How to change datetime.datetime(2012, 1, 1, 0, 0) to 1/1/2012 in Python?

» python » How to change datetime.datetime(2012, 1, 1, 0, 0) to 1/1/2012 in Python?

By : user3042504

Date : November 28 2020, 12:01 PM

I wish this helpful for you To remove the leading zeros on the month and day
For those that use windows:

code :

from datetime import datetime
datetime(2012, 1, 1, 0, 0).strftime("#%m/#%d/%Y")

from datetime import datetime
datetime(2012, 1, 1, 0, 0).strftime("-%m/-%d/%Y")

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How to handle a `datetime` value so to change it from `2012-04-27 00:00:00` to `27 april 2012`?

By : Bo Fo

Date : March 29 2020, 07:55 AM

help you fix your problem I am using Ruby on Rails v3.2.2 and I would like to display a data as-like . That is, I have a datetime database table column and I would like to change its contained values, for example, from 2012-04-27 00:00:00 to 27 april 2012. , The standard way is use rails localization.
model

code :

I18n.l(datetime, :format => :short)

I18n.l(datetime, :format => :long)

<%= l(datetime), :format => :custom %>

en:
  time:
    formats:
      long: "%A %B %d %Y | %I:%M %p GMT"
      short: "%d %B %Y, %H:%M GMT"
      custom: "your custom format"
  date:
    formats:
      short: "%d %B %Y"
      long: "%A %B %d %Y"
      custom: "your custom format"

how to switch strings from 14/Nov/2012 into 2012/11/14 without converting to datetime in Python?

By : MAxx

Date : March 29 2020, 07:55 AM

it should still fix some issue Maybe an odd question, but here we go: , This code would work:

code :

# Converts something like "14/Nov/2012:09:32:46"
# to "2012-11-14:09:32:46"
_MONTHS = ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]
def convert(s):
  p = s.split(':', 1)
  t = p[0].split('/')
  return "%s-%02d-%02d:" % (t[2], _MONTHS.index(t[1]) + 1, int(t[0])) + p[1]

Match time, Concatenate to datetime, and store as datetime column in SQL Server 2012

By : Jonas King-Holzsager

Date : March 29 2020, 07:55 AM

Does that help I assume the calculated FirstTimeReported should be in the past, derived from the DateEntered. This assumes of course that the row is entered within 24 hours of the ReportTime.

code :

WITH
    SupportContactsFirstTimeReported AS (
        SELECT 
              SupportID
            , DateEntered
            , DATEADD(minute, CAST(RIGHT(ReportedTime, 2) AS int), DATEADD(hour, CAST(LEFT(ReportedTime, 2) AS int), CAST(CAST(DateEntered AS date) AS datetime))) AS CalculatedReportedTime
        FROM supportContacts
    )
SELECT
      SupportID
    , CASE 
        WHEN CalculatedReportedTime <= DateEntered THEN CalculatedReportedTime
        ELSE DATEADD(day, -1, CalculatedReportedTime) 
      END AS FirstTimeReported
FROM SupportContactsFirstTimeReported;

How to convert timestamp (datetime.datetime(2012, 1, 1, 1, 0) into 2012-01-01 01:00:00

By : Josh Parks

Date : March 29 2020, 07:55 AM

will help you You have named your list datetime which overrides the datetime module. So datetime.strptime call fails.
Give a different name to your list.

AWS Lmbda TypeError: datetime.datetime(2012, 8, 8, 21, 46, 24, 862000) is not JSON serializable

By : Sulistyo Chandrianto

Date : March 29 2020, 07:55 AM

should help you out Haven't used AWS Lambda but this issue is python specific. datetime.datetime objects cannot be JSON serialized, you need to convert them to strings before JOSN serializing. You can use .strftime() to convert your datetime.datetime objects to strings.