Whats the correct format for django dateTime? - django

Im writting an app, and im using a datetime selector to allow users to select a date and time. After some formatting in javascript, im left with a dateTime like this:
2012-09-04 06:00 PM
and django throws and error saying:
[u"'2012-09-04 06:00 PM' value has an invalid format. It must be in YYYY-MM-DD HH:MM[:ss[.uuuuuu]][TZ] format."]
Ive also tried
2012-09-04 06:00PM
2012-09-04 06:00 P.M.
2012-09-04 06:00P.M.
but no luck. Can anyone see what Im doing here? Or know what django is looking for (in English)?


As said by the error msg, it expects YYYY-MM-DD HH:MM[:ss[.uuuuuu]][TZ], thus following are valid values:
2012-09-04 06:00
2012-09-04 06:00:00
2012-09-04 06:00:00.000000
# w/ optional TZ as timezone.
2012-09-04 06:00Z # utc
2012-09-04 06:00:00+0800
2012-09-04 06:00:00.000000-08:00


After talking around, what I figured out is that there is no AM/PM input. I had to just check if my dateTimePicker was returning an AM or PM suffix, then call time.split(":"); to break up hours and minutes, convert the hours to int from a string, then add 12 to it to convert the time to 24 hour time instead of 12 hour time. Theirs probably an easier way to do it, but thats what worked for me.
For example:
2012-09-04 06:00 PM
Needs to be
2012-09-04 18:00


I used this'2012-09-04 06:00:00.000000-08:00' and it worked perfectly

Related

Get datetime using qmake

We can get datetime using qmake _DATE_ which outputs
Sat Mar 12 17:29:00 2022
Can we format this output?

By using QDateTime by passing the value of DATE (in fromString), you may be able to format the output the way you want, by using toString.
You can try things like this :
toString("ddd MMMM d hh:mm:ss yy"); //Sat Mar 12 17:29:00 2022

Define and convert datetime in AWS Athena

I have a process where I need to to match UTC datetime and EDT datetime.
As you know, EDT can be changed between 4/5 hours from UTC.
How can I define one datetime to be in UTC and another to be in EDT and match the two?
Something like (datetime_A is my EDT timestamp, and datetime_B is my UTC):
Where CAST((datetime_A as EDT) to UTC)=datetime_B
Thanks!

Redshift - Adding timezone offset (Varchar) to timestamp column

as part of ETL to Redshift, in one of the source tables, there are 2 columns:
original_timestamp - TIMESTAMP: which is the local time when the record was inserted in whichever region
original_timezone_offset - Varchar: which is the offset to UTC
The data looks something like this:
original_timestamp original_timezone_offset
2011-06-22 11:00:00.000000 -0700
2014-11-29 17:00:00.000000 -0800
2014-12-02 22:00:00.000000 +0900
2011-06-03 09:23:00.000000 -0700
2011-07-28 03:00:00.000000 -0700
2011-05-01 01:30:00.000000 -0700
In my target table, I need to convert this to UTC (using the offset). How do I do it?
So far I have tried multiple things but dateadd() seems to be the closest solution. But the problem with dateadd() is, when I say:
SELECT original_timestamp, original_timezone_offset
,dateadd(H, original_timezone_offset, original_timestamp) as original_utc_time
it is adding/subtracting '700'/'800' hours instead of 7/8 hrs to the original timestamp because the offset is a VARCHAR and the values are like: -0700 etc.
Did anyone see this issue before? Appreciate any help/inputs. Thanks.

Just take the 'hours' part of the offset:
WITH t as (
SELECT '2011-06-22 11:00:00.000000'::timestamp as original_timestamp, '-0700' as original_timezone_offset
UNION ALL
SELECT '2014-11-29 17:00:00.000000'::timestamp,'-0800'
UNION ALL
SELECT '2014-12-02 22:00:00.000000'::timestamp,'+0900'
)
SELECT
original_timestamp,
original_timezone_offset,
DATEADD(hour, SUBSTRING(original_timezone_offset, 1, 3)::INT, original_timestamp)
FROM t
2011-06-22 11:00:00 -0700 2011-06-22 04:00:00
2014-11-29 17:00:00 -0800 2014-11-29 09:00:00
2014-12-02 22:00:00 +0900 2014-12-03 07:00:00
You'll need some additional fancy code if you have non-full-hour offsets (eg +0730).

