CODEWITHSUNDEEP
pythonpython-3.xdatetime
MarcoM
MarcoM

Reputation: 101

How to convert from days and hours to just hours?

I have only found question Convert Days and Time (Hours x Minutes x Seconds) to Time only on stackoverflow and that seems like it would help me out but it doesn't seem to totally apply to what I'm doing.

I'm writing a wage tracking program and need it to give me the total sum of all hour input. I've got a much smaller and abridged form of it to just work on this one aspect. It saves a lot of time as the main program requires all the individual start and end times to be input by the user. It is currently showing 2 days, 22:00:00 as on output whereas I ideally would prefer it to show 70:00, showing only the hours and minutes and getting rid of the unneeded second part.

import datetime


def time_diff(a, b):  # Calculates time difference between start and finish
    return b - a


start = '10:00'
mon_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
mon_fin = datetime.datetime.strptime(finish, '%H:%M')
mon_hours = time_diff(mon_start, mon_fin)

start = '10:00'
tue_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
tue_fin = datetime.datetime.strptime(finish, '%H:%M')
tue_hours = time_diff(tue_start, tue_fin)

start = '10:00'
wed_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
wed_fin = datetime.datetime.strptime(finish, '%H:%M')
wed_hours = time_diff(wed_start, wed_fin)

start = '10:00'
thu_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
thu_fin = datetime.datetime.strptime(finish, '%H:%M')
thu_hours = time_diff(thu_start, thu_fin)

start = '10:00'
fri_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
fri_fin = datetime.datetime.strptime(finish, '%H:%M')
fri_hours = time_diff(fri_start, fri_fin)

start = '10:00'
sat_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
sat_fin = datetime.datetime.strptime(finish, '%H:%M')
sat_hours = time_diff(sat_start, sat_fin)

start = '10:00'
sun_start = datetime.datetime.strptime(start, '%H:%M')
finish = '20:00'
sun_fin = datetime.datetime.strptime(finish, '%H:%M')
sun_hours = time_diff(sun_start, sun_fin)

total_hours = mon_hours + tue_hours + wed_hours + thu_hours + fri_hours + sat_hours + sun_hours
print(total_hours)

Upvotes: 0

Views: 645

Answers (3)

XAMES3
XAMES3

Reputation: 139

All the above solutions will work, but just wanted to share something from my side as well.

Code:

seconds = time_diff(a, b).total_seconds()


def convert(seconds):
    mins, secs = divmod(int(seconds), 60)
    hours, mins = divmod(mins, 60)
    return f'{hours:02d}:{mins:02d}:{secs:02d}'


print(convert(seconds))    # Will print in HH:MM:SS format

Upvotes: 0

Leemosh
Leemosh

Reputation: 905

Just use this print at the end:

print(
    "{}:{:02d}".format(
        total_hours.days * 24 + total_hours.seconds // 3600,
        (total_hours.seconds // 60) % 60,
    )

Upvotes: 1

Daweo
Daweo

Reputation: 36340

This:

def time_diff(a, b):  # Calculates time difference between start and finish
    return b - a

if a is datetime.datetime and b is datetime.datetime return datetime.timedelta object. You can use its .total_seconds method to get value in second, which then you might use for further calculations, for example

import datetime
total = datetime.timedelta(days=3,minutes=10)
total_minutes = int(total.total_seconds()//60)
hours, minutes = total_minutes // 60, total_minutes % 60
print(hours, minutes)  # 72 10

Upvotes: 2

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