Python 3: Unexpected EOF when parsing - python

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!

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.

Related

Python guessing game code keeps crashing after 1 guess. How would i fix this?

My code keeps crashing after I put in the 1st guess I make. I've looked at syntax and dont think that's a problem how do I make it so it goes past the 1st guess and executes it. When I put the guess in it just puts in all the prompts at once, and how do I call the function properly at the end? Any help would be appreciated.
Import time,os,random
def get_int(message):
while True:
user_input = input(message)
try:
user_input = int(user_input)
print('Thats an integer!')
break
except:
print('That does not work we need an integer!')
return user_input
def game_loop():
fin = False
while not fin:
a = get_int('give me a lower bound:')
b = get_int('give me a upper bound:')
if a < 0:
print("that doesn't work")
if a > b:
a, b = b, a
print(a,b)
os.system('clear')
print("The number you guess has to be between " + str(a) + "and " + str(b) + '.')
num_guesses = 0
target = random.randint(a,b)
time_in = time.time()
while True:
print('You have guessed ' + str(num_guesses) + " times.")
print()
guess_input = get_int('guess a number')
if guess_input == target:
print("Congrats! You guessed the number!")
time_uin = time.time()
break
elif guess_input < a or guess_input > b:
print("guess was out of range... + 1 guess")
elif guess_input < target:
print("Your guess was too low")
else:
print("Your guess was to high")
num_guesses = num_guesses + 1
if num_guesses<3:
print('Einstein?')
else:
print('you should be sorry')
time_t = time_uin - time_in
print('it took' + str(time_t) + 'seconds for you to guess a number')
print()
time_average = time_t / (num_guesses+1)
print('It took an average of' + str(time_average)+"seconds per question")
print()
while True:
play_again = input ('Type y to play again or n to stop')
print()
if play_again == 'n':
fin = True
print('Thank God')
time.sleep(2)
os.system('clear')
break
elif play_again == 'y':
print('here we go again')
time.sleep(2)
os.system('clear')
break
else:
print('WRONG CHOCICE')
break
game_loop()
If guess_input != target on the first iteration of the loop, time_uin is referenced before assignment. Hence the error:
UnboundLocalError: local variable 'time_uin' referenced before assignment
Solution is to run the following only if guess_input == target.
if num_guesses<3:
print('Einstein?')
else:
print('you should be sorry')
time_t = time_uin - time_in
print('it took' + str(time_t) + 'seconds for you to guess a number')
print()
time_average = time_t / (num_guesses+1)
print('It took an average of' + str(time_average)+"seconds per question")
print()
Please follow coppereyecat's link to learn how to debug a basic Python program.

Crashing when improper input given more than once

I'm trying to make a simple number guesser program, it works pretty well however if I enter 'a' twice instead of a valid int it crashes out. Can someone explain what I'm doing wrong here.
import random
def input_sanitiser():
guess = input("Please enter a number between 1 and 10: ")
while True:
if type(guess) != int:
guess = int(input("That isn't a number, try again: "))
elif guess not in range (1,11):
guess = int(input("This is not a valid number, try again: "))
else:
break
def main():
number = random.randrange(1,10)
guess = 0
input_sanitiser()
while guess != number:
if guess < number:
print("This number is too low!")
input_sanitiser()
if guess > number:
print("This number is too high!")
input_sanitiser()
else:
break
print ("Congratulations, you've guessed correctly")
if __name__ == "__main__":
main()
You want to check the input before trying to convert it to int:
int(input("This is not a valid number, try again: "))
I would write:
while True:
try:
guess = int(input("This is not a valid number, try again: "))
except ValueError:
pass
else:
break
Side note: the code isn't working as expected:
def main():
number = random.randrange(1,10)
guess = 0
input_sanitiser() # <<<<<<<<<<
while guess != number:
Note that input_sanitiser does not modify the variable guess in main, you need some other way round, like processing the input then returning the result from input_sanitiser, like this:
def input_sanitiser():
guess = input("Please enter a number between 1 and 10: ")
while True:
try:
guess = int(input("This is not a valid number, try again: "))
except ValueError:
continue # keep asking for a valid number
if guess not in range(1, 11):
print("number out of range")
continue
break
return guess
def main():
number = random.randrange(1,10)
guess = input_sanitiser()
while guess != number:
if guess < number:
print("This number is too low!")
guess = input_sanitiser()
if guess > number:
print("This number is too high!")
guess = input_sanitiser()
else:
break
print ("Congratulations, you've guessed correctly")

