Python 3: Unexpected EOF when parsing

Question

Using this code:

import random
import query
import sys
while True:
 try:
 number = int(input('Choose a number between 0 and 10:'))
 except ValueError:
 print("That is not a number.")
 continue
 if number > 10:
 print('Your number is too large.')
 continue
 elif number < 0:
 print('Your number is too small.')
 continue
 break
result = random.randint(0, 10)
print("You're number: " + str(number))
print("Our number: " + str(result))
if number == result:
 print('Congratulations!')
else:
 print('Close, but no cigar.')
while True:
 try:
 answer = query.query_yes_no('Do you wish to contunue?')
 if answer == "yes":
 while True:
 try:
 number = int(input('Choose a number between 0 and 10:'))
 except ValueError:
 print("That is not a number.")
 continue
 if number > 10:
 print('Your number is too large.')
 continue
 elif number < 0:
 print('Your number is too small.')
 continue
 break
 print("You're number: " + str(number))
 print("Our number: " + str(result))
 if number == result:
 print('Congratulations!')
 continue
 else:
 print('Close, but no cigar.')
 continue
 elif answer == "no":
 print('Goodbye.')
 break
 break
 break
exit()

I keep getting a SyntaxError: unexpected EOF while parsing. It says it is on line 60. I have tried removing the exit() and the breaks but that doesn't work. I'm sure it is something simple as I am still new to Python. Any help would be greatly appreciated!

Answer 1

You need to include another except for the first try inside your infinite while loop. That's why, it may be giving a syntax error.

• CannibalAngel: Thanks a lot! That fixed it! I took out the last 2 breaks and the exit() and added except:pass and it is working as intended now.