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Handle invalid string input without breaking out of a "while" loop [duplicate]

This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 10 months ago.
I'm working on a Python project to get better with passing parameters into functions, so built a small program that takes user input and appends it to a list.
I've got most of that working, but I'm stuck on trying to allow a user to add another input if they want. The program asks if they'd like to add another input, with an if/else to handle a "Y" or "N" answer. That's working for the most part. The problem comes when a user enters an invalid response (anything other than a non-case sensitive 'y' or 'n'). At this point, the program is just displaying the goodbye message and ending. I think it has to do with my while loop, since it's supposed to break out when the new_input() function isn't resolving to true anymore. I want that function to continue to cycle through until the user enters a valid response, which would then cause it to continue and take another input or end appropriately.
Is there a way to get this to work with my combination of an if/else statement or is there another method that would work better? I'll drop the relevant pieces of the program below.
def new_input():
answer = input("Would you like to add more wisdom?(y/n): ")
if answer == "y" or answer == "Y":
return True
elif answer == 'n' or answer == 'N':
return False
else:
print("Invalid input. Please try again!")
def main():
name = input('Please enter your name:')
take_input(name)
while new_input():
take_input(name)
print_lib()
print("Thanks for your brain cells, buddy! Bye bye!")
Thanks for your help!

You can always return True, unless N/n is pressed, this way the program will continue running:
def new_input():
answer = input("Would you like to add more wisdom?(y/n): ")
if answer == "y" or answer == "Y":
return True
elif answer == 'n' or answer == 'N':
return False
else:
print("Invalid input. Please try again!")
return True

Having trouble with while loops, input and strings in python

I am learning python and practicing my skills my making a simple text based adventure game.
In the game, I want to ask the player if they are ready to begin. I did this by creating a begin() function:
def begin():
print(raw_input("Are you ready to begin? > "))
while raw_input() != "yes":
if raw_input() == "yes":
break
print(start_adventure())
else:
print("Are you ready to begin? > ")
print(begin())
below this in my code is the function start_adventure()
def start_adventure():
print("Test, Test, Test")
When I run the program it starts up and I get to the point where it asks if I am ready to begin. Then it just loops infinitely and I can only exit the program if I completely close Powershell and restart Powershell. What am I doing wrong? How can I get the loop to stop once the player inputs "yes"?

What do you expect this to do? The solution to your problem is to try to understand what the code does, instead of just throwing stuff together. (Don't worry; at least 80% of us were at that stage at one point!)
As an aside, I strongly recommend using Python 3 instead of Python 2; they made a new version of Python because Python 2 was full of really strange, confusing stuff like input causing security vulnerabilities and 10 / 4 equalling 2.
What do you want this to do?
Repeatedly ask the user whether they are ready to begin until they answer "yes".
Call start_adventure().
Ok. Let's put what we've got so far into a function:
def begin():
while something:
raw_input("Are you ready to begin? > ")
start_adventure()
There are a lot of gaps in here, but it's a start. Currently, we're getting the user's input and throwing it away, because we're not storing it anywhere. Let's fix that.
def begin():
while something:
answer = raw_input("Are you ready to begin? > ")
start_adventure()
This is starting to take shape. We only want to keep looping while answer != "yes"...
def begin():
while answer != "yes":
answer = raw_input("Are you ready to begin? > ")
start_adventure()
Hooray! Let's see if this works!
Traceback (most recent call last):
File "example", line 2, in <module>
while answer != "yes":
NameError: name 'answer' is not defined
Hmm... We haven't set a value for answer yet. In order to make the loop run, it has to be something that isn't equal to "yes". Let's go with "no":
def begin():
answer = "no"
while answer != "yes":
answer = raw_input("Are you ready to begin? > ")
start_adventure()
This will work!

Python 3 Solution
You should not be calling raw_input() multiple times. Simply instantiate x and then wait until the user inputs Y to call your start_adventure function. This should get you started:
def start_adventure():
print('We have started!')
#do something here
def begin():
x = None
while x!='Y':
x = input('Are you ready to begin (Y/N)?')
if x=='Y':
start_adventure()
begin()

Your Raw input function (I'm assuming it works correctly) is never assigned to a variable. Instead you call it in your print statement, print the result of it and then you call it again in your while loop condition.
You never actually satisfy the while loop condition because your input isn't assigned to a variable. Assign Raw_input("Are you ready to begin? >") to a variable to store the input. Then while loop with the variable. Make sure in your while loop when the condition is met you reset the variable to something else.
Your program flow is wrong too, you need to call your raw input function inside the while loop. This will change the while loop condition so that when the condition is met (user types "yes") it won't loop infinitely. Hope this helps!
Example of what you need in code form:
//initialize the condition to no value
condition = None;
#check the condition
while condition != "yes"
#change the condition here based on user input **inside the loop**
condition = raw_input("are you ready to begin? >")
if condition == "yes":
#condition is met do what you need
else:
#condition not met loop again
#nothing needs to go here to print the message again

How do I use an elif statement properly?

