sum of digits in a number python

1number = 123 # the number you want summed up
2sum_of_digits = 0 
3for digit in str(number):
4  sum_of_digits += int(digit)
5  
6print(sum_of_digits) # printing the final sum of number

1def sum_digits(n):
2    s = 0
3    while n:
4        s += n % 10
5        n //= 10
6    return s

1# Here, first a complex input is used to determine the sum of all the integers in that input.
2
3# Rules : 1
4
5s = input()
6c = ""
7for i in s:
8    if i.isdigit():
9        c += i
10sum_d = sum(int(j) for j in c)
11print(sum_d)
12
13# Rules : 2
14
15s = input()
16c = ""
17sum = 0
18for i in s:
19    if i.isdigit():
20        c += i
21for j in c:
22  sum += int(c)%10
23  c = int(c)//10
24print(sum)
25
26# Sample input: murad#123#$%&*$%!@*34
27# Sample output: 13
28
29
30# Eval method using to problem:
31# Problem name: Python Evaluation
32# Code
33string = eval(input(""))
34print()
35
36# Sample Input: print(2 + 3)
37# Sample Output: 5
38
39

1digit_sum = lambda s: sum(int(digit) for digit in str(s)) #without recursion
2
3#sum of digits using recursion
4
5dsum = 0 # we define dsum outside of the function so its value isn't reset every time the function gets called recursivley
6
7def rdigit_sum(s):
8    global dsum # making dsum 'global' allows us to use it a function
9    if s: # checks if s has digits to add to dsum
10        dsum += s%10 # adds the current units digit to dsum
11        s = s//10 # removes the current units digit
12    else: # if there are no digits left
13        s = dsum  # this block reassigns s to dsum, then resets dsum to 0 so dsum doesn't already have a value if this function is called more than once in a program
14        dsum = 0
15        return s  
16    return rdigit_sum(s) # this is the 'recursive' part of the program that calls the function again

1number = 123 # the number you want to sum the digits of
2
3sum_of_digits = sum(int(digit) for digit in str(number))
4
5print(sum_of_digits)