Date = datetime.datetime.today().strftime("%d %B %Y")
x = datetime.datetime.strptime(Date , "%d %B %Y")
returns:
2018-05-09 00:00:00
instead of:
9 May 2018, what am I doing wrong?
Essentially I am trying to get the non zero padded version of strftime for which I searched around and I read that the way to go about it is strptime.
A possible solution is to use str.lstrip
Ex:
import datetime
Date = datetime.datetime.today().strftime("%d %B %Y")
print(Date.lstrip("0"))
Output:
9 May 2018
Related
Using datetime.datetime.now(), I receive some badly formatted timestamps.
Is there an intuitive way of creating a date timestamp in this format?
Wed Aug 7 13:38:59 2019 -0500
This is seen in git log.
You can use datetime.datetime.strftime() to format dates as shown below:
from datetime import datetime
d = '2019-08-07 13:38:59-0500'
d2 = datetime.strptime(d, '%Y-%m-%d %H:%M:%S%z')
d3 = d2.strftime('%a %b %d %H:%M:%S %Y %z')
print(d3)
This returns:
Wed Aug 07 13:38:59 2019 -050000
This website is a great resource for strftime formatting.
You can still use the datetime library. Using strftime, you can rewrite the datetime object into a nicely formatted string.
In your case, you are going for Wed Aug 7 13:38:59 2019 -0500, which translated to strftime formatting is "%a %b %d %H:%M:%S %Y %z".
Overall, it'd be
datetime.datetime.now().strftime("%a %b %d %H:%M:%S %Y %z")
Which will give a string that looks like 'Wed Aug 7 13:38:59 2019 -0500'.
I would do the following:
from time import gmtime, strftime
if __name__ == "__main__":
time = strftime("%a, %d %b %Y %H:%M:%S +0000", gmtime())
print(time)
This was found on the documentation page of the time module. There are also a lot of additional features you might be interested in using outlined here: https://docs.python.org/3/library/time.html#time.strftime
I want to convert my date into DateTime object for MySQL.
My string format is: Mon Aug 27 04:47:45 +0000 2018
Expected Output: 'YYYY-M-D H:mm:ss'
from datetime import datetime
t = datetime.strptime('Mon Aug 27 04:47:45 +0000 2008', '%a %b %d %H:%M:%S % z %Y')
t.strftime('%Y-%m-%d %H:%M:%S')
Refer section 8.1.8
here
If you are using python 3, this solution would work -
from datetime import datetime
x = 'Mon Aug 27 04:47:45 +0000 2018'
x = datetime.strftime(datetime.strptime(x, '%a %b %d %I:%M:%S %z %Y'), '%Y-%m-%d %H:%M:%S')
# OP '2018-08-27 04:47:45'
But for python 2, you might get a ValueError: 'z' is a bad directive.... In that case, you'll either have to use something like pytz or dateutil. The table that you need to look for all these conversions can be found here
Edit: You can't have Expected Output: 'YYYY-M-D H:mm:ss' if you convert your datetime string to datetime object. Datetime object has it's own format. Above gives you a string of the format that you want
from datetime import datetime
date_as_dt_object = datetime.strptime(dt, '%a %b %d %H:%M:%S %z %Y')
You can use date_as_dt_object in a raw query or an ORM. If used in a raw query pass it as a string like:
query = "select * from table where date >" + str(date_as_dt_object)
Check out this list for Python's strftime directives.
http://strftime.org/
How do i convert this datetime using python?
2017-10-16T08:27:16+0000
I tried to use strptime but getting ValueError: time data '2017-10-16T08:27:16+0000' does not match format 'The %d %B %Y at. %H:%M'
import datetime
datetime.datetime.strptime("2017-10-16T08:27:16+0000", "The %d %B %Y at. %H:%M")
'''
I want my output to look like this
The 16 october 2017 at. 08:27
'''
First parse the string correctly, then print it in the desired format:
import datetime
date = datetime.datetime.strptime("2017-10-16T08:27:16+0000", "%Y-%m-%dT%H:%M:%S%z")
print(date.strftime("The %d %B %Y at. %H:%M"))
https://blogs.harvard.edu/rprasad/2011/09/21/python-string-to-a-datetime-object/
You have to first strip your date using strptime() and then rebuild it using strftime()
import datetime
time = "2017-10-16T08:27:16+0000"
stripedTime = datetime.datetime.strptime(time, '%Y-%m-%dT%I:%M:%S%z')
rebuildTime = stripedTime.strftime('The %d %B %Y at. %H:%M')
print(rebuildTime)
I have two timestamp strings. I want to find the difference between them in seconds.
I've tried:
from time import gmtime, strptime
a = "Mon 11 Dec 2017 13:54:36 -0700"
b = "Mon 11 Dec 2017 13:54:36 -0000"
time1 = strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = strptime(b, "%a %d %b %Y %H:%M:%S %z")
time1-time2
Getting an error: TypeError: unsupported operand type(s) for -: 'time.struct_time' and 'time.struct_time'
So, how do I calculate the difference using package time?
I was successful using package datetime - in the code below, but I think I read that datetime ignores seconds in leap years, or something to that effect. Thus, I am trying to use 'time':
from datetime import datetime
time1 = datetime.strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = datetime.strptime(b, "%a %d %b %Y %H:%M:%S %z")
dif = time1 - time2
print(int(dif.total_seconds()))
Thank you very much!
First of all, you're using time.strptime, which returns <class 'time.struct_time'>, and it doesn't support the substract operator, one possible way to achieve what you want would be converting to datetime:
from datetime import datetime
from time import mktime
from time import gmtime, strptime
a = "Mon 11 Dec 2017 13:54:36 -0700"
b = "Mon 11 Dec 2017 13:54:36 -0000"
time1 = strptime(a, "%a %d %b %Y %H:%M:%S %z")
time2 = strptime(b, "%a %d %b %Y %H:%M:%S %z")
print(datetime.fromtimestamp(mktime(time1))-datetime.fromtimestamp(mktime(time2)))
Or even better, use datetime.datetime.strptime so you don't need intermediate conversions.
For a more detailed description of the supported operations of datetime please refer to the section supported operations in the docs here. Especially the section where it says:
If both are aware and have different tzinfo attributes, a-b acts as if
a and b were first converted to naive UTC datetimes first. The result
is (a.replace(tzinfo=None) - a.utcoffset()) - (b.replace(tzinfo=None)
- b.utcoffset()) except that the implementation never overflows.
In any case, maybe your best chance is considering an alternative method like the one proposed in this answer
I have next time value in unicode (<type 'unicode'>):
2017-08-09T15:02:58+0000.
How to convert it to friendly view (e.g. Day, Month of Year)?
This should do what you ask:
from datetime import datetime
a = '2017-08-09T15:02:58+0000'
datetime.strptime(a[:-5], '%Y-%m-%dT%H:%M:%S').strftime('%d, %b of %Y')
#09, Aug of 2017
strptime method throws error for timezone parameter that doesn't seem to interest you so I removed that part with a[:-5].
For the rest of the string you can just follow guidelines from datetime docs.
Using the same docs you can construct your datetime string using strftime() method like you wanted '%d, %b of %Y' or in plain words [day], [abbreviated month] of [Year]
try this
import datetime
today = datetime.date.today()
print today.strftime('We are the %d, %b %Y')
'We are the 22, Nov 2008'