I have one date-time in Ecto.DateTime and the 2nd one in DateTime. How can I convert them to each other? Isn't there a easy way without external dependencies? There's nothing in the documentation. One of them has to_erl, another from_unix, but there's no overlap in methods, such as to_unix/from_unix or to_erl/from_erl or something similar.
The equivalent of Ecto.DateTime is NaiveDateTime, since neither of them store a timezone, while DateTime does. Erlang datetimes also do not have a timezone, which is why there's no to_erl and from_erl in DateTime.
You can first convert to NaiveDateTime and then use DateTime.from_naive/2 along with the timezone your datetime is in (Elixir only supports Etc/UTC as of Elixir 1.4):
iex(1)> Ecto.DateTime.utc |> Ecto.DateTime.to_erl |> NaiveDateTime.from_erl! |> DateTime.from_naive!("Etc/UTC")
%DateTime{calendar: Calendar.ISO, day: 8, hour: 4, microsecond: {0, 0},
minute: 49, month: 2, second: 9, std_offset: 0, time_zone: "Etc/UTC",
utc_offset: 0, year: 2017, zone_abbr: "UTC"}
iex(2)> DateTime.utc_now |> DateTime.to_naive |> NaiveDateTime.to_erl |> Ecto.DateTime.from_erl
#Ecto.DateTime<2017-02-08 04:50:23>
If you were using Ecto.DateTime earlier though, you probably want to use NaiveDateTime now.
Related
Calling Timex.to_naive_datetime to convert to a naive datetime without the timezone subtracts an hour from the input datetime
I have tried other timezones and ruled out the possibility of daylight saving time conversion.
d = %DateTime{
year: 2000,
month: 2,
day: 29,
zone_abbr: "CET",
hour: 23,
minute: 0,
second: 7,
microsecond: {0, 0},
utc_offset: 3600,
std_offset: 0,
time_zone: "Europe/Warsaw"
}
#=> # DateTime<2000-02-29 23:00:07+01:00 CET Europe/Warsaw>
Timex.to_naive_datetime(d)
#=> ~N[2000-02-29 22:00:07]
d2 = %DateTime{
year: 2019,
month: 3,
day: 2,
zone_abbr: "PST",
hour: 23,
minute: 0,
second: 7,
microsecond: {0, 0},
utc_offset: 3600,
std_offset: 0,
time_zone: "America/Los_Angeles"
}
#=> #DateTime<2019-03-02 23:00:07+01:00 PST America/Los_Angeles>
Timex.to_naive_datetime(d2)
#=> ~N[2019-03-02 22:00:07]
I am expecting the first datetime to be converted to ~N[2000-02-29 23:00:07], but the output is ~N[2000-02-29 22:00:07].
Both of the structs you create contain utc_offset: 3600, which is UTC+1.
#DateTime<2000-02-29 23:00:07+01:00 CET Europe/Warsaw>
#DateTime<2019-03-02 23:00:07+01:00 PST America/Los_Angeles>
How did the +01:00 that get there for America/Los_Angeles? It's not valid for that timezone. If I generate the date using one of the standard functions:
d3 = Timex.to_datetime({{2019,3, 2}, {23, 0, 7}}, "America/Los_Angeles")
#=> #DateTime<2019-03-02 23:00:07-08:00 PST America/Los_Angeles>
Timex.to_naive_datetime(d3)
#=> ~N[2019-03-03 07:00:07]
I get the correct timezone offset. It seems Timex.to_naive_datetime/1 returns the UTC value of the timezone. If you just want to drop the timezone information, you can use DateTime.to_naive/1:
d1
#=> #DateTime<2000-02-29 23:00:07+01:00 CET Europe/Warsaw>
DateTime.to_naive(d1)
#=> ~N[2000-02-29 23:00:07]
How can I get the previous month without using a package or library in elixir?
For example, if the current date is 2018-01-25, I will get 2017-12-25.
Or If the current date is 2018-03-31, I will get 2018-02-28 (2018 is not a leap year)
The answer by #Sheharyar is almost there, the only difference you need to subtract the maximum of days in both months:
defmodule Dating do
def previous_month(%Date{day: day} = date) do
days = max(day, (Date.add(date, -day)).day)
Date.add(date, -days)
end
end
Works for all cases:
iex|1 ▶ Dating.previous_month(~D[2018-03-31])
#⇒ ~D[2018-02-28]
iex|2 ▶ Dating.previous_month(~D[2018-03-01])
#⇒ ~D[2018-02-01]
iex|3 ▶ Dating.previous_month(~D[2018-01-02])
#⇒ ~D[2017-12-02]
Use Timex library
iex(1)> ~D[2018-01-25] |> Timex.shift(months: -1)
~D[2017-12-25]
iex(2)> ~D[2018-03-31] |> Timex.shift(months: -1)
~D[2018-02-28]
I'm trying to format the Timex module to look a certain way. I'm trying to get today's date. but I want it formatted like this:
2017/12/12.
year/mn/day
In ruby I would go to the strftime class but I'm not sure how to do this with Elixir:
Current attempt:
Timex.local => #DateTime<2017-12-12 19:57:17.232916-05:00 EST America/Detroit>
How can I take that and format it how I specified?
Elixir 1.11 has Calendar.strftime/3 built-in for your strftime needs.