First, recognize that if your timestamps are already in local time of the given offset, then you need to subtract that offset to convert back to UTC. In that first example you gave, 2011-06-22 11:00:00 -0700 is equivalent to 2011-06-22 18:00:00 UTC.
However, rather than try to add or subtract these values yourself, you should let the AT TIME ZONE function do the work for you. It will create a timestamptz that is in your supplied offset, then you can use it again to convert to UTC.
(Note that you could use the CONVERT_TIMEZONE function instead, but that one is only understood by Redshift, where AT TIME ZONE works on regular PostgreSQL also.)
However, you have is that the time zone offsets you have aren't in a format understood by these functions. See time zone usage notes. So, before we try to convert, let's translate your offset strings to an understood format.
We will want -0700 to become +07:00. The colon is required, and the sign must be flipped because it will be interpreted with the POSIX-style time zone format. In that format, positive values lie west of GMT instead of the usual conventions specified in ISO 8601.
concat(translate(substring(original_timezone_offset, 1, 3), '-+', '+-'),':',substring(original_timezone_offset, 4, 2))
Then we will use that with AT TIME ZONE to do the conversion:
(original_timezone AT TIME ZONE <the above mess>) AT TIME ZONE 'UTC' AS utc_timestamp
Putting it all together...
WITH t as (
SELECT '2011-06-22 11:00:00.000000'::timestamp as original_timestamp, '-0700' as original_timezone_offset
UNION ALL
SELECT '2014-11-29 17:00:00.000000'::timestamp,'-0800'
UNION ALL
SELECT '2014-12-02 22:00:00.000000'::timestamp,'+0900'
)
SELECT
original_timestamp,
original_timezone_offset,
concat(translate(substring(original_timezone_offset, 1, 3), '-+', '+-'),':',substring(original_timezone_offset, 4, 2)) as modified_timezone_offset,
(original_timestamp AT TIME ZONE concat(translate(substring(original_timezone_offset, 1, 3), '-+', '+-'),':',substring(original_timezone_offset, 4, 2))) AT TIME ZONE 'UTC' AS utc_timestamptz
FROM t
Output:
2011-06-22 11:00:00 -0700 +07:00 2011-06-22 18:00:00
2014-11-29 17:00:00 -0800 +08:00 2014-11-30 01:00:00
2014-12-02 22:00:00 +0900 -09:00 2014-12-02 13:00:00
SQL Fiddle here.

Regex - Slice Date - Aug 22, 2017 02:00 PM EDT

I'm trying to take a date, for example Aug 22, 2017 02:00 PM EDT
and get the month, day, year from it.
month = re.findall(r'', date)[0]
day = re.findall(r'', date)[0]
year = re.findall(r'', date)[0]
I've started with something like this:
(.*)(?<=[a-zA-Z]{3}\s)
for the month. Is there a better way to do this?

You need to first convert to datetime and then extract the needed values like this (reusing the example):
from datetime import datetime
datetime_object = datetime.strptime('Jun 1 2005 1:33PM', '%b %d %Y %I:%M%p')
print(datetime_object.year, datetime_object.month, datetime_object.day)
From what I can see you probably won't need to specify the format but pass the string directly to the datetime.strptime function.

RubyMotion Day Number of year for today

I want to generate the current day number using RubyMotion code. I have looked at several IOS solutions but I'm not experienced enough to translate the code successfully to RubyMotion.
I am currently at the following point:
def today
NSDate.today
end
def day_number
NSDate.from_components (day: today)
end
When I run the above it gives me an return of 3852055-06-16 00:00:00 +0100. I thought that the 3852055 part was seconds but it doesn't seem to equate to either todays date or to 16th of June - and in any case why should it be returning 06-16 instead of 02-08?? Totally confused here.
I just want to get todays day number. As I write the date is 2nd August 2014 and the day number should be 214 so I'm obviously way out somewhere.
Any help would be greatly appreciated.
cheers

This will do the trick:
daynum = NSCalendar.currentCalendar.ordinalityOfUnit(NSDayCalendarUnit, inUnit:NSYearCalendarUnit, forDate:NSDate.date)
Now suppose you have the date in the form of a string, and you want to get the day number for it:
datestr = "2014-01-01 11:08:56 +0000"
First create an NSDateFormatter to convert the String into an NSDate
df = NSDateFormatter.new
df.dateFormat = "yyyy-MM-dd HH:mm:ss z"
mydate = df.dateFromString(datestr)
daynum = NSCalendar.currentCalendar.ordinalityOfUnit(NSDayCalendarUnit, inUnit:NSYearCalendarUnit, forDate:mydate)
If your date string is simpler:
datestr = "2014-01-01"
just use a simpler dateFormat string:
df.dateFormat = "yyyy-MM-dd"

I suggest you to take a look at motion-support gem and especially at core-exttime.
You can play with dates as you want:
=> Mon, 04 Aug 2014
(main)> Time
=> Time
(main)> Time.today
=> 2014-08-04 00:00:00 +0200
(main)> Date.today.day
=> 4
(main)> Time.today.day
=> 4
and a lot more.