I have the following date string ('US/Eastern'), which I need to convert to UTC:
date_src = '2014-07-07T23:10:00+0'
First I convert it to a "valid" format so I can operate it on later processes. I use the following to have an iso version of the date:
date = DateTime.parse(date_src).iso8601
At this point date is a nice '2014-07-07T23:10:00+00:00'. The last step on my process is to translate this date to UTC. I'm using the following:
TZInfo::Timezone.get('US/Eastern').local_to_utc(date)
The problem is this is giving me 20014 as output, instead of the UTC version of the original date. If I try:
TZInfo::Timezone.get('UTC').local_to_utc(date)
I get 2014, which is the correct year but still unexpected output.
Any ideas about what I'm doing wrong, and what I could use to solve the problem?
local_to_utc actually expects a Time or a DateTime instance:
TZInfo::Timezone.get('US/Eastern').local_to_utc(DateTime.parse(date_src))
# => #<DateTime: 2014-07-08T03:10:00+00:00 ((2456847j,11400s,0n),+0s,2299161j)>
From the documentation, you can have a hint on what actually happened:
All methods in TZInfo that operate on a time can be used with either Time or DateTime instances or with nteger timestamps (i.e. as returned by Time#to_i). The type of the values returned will match the the type passed in.
What actually happens is the local_to_utc calls to_i on the input parameter, which on a string returns the parsed integer from the beginning of the string (2014 in your case since date is the string 2014-07-07T23:10:00+00:00), and adds the time difference to it - 18000 for "US/Eastern" (5 hour difference), and 0 for UTC:
date.to_i
# => 2014
TZInfo::Timezone.get('US/Eastern').local_to_utc(date) - date.to_i
# => 18000
TZInfo::Timezone.get('UTC').local_to_utc(date) - date.to_i
# => 0
So the bottom line is - kind of serendipitously you saw this weird behavior, which stems from the compilation of some surprising quirks of the APIs you used...
Related
I am parsing a csv file and i need to change the dob to match a certain format of YYY-MM-DD I keep getting this error of parse': invalid date (ArgumentError)
it happens when it tries to parse this date of "6/6/99" How can i fix this so I don't get any errors for any of the dates I have?
list of all dates in csv, I am not sure if any of the other dates following the one above would error out as well.
"12/12/2010"
"1/1/1988"
"2/2/1966"
"6/6/99"
"1/4/88"
"4/4/1948"
"1/6/1988"
"1/7/1988"
"1/8/88"
"1/9/88"
"1988-02-12"
"1-11-88"
"1/12/88"
"1/13/88"
my code
require 'csv'
require 'time'
require 'date'
def parse_csv
table = CSV.parse(File.read("input.csv", encoding: 'bom|utf-8' ), headers: true, col_sep: ",")
formatted = table.map(&:to_h)
formatted.each do |x|
if x["dob"] =~ /^\d{4}-\d{2}-\d{2}$/
p "correct"
else
parsed = Date.parse(x["dob"], "%Y-%m-%d")
p parsed
end
end
end
parse_csv
From the documentation:
Date.parse
Parses the given representation of date and time, and creates a date
object.
This method does not function as a validator. If the input string
does not match valid formats strictly, you may get a cryptic result.
Should consider to use Date.strptime instead of this method as
possible.
Your current implementation doesn't really make sense; it's not doing what you think it is:
Date.parse('6/6/99', "%Y-%m-%d")
#=> Date::Error: invalid date
This isn't saying "convert "6/6/99" into "1999-06-06", it's saying "try to parse "6/6/99" into a Date object (and the second argument is essentially being ignored!).
If you're confident what format that date is supposed to be, then (as per the documentation referenced above!) you can use Date.strptime to try explicitly parsing it as this format. For example:
Date.strptime('6/6/99', "%m/%d/%y")
#=> #<Date: 1999-06-06 ((2451336j,0s,0n),+0s,2299161j)>
Or if you're not confident whether the first two values are supposed to represent month-day or day-month, then you'd need to handle this explicitly in the code and treat the error appropriately.
tl;dr:
Date.parse is unreliable for arbitrary inputs; it only makes a "best guess" for the date format. And unsurprisingly here, it fails for at least one of the ambiguous formats you're throwing at it.
Date.strptime is the correct way to parse each date, when you know which format you expect.
I want to check that a date object I have in a validator.rb file has a year field that is less than the year 10000.
required(:my_date_object).maybe(
:date?,
lt?: '10000-01-01'
)
When running system tests, the following error shows up:
ArgumentError:
comparison of Date with String failed
Should I look into converting the date field into a string using to_s or something similar and then doing a regexp format check? Or is there a more straightforward way of checking that the date is less than the year 10000?
You need to create a Date for the lt?.