Calendar.strftime(~U[2019-08-26 13:52:06.0Z], "%y-%m-%d %I:%M:%S %p")
"19-08-26 01:52:06 PM"
Timex is a third-party library that was created in the era when Elixir had no good support for dates/times. Nowadays, there is DateTime native class in the core, so I am unsure why do you want to use Timex at all.
In any case, DateTime is a struct:
iex|1 ▶ today = DateTime.utc_now
#⇒ #DateTime<2017-12-13 07:22:58.290075Z>
iex|2 ▶ [today.year, today.month, today.day]
#⇒ [2017, 12, 13]
iex|3 ▶ Enum.join [today.year, today.month, today.day], "/"
#⇒ "2017/12/13"
To pad with leading zeroes for "2018/1/1":
iex|4 ▶ with {:ok, today} <- Date.new(2018, 1, 1) do
...|4 ▶ [today.year, today.month, today.day]
...|4 ▶ |> Enum.map(&to_string/1)
...|4 ▶ |> Enum.map(&String.pad_leading(&1, 2, "0"))
...|4 ▶ |> Enum.join("/")
...|4 ▶ end
#⇒ "2018/01/01"
If you want to do this without an external library, you can use io_lib:format/2 to pad the integers with zeroes where necessary like this:
iex(1)> date = Date.utc_today
~D[2017-12-13]
iex(2)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"2017/12/13"
iex(3)> {:ok, date} = Date.new(2018, 1, 1)
{:ok, ~D[2018-01-01]}
iex(4)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"2018/01/01"
iex(5)> {:ok, date} = Date.new(1, 1, 1)
{:ok, ~D[0001-01-01]}
iex(6)> :io_lib.format("~4..0B/~2..0B/~2..0B", [date.year, date.month, date.day]) |> IO.iodata_to_binary
"0001/01/01"
You can do this to add zeros
Timex.local |> Timex.format!("{YYYY}/0{M}/0{D}") => "2017/01/01"
So it appears the Timex Module has a format!/2 function which will return a string of what ever date you pass to it.
Here is what I came up with:
Timex.local |> Timex.format!("{YYYY}/{M}/{D}") => "2017/12/12"
The answer to add 0 padding is incorrect and will always pad 0s even if it does not require it. The correct way to pad 0s is as follows:
Timex.local |> Timex.format!("{YYYY}/{0M}/{0D}") => "2017/01/01"
As of Elixir 1.11 you can do this sanely with a date formatter in the standard library's Calendar module:
https://hexdocs.pm/elixir/1.13.0/Calendar.html#strftime/3
DateTime.utc_now()
|> Calendar.strftime("%Y/%m/%d")
"2022/01/12"
I'm running into an issue when I'm trying to create a histogram of specific createdAt datetimes for orders. The issue is that even after created timezone aware datetimes, the .weekday() shows up as the same day, even though it should be a different time
The code I'm using to test this occurrence is as follows:
import datetime
import pytz
value = {
'createdAt': '2017-04-24T00:48:03+00:00'
}
created_at = datetime.datetime.strptime(value['createdAt'], '%Y-%m-%dT%H:%M:%S+00:00')
timezone = pytz.timezone('America/Los_Angeles')
created_at_naive = created_at
created_at_aware = timezone.localize(created_at_naive)
print(created_at_naive) # 2017-04-24 00:48:03
print(created_at_aware) # 2017-04-24 00:48:03-07:00
print(created_at_naive.weekday()) # 0 (Monday)
print(created_at_aware.weekday()) # 0 (should be Sunday)
The problem is that you need to actually change the datetime to the new timezone:
>>> timezone('UTC').localize(created_at)
datetime.datetime(2017, 4, 24, 0, 48, 3, tzinfo=<UTC>)
>>>timezone('UTC').localize(created_at).astimezone(timezone('America/Los_Angeles'))
datetime.datetime(2017, 4, 23, 17, 48, 3, tzinfo=<DstTzInfo 'America/Los_Angeles' PDT-1 day, 17:00:00 DST>)
The month format specifier doesn't seem to work.
from datetime import datetime
endDate = datetime.strptime('10 3 2011', '%j %m %Y')
print endDate
2011-01-10 00:00:00
endDate = datetime.strptime('21 5 1987', '%j %m %Y')
print endDate
1987-01-21 00:00:00
Now, according to the manual the manual:
%m = Month as a decimal number [01,12].
So, what am I missing, other than the hair I've pulled out trying to understand why my django __filter queries return nothing (the dates going in aren't valid!)? I've tried 03 and 05 to no avail.
Versions of things, platform, architecture et al:
$ python --version
Python 2.7
$ python3 --version
Python 3.1.2
$ uname -r
2.6.35.11-83.fc14.x86_64 (that's Linux/Fedora 14/64-bit).
You can't mix the %j with others format code like %m because if you look in the table that you linked %j is the Day of the year as a decimal number [001,366] so 10 correspondent to the 10 day of the year so it's 01 of January ...
So you have just to write :
>>> datetime.strptime('10 2011', '%j %Y')
datetime.datetime(2011, 1, 10, 0, 0)
Else if you you wanted to use 10 as the day of the mount you should do :
>>> datetime.strptime('10 3 2011', '%d %m %Y')
datetime.datetime(2011, 3, 10, 0, 0)
Isn't %j the "day of year" parser, which may be forcing strptime to choose January 21, overriding the %m rule?
%j specifies a day of the year. It's impossible for the 10th day of the year, January 10, to occur in March, so your month specification is being ignored. Garbage In, Garbage Out.