How to return time in RFC 3339 format (2014-06-01T12:00:00Z). I read docs about calendar module, but there was no explanation how to generate time format like this. My program should work in different time zones, so please give me advices.
The Erlang Central page Converting Between struct:time and ISO8601 Format has this example:
Unfortunately, no Erlang libraries provide this functionality. Luckily, the native Erlang date and time formats are very easy to format for display or transmission, even in ISO-8601 format:
-module(iso_fmt).
-export([iso_8601_fmt/1]).
iso_8601_fmt(DateTime) ->
{{Year,Month,Day},{Hour,Min,Sec}} = DateTime,
io_lib:format("~4.10.0B-~2.10.0B-~2.10.0B ~2.10.0B:~2.10.0B:~2.10.0B",
[Year, Month, Day, Hour, Min, Sec]).
format_iso8601() ->
{{Year, Month, Day}, {Hour, Min, Sec}} =
calendar:universal_time(),
iolist_to_binary(
io_lib:format(
"~.4.0w-~.2.0w-~.2.0wT~.2.0w:~.2.0w:~.2.0wZ",
[Year, Month, Day, Hour, Min, Sec] )).
Using the above module:
1> {{Year,Month,Day},{Hour,Min,Sec}} = erlang:localtime().
{{2004,8,28},{1,19,37}}
2> io:fwrite("~s\n",[iso_fmt:iso_8601_fmt(erlang:localtime())]).
2004-08-28 01:48:48
To make it output time in UTC, just pass it the return value of erlang:universaltime() instead of erlang:localtime().
Related
I am currently working on a plugin for grandMA2 lighting control using Lua. I need the current time. The only way to get the current time is the following function:
gma.show.getvar('TIME')
which always returns the current system time, which I then store in a variable. An example return value is "12h54m47.517s".
How can I separate the hours, minutes and seconds into 3 variables?
If os.date is available (and matches gma.show.getvar('TIME')), this is trivial:
If format starts with '!', then the date is formatted in Coordinated Universal Time. After this optional character, if format is the string "*t", then date returns a table with the following fields: year, month (1–12), day (1–31), hour (0–23), min (0–59), sec (0–61, due to leap seconds), wday (weekday, 1–7, Sunday is 1), yday (day of the year, 1–366), and isdst (daylight saving flag, a boolean). This last field may be absent if the information is not available.
local time = os.date('*t')
local hour, min, sec = time.hour, time.min, time.sec
This does not provide you with a sub-second precision though.
Otherwise, parsing the time string is a typical task for tostring and string.match:
local hour, min, sec = gma.show.getvar('TIME'):match('^(%d+)h(%d+)m(%d*%.?%d*)s$')
-- This is usually not needed as Lua will just coerce strings to numbers
-- as soon as you start doing arithmetic on them;
-- it still is good practice to convert the variables to the proper type though
-- (and starts being relevant when you compare them, use them as table keys or call strict functions that check their argument types on them)
hour, min, sec = tonumber(hour), tonumber(min), tonumber(sec)
Pattern explanation:
^ and $ pattern anchors: Match the full string (and not just part of it), making the match fail if the string does not have the right format.
(%d)+h: Capture hours: One or more digits followed by a literal h
(%d)+m: Capture minutes: One or more digits followed by a literal m
(%d*%.?%d*)s: Capture seconds: Zero or more digits followed by an optional dot followed by again zero or more digits, finally ending with a literal s. I do not know the specifics of the format and whether something like .1s, 1.s or 1s is occasionally emitted, but Lua's tonumber supports all of these so there should be no issue. Note that this is slightly overly permissive: It will also match . (just a dot) and an s without any leading digits. You might want (%d+%.?%d+)s instead to force digits appearing before & after the dot.
Lets do it with string method gsub()
local ts = gma.show.getvar('TIME')
local hours = ts:gsub('h.*', '')
local mins = ts:gsub('.*%f[^h]', ''):gsub('%f[m].*', '')
local secs = ts:gsub('.*%f[^m]', ''):gsub('%f[s].*', '')
To make a Timestring i suggest string method format()
-- secs as float
timestring = ('[%s:%s:%.3f]'):format(hours, mins, secs)
-- secs not as float
timestring = ('[%s:%s:%.f]'):format(hours, mins, secs)
I am trying to learn F# and was wondering if i have a json object which has time in microseconds as int. I want to get the day, date and time out of this and was wondering how to do it.