You can write it like follows:
required(:my_date_object) { lt?(Date.new(10000, 1, 1)) }
I am attempting to change an SQL datetime variable (2016-06-09 14:29:34) into a format that looks like this (00:00_20160601). I have tried to follow a couple of SO questions that will allow me to format a Time object.
This is what I have done so far:
start_datetime = "2016-06-09 14:29:34"
t =Time.new(start_datetime)
t.strftime("%H:%M_%Y%d%m")
This results in the time being formatted to 2016-01-01 00:00:00 +0000, which is obviously not what I want. I was wondering if someone could help me format the datetime object the way I specified?
You can do this with DateTime:
require 'datetime'
DateTime.parse("2016-06-09 14:29:34").strftime("%H:%M_%Y%d%m")
#=> "14:29_20160906"
The format you're feeding in is basically ISO-8601 so it's parsed by default.
Feeding that value into Time.new is completely incorrect. The first argument there is the year, the rest have to be supplied separately. That's why you get 2016-01-01, since everything else comes out as defaults.
Time.new is converting automatically and the result of "2016-06-09 14:29:34".to_i is 2016.
It's not entirely clear why your day value changes from 09 in the input to 01 in the desired output, so I'll use the normal thing and output the same as was input:
require 'time'
start_datetime = "2016-06-09 14:29:34"
t = Time.strptime(start_datetime, '%Y-%m-%d %H:%M:%S')
t.strftime('00:00_%Y%m%d') # => "00:00_20160609"
Since the hours and minutes are being thrown away there are a couple of other ways to go about this.
Ignore the hours and minutes when parsing:
t = Time.strptime(start_datetime, '%Y-%m-%d')
Or use a Date object instead of a Time object:
require 'date'
start_datetime = "2016-06-09 14:29:34"
t = Date.strptime(start_datetime, '%Y-%m-%d')
t.strftime('00:00_%Y%m%d') # => "00:00_20160609"
Making a GET request to a private (no public documentation) API returns data in JSON format.
The value for date looks as follows:
AanmeldDatum: "/Date(1262300400000+0100)/"
There's another variable called AangebodenSindsTekst which means OfferedSinceText and it's value is "8 augustus 2014". So the unknown Date format should get parsed into that specific value.
I'm wondering what kind of date format it is and how can I transform this to something like this 2014-08-08 with Ruby?
I've tried this:
require 'time'
t = '1262300400000+0100'
t2 = Time.parse(t)
# => ArgumentError: no time information in "1262300400000+0100"
Ruby's Time class is your friend, especially the strptime method:
require 'time'
foo = Time.strptime('1262300400000+0100', '%N') # => 2014-08-08 16:57:25 -0700
foo = Time.strptime('1262300400000+0100', '%N%z') # => 2014-08-08 08:57:25 -0700
%N tells Ruby to use nanoseconds. It's throwing away the precision after the 9th digit which is OK if you don't need the rest of the value. Nanosecond accuracy is good enough for most of us.
%z tells Ruby to find the timezone offset, which it then applies to the returned value.
While parse can often figure out how to tear apart an incoming string, it's not bullet-proof, nor is it all-knowing. For speed, I'd recommend learning and relying on strptime if your strings are consistent.
As the Tin Man pointed out in this answer, use the following instead:
Time.strptime('1262300400000+0100', '%Q%z')
it could be milliseconds since epoc, take off the last 3 zeros and plug it into a unix time stamp converter, comes out as Dec 31st 2009
TIME STAMP: 1262300400
DATE (M/D/Y # h:m:s): 12 / 31 / 09 # 11:00:00pm UTC
I have a bunch of strings with opening hours in this format:
Mon-Fri: AM7:00-PM8:00\nSat-Sun: AM8:00-PM6:00
I can deal with the "AM" part by just removing it, but I'd like to convert the PM by
Removing "PM"
Adding 12 to the number before the ":"
Taking care of the fact that PM is sometimes double-digits (e.g. PM11:00)
There can be zero or more PM times in the string.
I'm not sure how to manipulate the time as a number. I've gotten this far:
opening_hours.sub! /PM([\d]?[\d]):/, "***\1***"
Which outputs things like this:
AM7:15-***\u0001***00
The '\u0001` may be due to Japanese characters in the string.
You can take advantage of the fact that String#gsub accepts a block. Something like this will do for you?
s = "Mon-Fri: AM7:00-PM8:00\nSat-Sun: AM8:00-PM11:00"
s2 = s.gsub('AM', '').gsub(/PM(\d+)/) do |match|
(match.gsub('PM', '').to_i + 12).to_s
end
s2 # => "Mon-Fri: 7:00-20:00\nSat-Sun: 8:00-23:00"
Have a look at this question, ruby has a class called datatime.
Convert 12 hr time to 24 hr format in Ruby