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Yes or No answer from user with Validation and restart option?

(py) At the moment, the code below does not validate/output error messages when the user inputs something other than the two choices "y" and "n" because it's in a while loop.
again2=input("Would you like to calculate another GTIN-8 code? Type 'y' for Yes and 'n' for No. ").lower() #**
while again2 == "y":
print("\nOK! Thanks for using this GTIN-8 calculator!\n\n")
restart2()
break #Break ends the while loop
restart2()
I'm struggling to think of ways that will allow me to respond with an output when they input neither of the choices given. For example:
if again2 != "y" or "n"
print("Not a valid choice, try again")
#Here would be a statement that sends the program back to the line labelled with a **
So, when the user's input is not equal to "y" or "n" the program would return to the initial statement and ask the user to input again. Any ideas that still supports an efficient code with as little lines as possible? Thanks!

def get_choice(prompt="Enter y/n?",choices=["Y","y","n","N"],error="Invalid choice"):
while True:
result = input(prompt)
if result in choices: return result
print(error)
is probably a nice generic way to approach this problem
result = get_choice("Enter A,B, or C:",choices=list("ABCabc"),error="Thats not A or B or C")
you could of coarse make it not case sensitive... or add other types of criteria (e.g. must be an integer between 26 and 88)

A recursive solution:
def get_input():
ans = input('Y/N? ') #Use raw_input in python2
if ans.lower() in ('y', 'n'):
return ans
else:
print('Please try again.')
return get_input()
If they're really stubborn this will fail when it reaches maximum recursion depth (~900 wrong answers)

I need to figure out how to make my program repeat. (Python coding class)

I am a beginner student in a python coding class. I have the majority of the done and the program itself works, however I need to figure out a way to make the program ask if wants a subtraction or an adding problem, and if the user would like another question. I asked my teacher for assistance and he hasn't gotten back to me, so I'm simply trying to figure out and understand what exactly I need to do.
import random
x = int(input("Please enter an integer: "))
if x < 0:
x = 0
print('Negative changed to zero')
elif x == 0:
print('Zero')
elif x == 1:
print('Single')
else:
print('More')
maximum = 10 ** x;
maximum += 1
firstnum = random.randrange(1,maximum) # return an int from 1 to 100
secondnum = random.randrange(1, maximum)
compsum = firstnum + secondnum # adds the 2 random numbers together
# print (compsum) # print for troubleshooting
print("What is the sum of", firstnum, " +", secondnum, "?") # presents problem to user
added = int(input("Your answer is: ")) # gets user input
if added == compsum: # compares user input to real answer
print("You are correct!!!")
else:
print ("Sorry, you are incorrect")

You'll want to do something like this:
def foo():
print("Doing good work...")
while True:
foo()
if input("Want to do more good work? [y/n] ").strip().lower() == 'n':
break
I've seen this construct (i.e., using a break) used more often than using a sentinel in Python, but either will work. The sentinel version looks like this:
do_good_work = True
while do_good_work:
foo()
do_good_work = input("Want to do more good work? [y/n] ").strip().lower() != 'n'
You'll want to do more error checking than me in your code, too.

Asking users for input is straightforward, you just need to use the python built-in input() function. You then compare the stored answer to some possible outcomes. In your case this would work fine:
print('Would you like to test your adding or subtracting skills?')
user_choice = input('Answer A for adding or S for subtracting: ')
if user_choice.upper() == 'A':
# ask adding question
elif user_choice.upper() == 'S':
# ask substracting question
else:
print('Sorry I did not understand your choice')
For repeating the code While loops are your choice, they will repeatedly execute a statement in them while the starting condition is true.
while True: # Condition is always satisfied code will run forever
# put your program logic here
if input('Would you like another test? [Y/N]').upper() == 'N':
break # Break statement exits the loop
The result of using input() function is always a string. We use a .upper() method on it which converts it to UPPERCASE. If you write it like this, it doesn't matter whether someone will answer N or n the loop will still terminate.

If you want the possibility to have another question asked use a while loop and ask the user for an input. If you want the user to input whether (s)he want an addition or substraction you already used the tools to ask for such an input. Just ask the user for a string.

Python will not accept user inputs [duplicate]

This question already has answers here:
Why does "a == x or y or z" always evaluate to True? How can I compare "a" to all of those?
(8 answers)
Closed 7 years ago.
I managed to get this code to work before, but I've changed something accidentally and can't figure out what.
The code that will not work is:
while True:
answer = input ("Would you like to play this game? Type yes if you would like to. Type no to end the program")
if answer == 'no' or 'n' or 'No' or 'N':
sys.exit()
elif answer == 'yes' or 'y' or 'Yes' or 'Y':
code = input("Input a three digit code. Must be more than 001 and less than 100.")
When I run the code and put in one of the answers, the program will not run the next part and gives no error message.
In case it is necessary, I have put the code for the entire program below:
import random
import sys
while True:
answer = input ("Would you like to play this game? Type yes if you would like to. Type no to end the program")
if answer == 'no' or 'n' or 'No' or 'N':
sys.exit()
elif answer == 'yes' or 'y' or 'Yes' or 'Y':
code = input("Input a three digit code. Must be more than 001 and less than 100.")
try:
value = int(code)
except:
print ("Invalid code")
continue
if 1 <= value <= 100:
print (code)
print ("Valid code")
print ("I will now try to guess your number")
number = random.randint(1, 100)
while number > int(code) or number < int(code):
print ("Failed attempt. Number guessed is")
number = random.randint(1, 100)
print (number)
else:
if number == int(code):
print ("Your code is")
print (code)
else:
print ("Invalid code")
EDIT: Thank you so much, the yes option is working now, but the program will still not exit when selecting any of the no options, as it did before. The edited code is:
if answer in ('no', 'n', 'No', 'N'):
sys.exit()
elif answer in ('yes', 'y', 'Yes', 'Y'):
I checked by printing the answer value, and i believe it is registering the no input but not executing the command that follows for some reason.
EDIT: I'm still a bit fuzzy on the logic, but changing it to exit() fixed the problem. It asks for confirmation when closing now, when it didn't before, but otherwise sorted.

