[Answer]-Would DateTimeField() work if I have time in this format 1/7/11 9:15 ? If not what would?

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“What would be the best variable type/format to define in order to accept this date as it is?”

The DateTimeField.

“If not what would be the most efficient way to accomplish this task?”

You should use the datetime.strptime method from Python’s builtin datetime library:

>>> from datetime import datetime
>>> import json

>>> json_datetime = "1/7/11 9:15"  # still encoded as JSON
>>> py_datetime = json.loads(json_datetime)  # now decoded to a Python string
>>> datetime.strptime(py_datetime, "%m/%d/%y %I:%M")  # coerced into a datetime object
datetime.datetime(2011, 1, 7, 9, 15)

# Now you can save this object to a DateTimeField in a Django model.
👤Brian Dant

0👍

If you take a look at https://docs.djangoproject.com/en/dev/ref/models/fields/#datetimefield, it says that django uses the python datetime library which is docomented at http://docs.python.org/2/library/datetime.html.

Here is a working example (with many debug prints and step-by-step instructions:

from datetime import datetime

json_datetime = "1/7/11 9:15"
json_date, json_time = json_datetime.split(" ")
print json_date
print json_time
day, month, year = map(int, json_date.split("/")) #maps each string in stringlist resulting from split to an int
year = 2000 + year #be ceareful here! 2 digits for a year may cause trouble!!! (could be 1911 as well)
hours, minutes = map(int, json_time.split(":"))
print day
print month
print year
my_datetime = datetime(year, month, day, hours, minutes)
print my_datetime

#Generate a json date:
new_json_style = "{0}/{1}/{2} {3}:{4}".format(my_datetime.day, my_datetime.month, my_datetime.year, my_datetime.hour, my_datetime.minute)
print new_json_style
👤OBu

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