I actually happen to have needed to do this recently. You'll almost certainly want to use the .NET time objects (DateTime, DateTimeOffset, TimeSpan) in some capacity. Here's what I went with:
let TicksPerMicrosecond =
TimeSpan.TicksPerMillisecond / 1000L
let FromUnixTimeMicroseconds (us: int64) =
DateTimeOffset.FromUnixTimeMilliseconds 0L + TimeSpan.FromTicks(us * TicksPerMicrosecond)
From TimeSpan.TicksPerMillisecond we can calculate how many are in a microsecond (if I remember correctly it's 10, but this way it doesn't seem as "magic"). Then I can convert the microseconds value into ticks and add it to the epoch date.
To get the day of the week (assuming the time zone is UTC), you'd just use DateTimeOffset.DayOfWeek.
I am unable to understand the expiry date format of the JWT embedded in my application.
For example: 1473912000
What does this translate to? 1473912000 ms, some x date? Any help will be appreciated!
Like James has pointed out:
The number is the number of seconds since Jan 1 1970.
This is converted into the Date object in a quite straight-forward way (the *1000 part is here because in JS main time unit is millisecond):
const expiryDate = new Date(1473912000*1000);
Then you can use any Date method you please.
Likewise, in Ruby you can use Time.at(1473912000) to create a new Time instance like Maxim has shown.
The number is the number of seconds since Jan 1 1970. It is commonly used on unix systems to represent time. Your time is 2016-09-15 04:00 (UTC)
To convert you can try a web based system http://www.unixtimestamp.com/index.php
This is UNIX time in seconds:
➜ ~ irb
2.2.0 :001 > Time.at(1473912000)
=> 2016-09-15 07:00:00 +0300
I have a string of the following format:
"136:16:11.862504"
(hours:minutes:seconds:milliseconds)
Whenever I try to use Ruby's strptime to parse this string, it throws an ArgumentError: invalid strptime format - '%H:%M:%S'
I've actually searched quite extensively and cannot figure out an elegant way to parse this (besides the rather clunky solution of splitting the string by its colons and periods, and doing it all manually). Is there a way of doing this that I'm overlooking?
EDIT: I'm not looking to get a timestamp out of this, I'm looking to get a time duration.
What is your expected output? '136' is not a valid hour, and since you don't have a date portion, we can't simply turn those 'extra' hours into days. If you don't care about the date portion, this solution may work for you:
time = "136:16:11.862504"
hours, minutes, seconds = time.split(":").map(&:to_f)
hours %= 24
minutes %= 60
seconds %= 60
Time.new(0, 1, 1, hours, minutes, seconds, 0)
=> 0000-01-01 16:16:11 +0000
In case if nothing blocks you from using Regexp, you could use something based on this answer, for example:
/^(\d+):([0-5][0-9]):([0-5][0-9])\.\d+$/ =~ "136:16:11.862504"
puts "#{$1} : #{$2} : #{$3}"
136 : 16 : 11
I'm trying to parse a full ISO8601 datetime from JSON data in Lua.
I'm having trouble with the match pattern.
So far, this is what I have:
-- Example datetime string 2011-10-25T00:29:55.503-04:00
local datetime = "2011-10-25T00:29:55.503-04:00"
local pattern = "(%d+)%-(%d+)%-(%d+)T(%d+):(%d+):(%d+)%.(%d+)"
local xyear, xmonth, xday, xhour, xminute,
xseconds, xmillies, xoffset = datetime:match(pattern)
local convertedTimestamp = os.time({year = xyear, month = xmonth,
day = xday, hour = xhour, min = xminute, sec = xseconds})
I'm stuck at how to deal with the timezone on the pattern because there is no logical or that will handle the - or + or none.
