#mn)} and #{#sec} #{'second'.pluralize(#sec)} "
else
time_remains = time_remains + "#{#sec} #{'second'.pluralize(#sec)} "
end
end
use strftime -- see this page http://www.dzone.com/snippets/date-time-format-ruby it has the formats (though its similar to c/c++ and Java)
Also, you will likely want to do some research into timezone handling on the time class. there is a function: in_time_zone that can convert the timezone for you. So typically you store the times or the times fetched are in UTC then you can dynamically change the timezone before you strftime.
I can't seem to find an elegant way to do this...
Given a date how can I find the next Tuesday that is either the 2nd or the 4th Tuesday of the calendar month?
For example:
Given 2012-10-19 then return 2012-10-23
or
Given 2012-10-31 then return 2012-11-13
October November
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 1 2 3
7 8 9 10 11 12 13 4 5 6 7 8 9 10
14 15 16 17 18 19 20 11 12 13 14 15 16 17
21 22 23 24 25 26 27 18 19 20 21 22 23 24
28 29 30 31 25 26 27 28 29 30
Scroll to the bottom if you just want to see what the end result can look like..
Using code snippets from some date processing work I've done recently in ruby 1.9.3.
Some upgrades to DateTime:
require 'date'
class DateTime
ALL_DAYS = [ 'sunday', 'monday', 'tuesday',
'wednesday', 'thursday', 'friday', 'saturday' ]
def next_week
self + (7 - self.wday)
end
def next_wday (n)
n > self.wday ? self + (n - self.wday) : self.next_week.next_day(n)
end
def nth_wday (n, i)
current = self.next_wday(n)
while (i > 0)
current = current.next_wday(n)
i = i - 1
end
current
end
def first_of_month
self - self.mday + 1
end
def last_of_month
self.first_of_month.next_month - 1
end
end
method_missing Tricks:
I have also supplemented the class with some method missing tricks to map calls from next_tuesday to next_wday(2) andnth_tuesday(2)tonth_wday(2, 2)`, which makes the next snippet easier on the eyes.
class DateTime
# ...
def method_missing (sym, *args, &block)
day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
dindex = ALL_DAYS.include?(day) ? ALL_DAYS.index(day.downcase) : nil
if (sym =~ /^next_[a-zA-Z]+$/i) && dindex
self.send(:next_wday, dindex)
elsif (sym =~ /^nth_[a-zA-Z]+$/i) && dindex
self.send(:nth_wday, dindex, args[0])
else
super(sym, *args, &block)
end
end
def respond_to? (sym)
day = sym.to_s.gsub(/^(next|nth)_(?<day>[a-zA-Z]+)$/i, '\k<day>')
(((sym =~ /^next_[a-zA-Z]+$/i) || (sym =~ /^nth_[a-zA-Z]+$/i)) && ALL_DAYS.include?(day)) || super(sym)
end
end
Example:
Given a date:
today = DateTime.now
second_tuesday = (today.first_of_month - 1).nth_tuesday(2)
fourth_tuesday = (today.first_of_month - 1).nth_tuesday(4)
if today == second_tuesday
puts "Today is the second tuesday of this month!"
elsif today == fourth_tuesday
puts "Today is the fourth tuesday of this month!"
else
puts "Today is not interesting."
end
You could also edit method_missing to handle calls such as :second_tuesday_of_this_month, :fourth_tuesday_of_this_month, etc. I'll post the code here if I decide to write it at a later date.
Take a look at Chronic or Tickle, both are gems for parsing complex times and dates. Tickle in particular will parse recurring times (I think it uses Chronic as well).