User input Exit to break while loop

I'm doing an assignment for the computer to generate a random number and have the user input their guess. The problem is I'm supposed to give the user an option to input 'Exit' and it will break the While loop. What am I doing wrong? I'm running it and it says there's something wrong with the line guess = int(input("Guess a number from 1 to 9: "))
import random
num = random.randint(1,10)
tries = 1
guess = 0
guess = int(input("Guess a number from 1 to 9: "))
while guess != num:
if guess == num:
tries = tries + 1
break
elif guess == str('Exit'):
break
elif guess > num:
guess = int(input("Too high! Guess again: "))
tries = tries + 1
continue
else:
guess = int(input("Too low! Guess again: "))
tries = tries + 1
continue
print("Exactly right!")
print("You guessed " + str(tries) + " times.")
The easiest solution is probably to create a function that gets the displayed message as an input and returns the user input after testing that it fulfils your criteria:
def guess_input(input_message):
flag = False
#endless loop until we are satisfied with the input
while True:
#asking for user input
guess = input(input_message)
#testing, if input was x or exit no matter if upper or lower case
if guess.lower() == "x" or guess.lower() == "exit":
#return string "x" as a sign that the user wants to quit
return "x"
#try to convert the input into a number
try:
guess = int(guess)
#it was a number, but not between 1 and 9
if guess > 9 or guess < 1:
#flag showing an illegal input
flag = True
else:
#yes input as expected a number, break out of while loop
break
except:
#input is not an integer number
flag = True
#not the input, we would like to see
if flag:
#give feedback
print("Sorry, I didn't get that.")
#and change the message displayed during the input routine
input_message = "I can only accept numbers from 1 to 9 (or X for eXit): "
continue
#give back the guessed number
return guess
You can call this from within your main program like
#the first guess
guess = guess_input("Guess a number from 1 to 9: ")
or
#giving feedback from previous input and asking for the next guess
guess = guess_input("Too high! Guess again (or X to eXit): ")
You are trying the parse the string 'Exit' to an integer.
You can add a try/except around the casting line and handle invalid input.
import random
num = random.randint(1,9)
tries = 1
guess = 0
guess = input("Guess a number from 1 to 9: ")
try:
guess = int(guess) // try to cast the guess to a int
while guess != num:
if guess == num:
tries = tries + 1
break
elif guess > num:
guess = int(input("Too high! Guess again: "))
tries = tries + 1
continue
else:
guess = int(input("Too low! Guess again: "))
tries = tries + 1
continue
print("Exactly right!")
print("You guessed " + str(tries) + " times.")
except ValueError:
if guess == str('Exit'):
print("Good bye")
else:
print("Invalid input")

Guessing game: matching user input with randomly generated number

def var (guess):
return guess
guess = int(input("Guess a number 1 through 10: "))
import random
num = (random.randint(1,10))
while True:
try:
guess = num
print("you guessed the right number!")
break
except:
print("try again")
break
So for this program I am trying to figure out how to have the user input a number and to guess what number (1 through 10) the program generated. It seems that every time I input a value it always gives me the "you guess the right number!" string even if I input a value higher than 10.
EDIT: Why would someone downvote my question o_o
You need to get user's input inside while loop so that user's input got updated with each iteration.
import random
num = (random.randint(1,10))
while True:
try:
guess = int(input("Guess a number 1 through 10: "))
if guess == num:
print("you guessed the right number!")
break
else:
print("try again")
except:
print('Invalid Input')
try/except is for exception handling, Not matching values. What you are looking for is if statments, For example:
guess = int(input("Guess a number 1 through 10: "))
import random
num = (random.randint(1,10))
if guess == num:
print("You guessed the right number!")
else:
print("Try again")
I think you may have intended to continue looping until the right number is guessed, In which case, This will work:
import random
num = (random.randint(1,10))
while True:
guess = int(input("Guess a number 1 through 10: "))
if guess == num:
print("You guessed the right number!")
break
else:
print("Try again")

Checking if the user wants to continue, using while and continue in Python 3.3

This is a simple program to check if a number is odd or even.I want to check if the user wants to continue or not and when I run it I get an Invalid Syntax Error on the break line, what have I got wrong?
while True:
if cont != "no":
num = (int(input("Type a number. ")))
num_remainder = num % 2
if num_remainder == 0:
print ()
print (num, " is an even number.")
else:
print ()
print (num, " is an odd number.")
cont= (input("Would you like to continue?")
continue
else:
break
Thanks
I think this is simpler,
while True:
num = int(input("Type a number. "))
print () # blank line before assert whether is or not a even number
if num % 2 == 0:
print (num, " is an even number.")
else:
print (num, " is an odd number.")
if input("Would you like to continue?") == 'no': # Ohh, you want to break now.
break

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