how to generate 4 random alphabets and 4 random mumbers

Question

import random
alphabet = “ABCDEFGHIJKLMNOPQRSTUVQXYZabcdefghijklmnopqrstuvwxyz”
rand_index = random.randrange (0, len(alphabet))
first_slice = alphabet[:rand_index]
alphabet = first_slice

random_number = random.randrange(1, 999)

print(“Your password is: “, alphabet + str(random_number))

output should be 4charcters of alphabet and 4 characters of random number with extend of zero
So for example tziy0098 or gtjs0999

However the real result is abc715

I can’t figure out how to get only four random alphabet
And how to zero extend the number so i generate four characters of random number.
Please help

Answer 1

Although it's a bit weird way of doing this, just add .zfill(4) to your print statement to include a trailing zero.

so print ..... + str(random_number).zfill(4)

Answer 2

Use random.sample to get four random letters of the alphabet. If you can't use zfill, check the length of str(random_number) and add (4-len)*'0' (on the left side)

Answer 3

# ;)
print(alphabet + ('000' + str(random_number))[-4:])

Answer 4

or you can simply use a f str, that replaces % and str.format:
[print(f'-->{k:0>4d}<--') for k in [6,66,666,6666,66666]]
using str.format would be similar:
[print('-->{0:0>4d}<--'.format(k)) for k in [6,66,666,6666,66666]]

in your case print(alphabet+f'{random_number:0>4d}')

check out for more details:
https://docs.python.org/3/library/string.html#formatstrings

Answer 5

Oh i see thank youu

Answer 6

The thing is we cannot use .zfill() yet
We can only use concatenation

Answer 7

Thank you so muchh

Answer 8

Thank youuu so muchhhhh