Although I know lua doesn't support the timezone in the os.time function, at least I would know how it needed to be adjusted.
I've considered stripping off everything after the "." (milliseconds and timezone), but then i really wouldn't have a valid datetime. Milliseconds is not all that important and i wouldn't mind losing it, but the timezone changes things.
Note: Somebody may have some much better code for doing this and I'm not married to it, I just need to get something useful out of the datetime string :)
The full ISO 8601 format can't be done with a single pattern match. There is too much variation.
Some examples from the wikipedia page:
There is a "compressed" format that doesn't separate numbers: YYYYMMDD vs YYYY-MM-DD
The day can be omited: YYYY-MM-DD and YYYY-MM are both valid dates
The ordinal date is also valid: YYYY-DDD, where DDD is the day of the year (1-365/6)
When representing the time, the minutes and seconds can be ommited: hh:mm:ss, hh:mm and hh are all valid times
Moreover, time also has a compressed version: hhmmss, hhmm
And on top of that, time accepts fractions, using both the dot or the comma to denote fractions of the lower time element in the time section. 14:30,5, 1430,5, 14:30.5, or 1430.5 all represent 14 hours, 30 seconds and a half.
Finally, the timezone section is optional. When present, it can be either the letter Z, ±hh:mm, ±hh or ±hhmm.
So, there are lots of possible exceptions to take into account, if you are going to parse according to the full spec. In that case, your initial code might look like this:
function parseDateTime(str)
local Y,M,D = parseDate(str)
local h,m,s = parseTime(str)
local oh,om = parseOffset(str)
return os.time({year=Y, month=M, day=D, hour=(h+oh), min=(m+om), sec=s})
end
And then you would have to create parseDate, parseTime and parseOffset. The later should return the time offsets from UTC, while the first two would have to take into account things like compressed formats, time fractions, comma or dot separators, and the like.
parseDate will likely use the "^" character at the beginning of its pattern matches, since the date has to be at the beginning of the string. parseTime's patterns will likely start with "T". And parseOffset's will end with "$", since the time offsets, when they exist, are at the end.
A "full ISO" parseOffset function might look similar to this:
function parseOffset(str)
if str:sub(-1)=="Z" then return 0,0 end -- ends with Z, Zulu time
-- matches ±hh:mm, ±hhmm or ±hh; else returns nils
local sign, oh, om = str:match("([-+])(%d%d):?(%d?%d?)$")
sign, oh, om = sign or "+", oh or "00", om or "00"
return tonumber(sign .. oh), tonumber(sign .. om)
end
By the way, I'm assuming that your computer is working in UTC time. If that's not the case, you will have to include an additional offset on your hours/minutes to account for that.
function parseDateTime(str)
local Y,M,D = parseDate(str)
local h,m,s = parseTime(str)
local oh,om = parseOffset(str)
local loh,lom = getLocalUTCOffset()
return os.time({year=Y, month=M, day=D, hour=(h+oh-loh), min=(m+om-lom), sec=s})
end
To get your local offset you might want to look at http://lua-users.org/wiki/TimeZone .
I hope this helps. Regards!
There is also the luadate package, which supports iso8601. (You probably want the patched version)
Here is a simple parseDate function for ISO dates. Note that I'm using "now" as a fallback. This may or may not work for you. YMMV 😉.
--[[
Parse date given in any of supported forms.
Note! For unrecognised format will return now.
#param str ISO date. Formats:
Y-m-d
Y-m -- this will assume January
Y -- this will assume 1st January
]]
function parseDate(str)
local y, m, d = str:match("(%d%d%d%d)-?(%d?%d?)-?(%d?%d?)$")
-- fallback to now
if y == nil then
return os.time()
end
-- defaults
if m == '' then
m = 1
end
if d == '' then
d = 1
end
-- create time
return os.time{year=y, month=m, day=d, hour=0}
end
--[[
--Tests:
print( os.date( "%Y-%m-%d", parseDate("2019-12-28") ) )
print( os.date( "%Y-%m-%d", parseDate("2019-12") ) )
print( os.date( "%Y-%m-%d", parseDate("2019") ) )
]]