Check out this gem, you might be able to figure out your answer
https://github.com/mojombo/chronic/
Since you already use Rails, you don't need the includes, but this works in pure Ruby as well for reference.
require 'rubygems'
require 'active_support/core_ext'
d = DateTime.parse('2012-10-19')
result = nil
valid_weeks = [d.beginning_of_month.cweek + 1, d.beginning_of_month.cweek + 3]
if valid_weeks.include?(d.next_week(:tuesday).cweek)
result = d.next_week(:tuesday)
else
result = d.next_week.next_week(:tuesday)
end
puts result
I think you should probably use a library if you're needing to branch out into more interesting logic, but if what you've described is all you need, the code below should help
SECONDS_PER_DAY = 60 * 60 * 24
def find_tuesday_datenight(now)
tuesdays = [*-31..62].map { |i| now + (SECONDS_PER_DAY * i) }
.select { |d| d.tuesday? }
.group_by { |d| d.month }
[tuesdays[now.month][1], tuesdays[now.month][-2], tuesdays[(now.month + 1) % 12][1]]
.find {|d| d.yday > now.yday }
end
Loop through the last month and next month, grab the tuesdays, group by month, take the 2nd and the 2nd last tuesday of the current month (If you actually do want the 4th tuesday, just change -2 to 3) and the 2nd tuesday of the next month and then choose the first one after the provided date.
Here's some tests, 4 tuesdays in month, 5 tuesdays in month, random, and your examples:
[[2013, 5, 1], [2013, 12, 1], [2012, 10, 1], [2012, 10, 19], [2012, 10, 31]].each do |t|
puts "#{t} => #{find_tuesday_datenight(Time.new *t)}"
end
which produces
[2013, 5, 1] => 2013-05-14 00:00:00 +0800
[2013, 12, 1] => 2013-12-10 00:00:00 +0800
[2012, 10, 1] => 2012-10-09 00:00:00 +0800
[2012, 10, 19] => 2012-10-23 00:00:00 +0800
[2012, 10, 31] => 2012-11-13 00:00:00 +0800
I'm sure it could be simplified, and I'd love to hear some suggestions :) (way too late &tired to even bother figuring out what the actual range should be for valid dates i.e. smaller than -31..62)
so here is the code that will resolve a weekday for a given week in a month (what you asked for with little sugar). You should not have problems if you are running inside rails framework. Otherwise make sure you have active_support gem installed. Method name is stupid so feel free to change it :)
usage: get_next_day_of_week(some_date, :friday, 1)
require 'rubygems'
require 'active_support/core_ext'
def get_next_day_of_week(date, day_name, count)
next_date = date + (-date.days_to_week_start(day_name.to_sym) % 7)
while (next_date.mday / 7) != count - 1 do
next_date = next_date + 7
end
next_date
end
I use the following to calculate Microsoft's patch Tuesday date. It was adapted from some C# code.
require 'date'
#find nth iteration of given day (day specified in 'weekday' variable)
findnthday = 2
#Ruby wday number (days are numbered 0-7 beginning with Sunday)
weekday = 2
today = Time.now
todayM = today.month
todayY = today.year
StrtMonth = DateTime.new(todayY,todayM ,1)
while StrtMonth.wday != weekday do
StrtMonth = StrtMonth + 1;
end
PatchTuesday = StrtMonth + (7 * (findnthday - 1))
I'm trying to find the best way to generate the following output
<name> job took 30 seconds
<name> job took 1 minute and 20 seconds
<name> job took 30 minutes and 1 second
<name> job took 3 hours and 2 minutes
I started this code
def time_range_details
time = (self.created_at..self.updated_at).count
sync_time = case time
when 0..60 then "#{time} secs"
else "#{time/60} minunte(s) and #{time-min*60} seconds"
end
end
Is there a more efficient way of doing this. It seems like a lot of redundant code for something super simple.
Another use for this is:
<title> was posted 20 seconds ago
<title> was posted 2 hours ago
The code for this is similar, but instead i use Time.now:
def time_since_posted
time = (self.created_at..Time.now).count
...
