We Keep Coding

How to make a count for each day passed from start date in python

I'm looking for a way to count each day passed from a start date in python. So if the start date was 21/02/2020 and count equals to 0, when the next day starts count should increment by 1.
Edit: After using Rusty's code I am able to show you a minimal reproducible example.
import datetime
start = datetime.datetime.strptime(input("Choose a start date (mm/dd/yyyy): "), '%m/%d/%Y')
current = datetime.datetime.now()
delta = current - start
count = delta.days
print(count)

import datetime
import time
count = 0
# "...from today..."
today = datetime.datetime.today()
# "...to infinity..."
while True:
now = datetime.datetime.today()
# "...as soon as the next day starts..."
if today.day != now.day:
# "...it would increment count by 1..."
count = count + 1
print(count)
today = now
time.sleep(1)

import datetime
today = datetime.datetime.strptime('03/21/2020', '%m/%d/%Y')
tomorrow = datetime.datetime.strptime('03/22/2020', '%m/%d/%Y')
next_saturday = datetime.datetime.strptime('03/28/2020', '%m/%d/%Y')
delta = tomorrow - today
count = delta.days
print(count)
delta = next_saturday - today
count = delta.days
print(count)

if date is weekend - auto rerun the function

So I can generate random days in a given start-end date relationship, but, if the date happens to be a weekend - currently all I can get working is to print to the user 'it is a weekend'. What I would like to do is, if the random day IS a weekend, rerun the function so the user does not have to manually. Basically - only print out weekdays - currently, if the random day is a weekend, it prints a blank space or None value. Only return/print weekdays is the main goal.
Here is the code so far:
from datetime import datetime, timedelta
from random import randrange
def random_date(start, end):
delta = end - start
random_day = randrange(delta.days)
myresult = start + timedelta(days=random_day)
return myresult
d1 = datetime.strptime('9/1/2018', '%m/%d/%Y')
d2 = datetime.strptime('9/30/2018', '%m/%d/%Y')
myresult = random_date(d1, d2)
if myresult.weekday() not in (5, 6):
print myresult.strftime('%m-%d-%Y')
else:
print "hit a weekend"

An option:
def random_weekday(start, end):
date = None
while (not date or date.weekday() in (5, 6)):
days = randrange((end - start).days)
date = start + timedelta(days=days)
return date
start = datetime.strptime('9/1/2018', '%m/%d/%Y')
end = datetime.strptime('9/30/2018', '%m/%d/%Y')
for i in range(20):
print(random_weekday(start, end).strftime('%m-%d-%Y'))

So, you need a while-loop to keep getting dates until you get one that's not a weekend, like this:
from datetime import datetime
from random import randrange
from datetime import timedelta
def random_date(start, end):
delta = end - start
random_day = randrange(delta.days)
myresult = start + timedelta(days=random_day)
return myresult
while True:
d1 = datetime.strptime('9/1/2018', '%m/%d/%Y')
d2 = datetime.strptime('9/30/2018', '%m/%d/%Y')
myresult = random_date(d1, d2)
if myresult.weekday() not in (5,6):
break
print myresult.strftime('%m-%d-%Y')

Datetime format when adding time durations

time_1 = datetime.timedelta(hours = hours_1, minutes = minutes_1, seconds = seconds_1)
time_2 = datetime.timedelta(hours = hours_2, minutes = minutes_2, seconds = seconds_2)
when added it gives the answer in days and hours but I really just want the total number of hours.

This should help.
import datetime
time_1 = datetime.timedelta(hours = 10, minutes = 2, seconds = 5)
time_2 = datetime.timedelta(hours = 30, minutes = 5, seconds = 7)
print(time_1 + time_2)
print(((time_1 + time_2).total_seconds()/60)/60)
Output:
1 day, 16:07:12
40.12

Adding x seconds worth of time onto a string variable time of HH:MM

I'm trying to work out a way to make a new variable AUTOMATIC_END_TIME based on adding the minimum amount of time onto the start time but I can't figure out the way to allow START_TIME to be turned into a time that can then have time added onto it.
So far my script has the following:
from time import sleep
from datetime import datetime, timedelta
START_TIME = "19:18"
END_TIME = "19:25"
LOGA = ["one", "two"]
TIME_DIFFERENCE = datetime.strptime(END_TIME, "%H:%M") - datetime.strptime(START_TIME, "%H:%M")
TIME_DIFFERENCE = TIME_DIFFERENCE.seconds
if len(LOGA) * (60 * (5 + 1)) >= TIME_DIFFERENCE:
print "Show minimum end time"
AUTOMATIC_END_TIME = "" # Should come out as 19:30
The current script shouldn't change at all except for AUTOMATIC_END_TIME which should be START_TIME + (60 * (5 + 1) It should come out as 19:30

>>> (datetime.strptime(START_TIME, "%H:%M") + timedelta(minutes=12)).strftime('%H:%M')
'19:30'

from time import sleep
from datetime import datetime, timedelta
START_TIME = "19:18"
END_TIME = "19:25"
LOGA = ["one", "two"]
TIME_DIFFERENCE = datetime.strptime(END_TIME, "%H:%M") - datetime.strptime(START_TIME, "%H:%M")
TIME_DIFFERENCE = TIME_DIFFERENCE.seconds
if len(LOGA) * (60 * (5 + 1)) >= TIME_DIFFERENCE:
print "Show minimum end time"
AUTOMATIC_END_TIME = (datetime.strptime(START_TIME, "%H:%M") + timedelta(minutes=12)).strftime('%H:%M')
print AUTOMATIC_END_TIME

Convert seconds to end to date format

SOAP client return seconds to end event.
How can I get from this seconds date in format "yyy-mm-dd hh:ii:ss"

A quick example (add 50000 seconds from now with datetime.timedelta):
>>> import datetime
>>> time_now = datetime.datetime.now()
>>> time_event = time_now + datetime.timedelta(seconds=50000)
>>> time_event.strftime("%Y-%m-%d %H:%M:%S")
'2010-05-08 12:07:05'