Python：仅循环打印最后一个值

Question

My program to print the odd numbers up to a certain number mathematically works, however, I'm having trouble printing every odd number.

For some reason the program only prints the final odd number:

x=0
N=input('What is your number?')
N=float(N)
check=(N/2)
if (check).is_integer()==1:
    print('Your number is even')
    index=N-N/2-1
    while x<=index:
        x=2*index+1
        print(x)
        index=index+1
else:
    print('Your number is odd')
    index=(N-1)/2
    while x<=index:
        x=2*index+1
        print(x)
        index=index+1

Answer 1

Your counting of the odd numbers is very.. odd . I cannot make out what you were trying to do. Why not simply start x at 1 and increment by 2 each time until you get to N ?

x = 1
while x <= N:
    print(x)
    x += 2

You don't need to use float values here; stick with int and use the % modulus (remainder) operator; this has the advantage you can then use a range() to produce all odd numbers up to N :

N = int(input('What is your number?'))
if N % 2 == 0:
    print('Your number is even')
else:
    print('Your number is odd')

for x in range(1, N + 1, 2):
    print(x)

Answer 2

There are a few suggestions I'd make.

First , you're using a float but unless you're going to be using floating point numbers (numbers with a decimal) I'd recommend you stick with integers.

Second , if you're trying to check whether a number is even or odd, you should check out the modulo (%) operator.

Third , always keep things simple. Don't repeat your code unless you need to.

Here is an example of a simpler version of your program.

user_input=input('What is your number?')
user_input=int(user_input)
if user_input % 2 == 0:
    print('Your number is even')
else:
    print('Your number is odd')
for index in range(1, user_input+1, 2):
    print(index)

Finally, be aware that your script will error if you put in garbage input, like "y7" - to fix this you should/can use exception handling.