...
end
If you need something more "precise" than distance_of_time_in_words, you can write something along these lines:
def humanize(secs)
[[60, :seconds], [60, :minutes], [24, :hours], [Float::INFINITY, :days]].map{ |count, name|
if secs > 0
secs, n = secs.divmod(count)
"#{n.to_i} #{name}" unless n.to_i==0
end
}.compact.reverse.join(' ')
end
p humanize 1234
#=>"20 minutes 34 seconds"
p humanize 12345
#=>"3 hours 25 minutes 45 seconds"
p humanize 123456
#=>"1 days 10 hours 17 minutes 36 seconds"
p humanize(Time.now - Time.local(2010,11,5))
#=>"4 days 18 hours 24 minutes 7 seconds"
Oh, one remark on your code:
(self.created_at..self.updated_at).count
is really bad way to get the difference. Use simply:
self.updated_at - self.created_at
There are two methods in DateHelper that might give you what you want:
time_ago_in_words
time_ago_in_words( 1234.seconds.from_now ) #=> "21 minutes"
time_ago_in_words( 12345.seconds.ago ) #=> "about 3 hours"
distance_of_time_in_words
distance_of_time_in_words( Time.now, 1234.seconds.from_now ) #=> "21 minutes"
distance_of_time_in_words( Time.now, 12345.seconds.ago ) #=> "about 3 hours"
chronic_duration parses numeric time to readable and vice versa
If you want to show significant durations in the seconds to days range, an alternative would be (as it doesn't have to perform the best):
def human_duration(secs, significant_only = true)
n = secs.round
parts = [60, 60, 24, 0].map{|d| next n if d.zero?; n, r = n.divmod d; r}.
reverse.zip(%w(d h m s)).drop_while{|n, u| n.zero? }
if significant_only
parts = parts[0..1] # no rounding, sorry
parts << '0' if parts.empty?
end
parts.flatten.join
end
start = Time.now
# perform job
puts "Elapsed time: #{human_duration(Time.now - start)}"
human_duration(0.3) == '0'
human_duration(0.5) == '1s'
human_duration(60) == '1m0s'
human_duration(4200) == '1h10m'
human_duration(3600*24) == '1d0h'
human_duration(3600*24 + 3*60*60) == '1d3h'
human_duration(3600*24 + 3*60*60 + 59*60) == '1d3h' # simple code, doesn't round
human_duration(3600*24 + 3*60*60 + 59*60, false) == '1d3h59m0s'
Alternatively you may be only interested in stripping the seconds part when it doesn't matter (also demonstrating another approach):
def human_duration(duration_in_seconds)
n = duration_in_seconds.round
parts = []
[60, 60, 24].each{|d| n, r = n.divmod d; parts << r; break if n.zero?}
parts << n unless n.zero?
pairs = parts.reverse.zip(%w(d h m s)[-parts.size..-1])
pairs.pop if pairs.size > 2 # do not report seconds when irrelevant
pairs.flatten.join
end
Hope that helps.
There is problem with distance_of_time_in_words if u ll pass there 1 hour 30 min it ll return about 2 hours
Simply add in helper:
PERIODS = {
'day' => 86400,
'hour' => 3600,
'minute' => 60
}
def formatted_time(total)
return 'now' if total.zero?
PERIODS.map do |name, span|
next if span > total
amount, total = total.divmod(span)
pluralize(amount, name)
end.compact.to_sentence
end
Basically just pass your data in seconds.
Rails has a DateHelper for views. If that is not exactly what you want, you may have to write your own.
#Mladen Jablanović has an answer with good sample code. However, if you don't mind continuing to customize a sample humanize method, this might be a good starting point.
def humanized_array_secs(sec)
[[60, 'minutes '], [60, 'hours '], [24, 'days ']].inject([[sec, 'seconds']]) do |ary, (count, next_name)|
div, prev_name = ary.pop
quot, remain = div.divmod(count)
ary.push([remain, prev_name])
ary.push([quot, next_name])
ary
end.reverse
end
This gives you an array of values and unit names that you can manipulate.
If the first element is non-zero, it is the number of days. You may want to write code to handle multiple days, like showing weeks, months, and years. Otherwise, trim off the leading 0 values, and take the next two.
def humanized_secs(sec)
return 'now' if 1 > sec
humanized_array = humanized_array_secs(sec.to_i)
days = humanized_array[-1][0]
case
when 366 <= days
"#{days / 365} years"
when 31 <= days
"#{days / 31} months"
when 7 <= days
"#{days / 7} weeks"
else
while humanized_array.any? && (0 == humanized_array[-1][0])
humanized_array.pop
end
humanized_array.reverse[0..1].flatten.join
end
end
The code even finds use for a ruby while statement.