How to parse timezone with colon

Question

Is there a way to parse timezone in \"+00:00\" format with datetime.strptime? For instance:

Python 3.4.3 (v3.4.3:9b73f1c3e601, Feb 24 2015, 22:4         


        

Answer 1

Adapted from a rejected edit:

Python >= 3.7:

from datetime import datetime
d = "2019-12-25T23:59:59+00:00"
print(datetime.strptime(d, "%Y-%m-%dT%H:%M:%S%z"))

Changed in version 3.7: When the %z directive is provided to the strptime() method, the UTC offsets can have a colon as a separator between hours, minutes and seconds. For example, '+01:00:00' will be parsed as an offset of one hour. In addition, providing 'Z' is identical to '+00:00'.

https://docs.python.org/3.7/library/datetime.html

Python <= 3.6

There is no built-in way, but the best solution is:

from datetime import datetime
d = "2019-12-25T23:59:59+00:00"
if ":" == d[-3]:
    d = d[:-3]+d[-2:]
print(datetime.strptime(d, "%Y-%m-%dT%H:%M:%S%z"))

Explanations: https://bugs.python.org/issue15873 https://bugs.python.org/msg169952

Answer 2

Another solution is to use the dateutil library:

from dateutil import parser

mystr = '2015-04-30T23:59:59+00:00'

x = parser.parse(mystr)

# 2015-04-30 23:59:59+00:00

Answer 3

Currently, there is no cure for this, and here is and explanation: https://bugs.python.org/issue15873 more precisely, here: https://bugs.python.org/msg169952 . But you can override this issue, this way:

from datetime import datetime
d = "2015-04-30T23:59:59+00:00"
if ":" == d[-3:-2]:
    d = d[:-3]+d[-2:]
print(datetime.strptime(d, "%Y-%m-%dT%H:%M:%S%z"))