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How to convert a date string zone oriented and hour string into datetime

I have two strings one: date='2021-12-30T23:00Z' where Z means UTC timezone and 23:00 means hour. I also have an hour string hour='3'. What I want is to convert date to datetime object and add this hour string to date as a delta. In result I would get a datetime object with hour: '2021-12-31T02:00Z' I tried function datetime.datetime.fromisoformat() with no luck.

Use strftime with their format.
from datetime import datetime, timedelta
date='2021-12-30T23:00Z'
date = datetime.strptime(date, '%Y-%m-%dT%H:%MZ')
new_date = date + timedelta(hours=3)
new_date = new_date.strftime('%Y-%m-%dT%H:%MZ')
print(new_date)
Output:
2021-12-31T02:00Z

You could do something like this:
from datetime import datetime
from datetime import timedelta
date = "2021-12-30T23:00Z"
hour = "3"
d = datetime.strptime(date, "%Y-%m-%dT%H:%M%z") + timedelta(hours=int(hour))
print(d)
output:
2021-12-31 02:00:00+00:00

Is there a function in python that could generate date 4 weeks from current date/given date?

Is there any function in python that can generate date for example 4 weeks from now or given date?
I've gone through documentation from datetime modeule but couldnt find any example that can support my question.

four_weeks = datetime.timedelta(days=4*7)
dt = datetime.datetime.now()
print(dt + four_weeks)

Here you go:
from datetime import timedelta
from datetime import datetime
today = datetime.today()
print(today + timedelta(weeks=1))

I think the thing you're looking for is timedelta.
from datetime import timedelta
def add_weeks(dt, n_weeks):
n_days = 7 * n_weeks
return dt + timedelta(days=n_days)

In python datetime module has a class called datetime which represents a date + time, an point on time line. There is another class called timedelta that represents difference between two dates (datetiems).
You can add a date with a timedelta.
example code:
from datetime import datetime, timedelta
now = datetime.now()
duration = timedelta(days=28)
target = now + duration
print(target)

Unable to subtract a day from any specific date format

I'm trying to subtract a day from this date 06-30-2019 in order to make it 06-29-2019 but can't figure out any way to achive that.
I've tried with:
import datetime
date = "06-30-2019"
date = datetime.datetime.strptime(date,'%m-%d-%Y').strftime('%m-%d-%Y')
print(date)
It surely gives me back the date I used above.
How can I subtract a day from a date in the above format?

try this
import datetime
date = "06/30/19"
date = datetime.datetime.strptime(date, "%m/%d/%y")
NewDate = date + datetime.timedelta(days=-1)
print(NewDate) # 2019-06-29 00:00:00

Your code:
date = "06-30-2019"
date = datetime.datetime.strptime(date,'%m-%d-%Y').strftime('%m-%d-%Y')
Check type of date variable.
type(date)
Out[]: str
It is in string format. To perform subtraction operation you must convert it into date format first. You can use pd.to_datetime()
# Import packages
import pandas as pd
from datetime import timedelta
# input date
date = "06-30-2019"
# Convert it into pd.to_datetime format
date = pd.to_datetime(date)
print(date)
# Substracting days
number_of_days = 1
new_date = date - timedelta(number_of_days)
print(new_date)
output:
2019-06-29 00:00:00
If you want to get rid of timestamp you can use:
str(new_date.date())
Out[]: '2019-06-29'

use timedelta
import datetime
date = datetime.datetime.strptime("06/30/19" ,"%m/%d/%y")
print( date - datetime.timedelta(days=1))

Get year,month and day from python variable

I'd like to get the break of a variable by year, month and day. Here's what I got:
import datetime
from datetime import date, timedelta
yesterday = date.today() - timedelta(1)
print (yesterday)
year = datetime.date.yesterday.year
month = datetime.date.yesterday.month
day=datetime.date.yesterday.day
print (year)
print (month)
print (day)
I'm getting an error that datetime.date has no attribute. I'm a total noob at python and I'm stuck, any help is appreciated

you were close
import datetime
from datetime import date, timedelta
yesterday = date.today() - timedelta(1)
print (yesterday)
year = yesterday.year
month = yesterday.month
day=yesterday.day
print (year)
print (month)
print (day)
result is
2019-03-10
2019
3
10

You can use strftime method
A simple example:
>>> from datetime import datetime
>>> now = datetime.utcnow()
>>> year_month_day_format = '%Y-%m-%d'
>>> now.strftime(year_month_day_format)
'2020-11-06'
>>> hour_minute_format = '%H:%M'
>>> now.strftime(hour_minute_format)
'22:54'
Hopping, it will help someones

You can also simplify your import statements like so:
from datetime import datetime, timedelta
yesterday = datetime.today() - timedelta(1)
print(yesterday)
year = yesterday.year
month = yesterday.month
day = yesterday.day
print(year)
print(month)
print(day)
You will get the output:
2019-03-10 21:19:36.695577
2019
3
10

For current day
import datetime
current_datetime=datetime.datetime.now()
print("current_year:{}".format(current_datetime.year))
print("current_month:{}".format(current_datetime.month))
print("current_day:{}".format(current_datetime.day))

If you want in this format for example "10-Oct-2018". You can try this code for current day.
from datetime import datetime, timezone
now_utc = datetime.now(timezone.utc)
year = now_utc.strftime("%Y")
month = now_utc.strftime("%b")
day = now_utc.strftime("%d")
result = day+"-"+month+"-"+year
print(result)

How to subtract from date given as user input?

So I'm trying to subtract one day from a users input of 2018-02-22 for example. Im a little bit stuck with line 5 - my friend who is is leaps and bounds ahead of me suggested I try this method for subtracting one day.
In my online lessons I haven't quite got to datetime yet so Im trying to ghetto it together by reading posts online but got a stuck with the error :
TypeError: descriptor 'date' of 'datetime.datetime' object needs an argument
from datetime import datetime, timedelta
date_entry = input('Enter a date in YYYY-MM-DD format')
year, month, day = map(int, date_entry.split('-'))
date1 = datetime.date()
new = date1.replace(day=date1.day-1, hour=1, minute=0, second=0, microsecond=0)
print (new)
So my aim is the get the out put 2018-02-21 if I put in 2018-02-22.
Any help would be amazing :)

First of all a timedeltais best used for the purpose of doing date arithmetcs. You import timedelta but don't use it.
day = timedelta(days=1)
newdate = input_datetime - day
Second problem is you're not initializing a date object properly. But in this case it would be better to use datetime as well as strptime to parse the datetime from the input string in a certain format.
date_entry = input('Enter a date in YYYY-MM-DD format')
input_date = datetime.strptime(date_entry, "%Y-%m-%d")
day = timedelta(days=1)
newdate = input_datetime - day

from datetime import datetime, timedelta
date_entry = input('Enter a date in YYYY-MM-DD format')
date1 = datetime.strptime(date_entry, '%Y-%m-%d')
print date1+timedelta(1)

Maybe you need something like this
from datetime import datetime, timedelta
date_entry = input('Enter a date in YYYY-MM-DD format ')
# convert date into datetime object
date1 = datetime.strptime(date_entry, "%Y-%m-%d")
new_date = date1 -timedelta(days=1) # subtract 1 day from date
# convert date into original string like format
new_date = datetime.strftime(new_date, "%Y-%m-%d")
print(new_date)
Output:
Enter a date in YYYY-MM-DD format 2017-01-01
'2016-12-31'