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How do I get rid of the year, month, and day? [duplicate]

This question already has answers here:
Compare only time part in datetime - Python
(3 answers)
Closed 1 year ago.
I was trying to write a program that converts a string to datetime type. The code looks like this.
import datetime
time="20:36"
p=datetime.datetime.strptime(time, "%H:%M")
print(p)
and the output was
1900-01-01 20:36:00
How do I get rid of the '1990-01-01' ?

It's unfortunate that you cannot simply write
p = datetime.time.strptime(time, "%H:%M")
because time.strptime is not defined. You are forced to use datetime.strptime, which returns a datetime object no matter what fields are actually being parsed.
On the plus side, you can get the time object you want relatively easily.
t = p.time()
print(t) # Outputs 20:36:00

You're creating a datetime object from your string and displaying the datetime as a whole.
What you should do:
display just the time part:
use strftime:
The code
import datetime
time="20:36"
p=datetime.datetime.strptime(time, "%H:%M")
print(p.strftime("%H:%M"))
use time:
The code:
import datetime
time="20:36"
p=datetime.datetime.strptime(time, "%H:%M")
print(p.time())
What you really should do:
Create a time object from your data. For that you can use datetime.time
import datetime
time="20:36"
hour, minute = time.split(":")
t = datetime.time(hour=int(hour), minute=int(minute))
print(t)

Converting str to datetime. time [duplicate]

This question already has answers here:
Convert string into datetime.time object
(4 answers)
Closed 1 year ago.
I have a string in the str format
PATTERN_OUT = "%H:%M"
date_time = (datetime.strftime(enddateandtime, PATTERN_OUT))
I need to convert it to datetime.time. How can this be done?

You can utilize the fact that time string is \d\d:\d\d.
Look at the following snippet.
from datetime import time
time_str = "10:01"
time(*map(int, time_str.split(':')))
Add exception handler if required.

Convert date string format to a datetime Python Object [duplicate]

This question already has answers here:
Convert string "Jun 1 2005 1:33PM" into datetime
(26 answers)
Closed 2 years ago.
i've one doubt.
I'm doing request to an API which return me event date. I was hope that date will be a timestamp, but i get this value:
{"date":"2020-08-24T21:15:00+00:00"}
I want to get a python datetime object.
How can I do that?

from datetime import datetime
dates = {"date":"2020-08-24T21:15:00+00:00"}
date = dates.get("date")
day = datetime.strptime(date, "%Y-%m-%dT%H:%M:%S+00:00")
Your looking for strptime.
Heres a good article:
https://www.programiz.com/python-programming/datetime/strptime

Use dateutil.parser which smartly parse date string:
import json
import dateutil.parser
result = '{"date":"2020-08-24T21:15:00+00:00"}'
x = json.loads(result)
dt = dateutil.parser.parse(x['date'])
# 2020-08-24 21:15:00+00:00
print(dt)
# <class 'datetime.datetime'>
print(type(dt))

I think you can do it respecting the format while parsing the string:
You have to try to follow the structure of the string and assign each value to the correct time value. For example:
str = '2018-06-29 08:15:27.243860'
time_obj = datetime.datetime.strptime(date_time_str, '%Y-%m-%d %H:%M:%S.%f')
Note that your case is pretty different.
It could be similar to '%Y-%m-%dT%H:%M:%S.f'

Converting month in number to word [duplicate]

This question already has answers here:
Parse date string and change format
(10 answers)
Closed 4 years ago.
i have string
date = "2018-09-12"
i want to get output like 2018-September-12
and i try like this
from datetime import datetime
date3 = datetime.strptime(date, '%Y-%m%B-%d')
or date3 = datetime.strptime(date, '%Y-%B-%d')
but always get time data '2018-09-12' does not match format '%Y-%m%B-%d'

Use strftime
Ex:
from datetime import datetime
date = "2018-09-12"
date3 = datetime.strptime(date, '%Y-%m-%d').strftime("%Y-%B-%d")
print(date3)
Output:
2018-September-12
strptime to convert string datetime to datetime object.
strftime to convert datetime object to required string format.

How to convert a string to datetime object [duplicate]

This question already has answers here:
How do I parse an ISO 8601-formatted date?
(29 answers)
Closed 5 years ago.
With the time value being:
value = '2017-08-31T02:24:29.000Z'
I try to convert it to a datetime object with:
import datetime
datetime_object = datetime.datetime.strptime(value, '%Y-%b-%d %I:%M%p')
But the command crashes with the exception:
ValueError: time data '2017-08-31T02:24:29.000Z' does not match format '%Y-%b-%d %I:%M%p'

You should be using a builtin Python's datautil module instead of date time:
import dateutil.parser
value = '2017-08-31T02:24:29.000Z'
result = dateutil.parser.parse(value)

First of all, you are missing the formatter for the microsecond.
Second of all, there is no second colon for dividing the minute and second.
Third, the %b operator is for the monthname (Jan,Feb,etc.). You want to use %m.
Final format is '%Y-%m-%dT%I:%M:%S.%fZ'.
This is your code:
datetime_object = datetime.datetime.strptime(value, '%Y-%m-%dT%I:%M:%S.%fZ')
You should get 2017-08-31 02:24:29 as the value of datetime_object.