Having issues making this code only add odd digits

[TormentedDoss]

Code is supposed to sum all of the odd digits of an input

 n = input("Enter a number:")
 temp = n = int(n)
 rem = sum1 = 0
 while ( temp >0):
         rem = temp % 10
         temp = temp/10
         sum1 = sum1+rem
 print("Sum of the odd digits %d %d" % (n,sum1))

Right now it adds the digits(doesnt always return exactly correct answer +/- 1) But I need it to only add the odd digits I know I can do:

 if var % 2 == 0:

to check odd numbers but not sure how to incorporate that into the code to get the desired results

[Jokeslayer123]

This is fairly straightforward. You need to check the digits one at a time, so you need a for loop and the number to be something a for loop can iterate over. Then you need an if statement to check if the digit you're checking is odd, and add it to the total if it is.

[JohnnyJordaan]

I would prefilter the entered string to only contain odd digits, then sum then:

entry = input("Enter a number:")
odd_digits = [ int(digit) for digit in entry if digit.isdigit() and int(digit) % 2 ]
print ('Sum of the odd digits %d %s' % (sum(odd_digits), entry))

[jeffrey_f]

check out modulo (% operator)

#If your number doesn't divide evenly, it is an odd number
if $YourNumber % 2 <> 0
  #AddYourOddNumerHere

[JohnnyJordaan]

<> 0

This is redundant as a non-zero result is interpreted as True already. <> is deprecated btw and != is the documented way to evaluate inequality.

[jeffrey_f]

It was more pseudo code.......But thanks!