How to remove 0:00:00 from the output of datetime.date() in python?

Question

How to remove 0:00:00 from the output of datetime.date() function? For example:

import datetime
now = datetime.datetime.now()
year1 = now.strftime("%Y")
month2 = now.strftime("%m")
day3 = now.strftime("%d")
year = int(year1)
month = int(month2)
day = int(day3)
first_day = datetime.date(2021,8,1)
second_day = datetime.date(year,month,day)
daysleft = first_day - second_day
print(daysleft)

I get the output:

9 days, 0:00:00

If you didn't understand the question title, My main goal is to remove the 0:00:00 before the period (.). I've seen many other questions like this (in stack overflow/exchange and other websites), but it was nothing in python coding language.

and if you need to display in label in tkinter - `label["text"] = f'{daysleft.days} days'` — furas, Jul 24 '21 at 03:21
shorter `daysleft = first_day - now.date()` or `daysleft = first_day - datetime.date.today()` — furas, Jul 24 '21 at 03:22
BTW: instead of `year1 = now.strftime("%Y")` `year = int(year1)` you can get `now.year` — furas, Jul 24 '21 at 03:24
Does this answer your question? [Format timedelta to string](https://stackoverflow.com/questions/538666/format-timedelta-to-string) — FObersteiner, Jul 25 '21 at 16:21

score 1 · Answer 1 · answered Jul 24 '21 at 00:02

1

You can get just the days by asking for the .days attribute of a datetime.timedelta object. E.g.:

print('{} days'.format(daysleft.days))

answered Jul 24 '21 at 00:02

larsks

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actually i need for printing in tkinter if curious – Jul 24 '21 at 02:58
I'm telling how many days ***till*** 2021-8-1 – Jul 24 '21 at 03:02

score 0 · Answer 2 · edited Jul 24 '21 at 00:50

0

If your only goal is printing the output and you don't care about maintaining daysleft as a datetime object, you can always convert it to a string and .split() it like this:

print(str(daysleft).split(",")[0])

edited Jul 24 '21 at 00:50

not_speshal

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answered Jul 24 '21 at 00:18

Jack

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actually i need for printing in tkinter if curious – Jul 24 '21 at 02:58
using in tkinter so i think it should be daysleft2 = str(daysleft).split(",")[0])| lbl = tk.Label(text=daysleft2)| lbl.pack()| the character | is for separating lines in the code – Jul 24 '21 at 03:03
Yup, that looks good. My code above is only printing the days left as you intended, but the way you're storing it into a new variable should work fine. If you don't need the original string (daysleft) you can always reassign it like `daysleft = str((str(daysleft).split(",")[0])` so you don't have any extra vars. – Jack Jul 24 '21 at 17:44

furas · Answer 3 · 2021-07-24T04:08:03.240

datetime has many useful functions and values

To get only days you can daysleft.days

print(f'{daysleft.days} days')

but you can also reduce other code to

now = datetime.datetime.now()
daysleft = first_day - now.date()

or

today = datetime.date.today()
daysleft = first_day - today

BTW:

If you need year, month, day as numbers then you can get it as

year   = now.year
month  = now.month
day    = now.day
hour   = now.hour
minute = now.minute
second = now.second
ms     = now.microsecond
weekday = now.weekday()    # 0 = monday, ..., 6 = sunday
weekday = now.isoweekday() # 1 = monday, ..., 7 = sunday

Other useful function timedelta

one_day = datetime.timedelta(days=1)

today = datetime.date.today()

yesterday = today - one_day
tomorrow  = today + one_day

day_after_tomorrow = today + 2*one_day

Minimal working code with tkinter

import datetime

first_day = datetime.date(2021, 8, 1)

now = datetime.datetime.now()
daysleft = first_day - now.date()

today = datetime.date.today()
daysleft = first_day - today

print(f'{daysleft.days} days')

# ---

import tkinter as tk

root = tk.Tk()
lbl = tk.Label(root, text=f'{daysleft.days} days till {first_day}')
lbl.pack()
root.mainloop()

How to remove 0:00:00 from the output of datetime.date() in python?

3 Answers3