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How to limit the input in Python for only dates?

Im currently working on my project.
I want the user to only be allowed to input a date (ex, January 2). If he enters anything else than a date a message should appear like "This is not a date, try again" repeatedly until a real date is given. How do i do this?
My initial idea was to create a .txt file were i write all the 365 dates and then somehow code that the user is only allowed to enter a string that matches one of the elements in the file, else try again.
I would really apreciate your help

Use dateutil.parser to handle dates of arbitrary formats.
Code
import dateutil.parser
def valid_date(date_string):
try:
date = dateutil.parser.parse(date_string)
return True
except ValueError:
return False
Test
for date in ['Somestring', 'Feb 20, 2021', 'Feb 20', 'Feb 30, 2021', 'January 25, 2011', '1/15/2020']:
print(f'Valid date {date}: {valid_date(date)}')
Output
Valid date Somestring: False # detects non-date strings
Valid date Feb 20, 2021: True
Valid date Feb 20: True
Valid date Feb 30, 2021: False # Recognizes Feb 30 as invalid
Valid date January 25, 2011: True
Valid date 1/15/2020: True # Handles different formats

There is no need to store all possible valid dates in a file.
Use datetime.strptime() to parse a string (entered by the user) into a datetime object according to a specific format.
strptime will raise an exception the input specified does not adhere to the pattern, so you can catch that exception and tell the user to try again.
Wrap it all in a while loop to make it work forever, until the user gets it right.
You can start with this:
from datetime import datetime
pattern = '%B %d, %Y' # e.g. January 2, 2021
inp = ''
date = None
while date is None:
inp = input('Please enter a date: ')
try:
date = datetime.strptime(inp, pattern)
break
except ValueError:
print(f'"{inp}" is not a valid date.')
continue
For a full list of the %-codes that strptime supports, check out the Python docs.

Provide you with several ways to verify the date, these are just simple implementations, and there is no strict check, you can choose one of the methods and then supplement the detailed check by yourself.
Use date(year,month,day)
def isValidDate(year, month, day):
try:
date(year, month, day)
except:
return False
else:
return True
Use date.fromisoformat()
def isValidDate(datestr):
try:
date.fromisoformat(datestr)
except:
return False
else:
return True
Use strptime
def check_date(i):
valids = ['%Y-%m-%d', '%Y%M']
for valid in valids
try:
return strptime(i, valid)
except ValueError as e:
pass
return False
Use regex
def check_date(str):
reg = /^(\d{4})-(\d{2})-(\d{2})$/;
return reg.test(str)

How can I adjust 'the time' in python with module Re

this is a funny question.
I try to find out the right time in some phrases.
I use try-except module and re module
but there is something wrong in my code that can't deal with some tough phrase
As is depicted belong, I input the rediculous time 1997-25-52 or 1996-42-120
it still can output an answer.
def regular_time(time):
"""
部分电影日期带有国家, 例如：'1994-09-10(加拿大)'
正则提取日期
"""
import re
pattern = '^(([1-2]\d{3})-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]))'
try:
matches = re.match(pattern, time, flags=0).group()
return matches
except Exception as e:
try:
pattern = '^(([1-2]\d{3})-(0[1-9]|1[0-2]))'
matches = re.match(pattern, time, flags=0).group()+'-01'
return matches
except:
try:
pattern = '^(([1-2]\d{3}))'
matches = re.match(pattern, time, flags=0).group() + '-01-01'
return matches
except:
print('errors')
time='1996-12-58'
regular_time(time)
How can I deal with this problem? Many thanks if you could do me a favor

Question: Default date from invalid datestring
Using datetime handles also leap years!
datetime.datetime.strptime
datetime.date.strftime
For example:
import re
from datetime import datetime
def regular_time(time):
_t = time.split('-')
# allways 3 itmes
while len(_t) < 3:
_t.append('01')
# year month and day ranges
ymd = [(range(1900, 2099), '1900'),
(range(1, 13), '01'),
(range(1, 32), '01')
]
# validate ranges
for n in range(3):
if not int(_t[n]) in ymd[n][0]:
_t[n] = ymd[n][1]
_time = '-'.join(_t)
try:
date = datetime.strptime(_time, '%Y-%m-%d')
print('VALID:{} => {}'
.format(time, date.strftime('%Y-%m-%d')))
except ValueError as e:
if "day is out of range for month" in e:
print('{} for {}, change to 01'.format(e, time))
_t[2] = '01'
regular_time('-'.join(_t))
else:
print('INVALID[{}]:{}'.format(_time, e))
for time in ['1996', '1996-18', '2019-09-31', '2019-01-31',
'1996-12-58', '1997-25-52', '1996-42-120']:
regular_time(time)
Output:
VALID:1996 => 1996-01-01
VALID:1996-18 => 1996-01-01
day is out of range for month for 2019-09-31, change to 01
VALID:2019-09-01 => 2019-09-01
VALID:2019-01-31 => 2019-01-31
VALID:1996-12-58 => 1996-12-01
VALID:1997-25-52 => 1997-01-01
VALID:1996-42-120 => 1996-01-01
Tested with Python 3.6

Your test case returns "1996-12-01", which is that it hits second-level "try-except", since it matches pattern of correct year and month (first failed cause day is unrealistic), then it just simplifies it to the first day of the month by adding "-01".
If you want to keep all parts of the date realistic - don't overwrite original "pattern". But fail it in first step.

How to convert a downloaded string to datetime format?