Problem causing silent exit:
if answer == 'no' or 'n' or 'No' or 'N':
sys.exit()
That test is testing answer == 'no' as one test, then 'n' as a separate test, and so on. or chains return when any test returns a "truthy" value (or the last evaluated value if none are truthy), so the test always ends up evaluating as "truthy" because a non-empty string like 'n' is truthy. If you're trying to test for any one of those values, you'd do an "is contained in" test to see if answer is one of a recognized group of values, e.g.:
if answer in ('no', 'n', 'No', 'N'):

The reason is due to this expression:
if answer == 'no' or 'n' or 'No' or 'N':
In python, the above is exactly the same as this:
if (answer == 'no') or ('n' != '') or ('No' != '') or ('N' != ''):
Since all but the first expression evaluates to true, the whole expression is true.
The simplest solution is to convert your input to lowercase and trim off any extra space, then check if the answer is in a list of allowable answers so that you can easily compare for "n", "N", "no", "NO", "No", "nO".
if answer.strip().lower() in ("n", "no"):

python - checking string for certain words

What I'm trying to do is that, if I ask a question only answerable by Yes and
No, I want to make sure that yes or no is the answer. Otherwise, it will stop.
print("Looks like it's your first time playing this game.")
time.sleep(1.500)
print("Would you like to install the data?")
answer = input(">").lower
if len(answer) > 1:
if answer == "no":
print("I see. You want to play portably. We will be using a password system.")
time.sleep(1.500)
print("Your progress will be encrypted into a long string. Make sure to remember it!")
else:
print("Didn't understand you.")
elif len(answer) > 2:
if word == "yes":
print("I see. Your progress will be saved. You can back it up when you need to.")
time.sleep(1.500)
else:
print("Didn't understand you.")

First:
answer = input(">").lower
should be
answer = input(">").lower()
Second, len(answer) > 1 is true for both "no" and "yes" (and anything larger than one character, for that matter). The elif block will never be evaluated. Without modifying significantly the logic of your current code, you should do instead:
if answer == 'no':
# do this
elif answer == 'yes':
# do that
else:
print("Didn't understand you.")

Something like:
if word.lower() in ('yes', 'no'):
would be the simplest approach (assuming case doesn't matter).
Side-note:
answer = input(">").lower
Is assigning a reference to the lower method to answer, not calling it. You need to add parens, answer = input(">").lower(). Also, if this is Python 2, you need to use raw_input, not input (In Python 3, input is correct).

How to check if user input has these lines?

It's my second program in Python. I seem to be happy with my progress. The question is I am trying to make is:
choice=eg.ynbox(msg='wat to do')
for["true","1"] in choice:
print("yes")
for["false","0"] in choice:
print("no")
The problem is that this for condition is not working. I have seen such code while I was looking up answers for my previous question but I forget. I tried googling but I dont know how to put this in words..A little syntax help is necessary
BTW: its a gui program with easygui 0.96..

choice = eg.ynbox(msg='wat to do')
if any(word in choice for word in ["true","1"]):
print('yes')
elif any(word in choice for word in ["false","0"]):
print('no')
else:
print('invalid input')
or, if the list is short:
choice = eg.ynbox(msg='wat to do')
if 'true' in choice or '1' in choice::
print('yes')
if 'false' in choice or '0' in choice::
print('no')
else:
print('invalid input')

I assume by eg.ynbox(msg='wat to do') you mean you are creating a Yes/No dialog box. This means that the value stored in choice is an Integer where 1 represents True and 0 represents False. This is correct in both Python 2.x and Python 3.x as long as True and False have not been reassigned in Python 2.x True and False are reserved keywords in Python 3.x and thus are guaranteed not to change. Therefore, you simply have an if statement that works using this value:
if choice:
print 'Yes'
else:
print 'No'
You do not need to match on 1 and 0 since they represent both True and False.

You might try the following code in place of yours:
def is_accept(word):
return word.lower() in {'true', '1', 'yes', 'accept'}
def is_cancel(word):
return word.lower() in {'false', '0', 'no', 'cancel'}
def get_choice(prompt=''):
while True:
choice = eg.ynbox(msg=prompt)
if is_accept(choice):
print('Yes')
return True
if is_cancel(choice):
print('No')
return False
print('I did not understand your choice.')