I am trying to check if today's date < date downloaded from text file online. Here is my code :
import datetime
import requests
URL = "http://directlinktotextfile.com/text.txt"
result = requests.get(URL)
today = datetime.datetime.now().date()
Url_date = result.text
Url_date.strip()
Url_date = datetime.date(Url_date)
if today < Url_date :
print "Today is less than future date"
raw_input()
else:
print "Today is greater than or = to future date"
raw_input()
The result that comes back is just this : 2018,02,14. I use .strip() in case there might be blank spaces or extra lines. I've printed out result.text after strip() and it shows the correct details. Why is it that I can't check if today < Url_date. It works fine if I enter manually a date into datetime.date(2018,02,14), but when I'm downloading the string it won't work. Any suggestions?

You pass string to datetime.date() which should be each an integer.
Url_list = []
Url_list = Url_date.split(",")
yr = int(Url_list[0])
mn = int(Url_list[1])
d = int(Url_list[2])
Now pass these integers to datetime.date
Url_date = datetime.date(yr, mn, d)

The arguments you pass to datetime.date(arg1, arg2, arg3) are not strings as a whole. When you pass it from url, what you are actually doing is
datetime.date("2018,2,14")
Note that you are passing only one string argument and not 3 different integers. You should split the date string using comma and then convert each into integers and then pass them as arguments to datetime.date.

Here is what your code is trying to do :
Url_date = datetime.date("2018,02,14")
But he wants to have:
Url_date = datetime.date(2018,02,14)
Do
Url_date.split(',') # Result: ['2018','02','14']
And then convert all the string in the array in integers
It should be ok :)

Use strptime:
import datetime
today = datetime.datetime.now().date()
parsed = datetime.datetime.strptime("2018,02,14", "%Y,%m,%d").date()
print(today < parsed) # True

How to translate datetime string to sql format?

Is there any shorter method to translate this ISO 8601 compatible UTC time to the SQL DATETIME format?
>>> str = "2016-03-28T20:23:46+0800"
>>> temp = str.split('T')
>>> temp[1] = temp[1].rstrip('+')
>>> temp[1]
'20:23:46+0800'
>>> temp[1] = temp[1].split('+')[0]
>>> result = " ".join(temp)
>>> result
'2016-03-28 20:23:46'
Thanks!

You can simply switch formats:
>>> date_str = "2016-03-28T20:23:46+0800"
>>> format = '%Y-%m-%dT%H:%M:%S%z'
>>> new_format = '%Y-%m-%d %H:%M:%S'
>>> datetime.datetime.strptime(date_str, format).strftime(new_format)
'2016-03-28 20:23:46'
This will not work in python 2.x as it does not support the %z flag. See timezone python 2 support for a workaround

There is no easy way to do this.
Checkout this post for more details on possible solutions.
If you're looking for a quick hack try this:
st = '2016-03-28T20:23:46+0800'
st[:19].replace("T", " ")
Or, if you need the date in datetime:
datetime.datetime.strptime(st[:19], '%Y-%m-%dT%H:%M:%S')

python convert string to datetime

i have a loop where i try to process set of data where one action is to convert ordinary string to datetime. everything works fine except sometimes happend a weird thing ... here is what i know
there are exactly the same parameters entering the function always
those parameters are the same type always
first time i run it, it always get trought
when it gets to second element in the loop in appx 80% throw and value error (time data did not match format)
but after i run it again, everything is ok, and it gets stuck on next emelement ...
because my function is pretty big and there are many things happing i decide to provide you with some saple code whitch i wrote right here, just for clarification:
data = ['January 20 1999', 'March 4 2010', 'June 11 1819']
dformat = '%B %d %Y'
for item in data:
out = datetime.datetime.strptime(item, dformat)
print out
although this clearly works in my program it doesnt ... i have try everything i have came up with but havent succeeded yet therefore i would be glad with any idea you provide thanks
btw: the error i always get looks like this
ValueError: time data did not match format: data=March 4 2010 fmt=%B %d %Y

You probably have a different locale set up. %B is March in locales that use English, but in other locales it will fail.
For example:
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'sv_SE.utf8')
'sv_SE.utf8'
>>> import datetime
>>>
>>> data = ['January 20 1999', 'March 4 2010', 'June 11 1819']
>>> for item in data:
... print datetime.datetime.strptime(item, '%B %d %Y')
...
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "/usr/lib/python2.6/_strptime.py", line 325, in _strptime
(data_string, format))
ValueError: time data 'January 20 1999' does not match format '%B %d %Y'
Here you see that even though the format does match, it claims it doesn't. And that's because the month names doesn't match. Change it to Swedish locale names, and it works again:
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'sv_SE.utf8')
'sv_SE.utf8'
>>> import datetime
>>>
>>> data = ['Januari 20 1999', 'Mars 4 2010', 'Juni 11 1819']
>>> for item in data:
... print datetime.datetime.strptime(item, '%B %d %Y')
...
1999-01-20 00:00:00
2010-03-04 00:00:00
1819-06-11 00:00:00
(Note that the above locale 'sv_SE.utf8' might not work for you, because you have to have that specific locale installed. To see which ones that are installed on a Unix machine, run this command from the command line:
$ locale -a
C
en_AG
en_AU.utf8
en_BW.utf8
en_CA.utf8
en_DK.utf8
en_GB.utf8
en_HK.utf8
en_IE.utf8
en_IN
en_NG
en_NZ.utf8
en_PH.utf8
en_SG.utf8
en_US.utf8
en_ZA.utf8
en_ZW.utf8
POSIX
sv_FI.utf8
sv_SE.utf8
)

Pretty weird though... In the same run locale usually doesn't change. However, if your program keeps doing this, you might want to call 'setlocale' everytime the code enters into the loop (ugly solution, I know).