Please don't judge me.. I've only been working with Python for a month now.
While laying in my bed I thought of making this and created it in a few minutes but I made to many else and if statements and my code just looks so messy, I kept adding things that weren't needed..(For fun :D)
Anyways, here is my code.. If you could tell me how to use the "elif" statements properly that'd be awesome.(I'm still learning python)
Question: I've tried using an elif statement multiple times and I keep getting an error. How do I fix this?
key = True # Game key, if this is false the program won't work.
print("Please type a password: ") # Asking for users password
Pass = input()
print("Thank you for typing your password, please make sure it's secure by trying again..") # Ask them to confirm their password by re-typing it
Again = input()
if Pass == Again:
print("Thank you for choosing a working password, please create your character")
print("Please type your username without numbers")
else:
print("Something's wrong with your password or username!")
# Has user confirm if his information is correct
User = input()
print("checking..")
if User.isalpha() and key == True:
print("So your Username is " + User + " and your chosen password is: " + str(Pass))
else:
print("Either your key is broken or something is wrong..")
if len(User) >= 4: # Checking if the chosen username has 4 or more characters in it
print("The length of your Username is: ")
print(str(len(User)))
print("If this information is correct please type 'true' or 'false'")
else:
print("Please type a username longer than 4 characters!")
answer = input() # I kinda fucked up because my coding is dirty and unorganized lol..
if answer == str(True):
print("Thank you, we're setting up your account! :D")
else:
print("Please re-run the program and fix your information!")

We can't debug code you haven't posted, and (as is to be expected - you are on a learning exercise here, and there's a lot to think about) your program structure isn't very helpful. For example, when the user enters non-matching passwords you tell them about it, but nevertheless continue to ask them for their username. Don't worry about this, you will soon learn how to fix it.
Since you ask about the elif, it is basically a syntax abbreviation for else if that avoids going to multiple indentation levels. Suppose you wanted a value of '1' or '2' to take different actions, and to declare other values invalid. You could write
if value == '1':
#take action appropriate to 1
else:
if value == '2':
# take action appropriate to 2
else:
raise ValueError("Allowed inputs are '1' or '2'")
Note that the different actions are at different indentation levels. The more cases you have to consider, the more levels of indentation you have to introduce. So it's generally felt to be more readable to write
if value == '1':
# Take action appropriate to 1
elif value == '2':
# Take action appropriate to 2
else:
raise ValueError("Allowed inputs are '1' or '2'")
Now all the actions and decisions are at the same indentation levels. That's pretty much all there is to it. If you leave the else case off then you won't take any actions at all, so it's normally used to specify the default action, in this case raising an exception.
PS: If you want to be sure the user has entered two matching passwords before you proceed, look at the while loop, which allows you to repeat a set of actions until some condition (in this case the passwords being equal) is true.

Here is an example if if/elif/else statement in python3:
test = 'mytest'
if test == 'not_my_test':
print('nope')
elif test == 'mytest':
print('yay')
else:
print('something else')
You can find more information here : https://docs.python.org/3/tutorial/controlflow.html
EDIT:
As a general remark, you should not define variable using capital letters (PEP convention: https://www.python.org/dev/peps/pep-0008/?)

So using Elif
if Pass == Again:
print("Thank you for choosing a working password, please create your character")
print("Please type your username without numbers")
elif Pass != Again:
print("Something's wrong with your password or username!")
Though what is this error you're getting, we can't really help you without it.

converting logic tests into try statements (python3)

I have been trying to convert some code into a try statement but I can't seem to get anything working.
Here is my code in pseudo code:
start
run function
check for user input ('Would you like to test another variable? (y/n) ')
if: yes ('y') restart from top
elif: no ('n') exit program (loop is at end of program)
else: return an error saying that the input is invalid.
And here is my code (which works) in python 3.4
run = True
while run == True:
spuriousCorrelate(directory)
cont = True
while cont == True:
choice = input('Would you like to test another variable? (y/n) ')
if choice == 'y':
cont = False
elif choice == 'n':
run = False
cont = False
else:
print('This is not a valid answer please try again.')
run = True
cont = True
Now what is the proper way for me to convert this into a try statement or to neaten my code somewhat?
This isn't a copy of the mentioned referenced post as I am trying to manage two nested statements rather than only get the correct answer.

If you want to make your code neater, you should consider having
while run:
instead of
while run == True:
and also remove the last two lines, because setting run and cont to True again isn't necessary (their value didn't change).
Furthermore, I think that a try - except block would be useful in the case of an integer input, for example:
num = input("Please enter an integer: ")
try:
num = int(num)
except ValueError:
print("Error,", num, "is not a number.")
In your case though I think it's better to stick with if - elif - else blocks.

Ok so as a general case I will try to avoid try...except blocks
Don't do this. Use the right tool for the job.
Use raise to signal that your code can't (or shouldn't) deal with the scenario.
Use try-except to process that signal.
Now what is the proper way for me to convert this into a try statement?
Don't convert.
You don't have anything that raises in your code, so there is no point of try-except.
What is the proper way to neaten my code somewhat?
Get rid of your flag variables (run, cont). You have break, use it!
This is prefered way of imlementing do-while, as Python docs says; unfortunately, I cannot find it to link it right now.
If someone finds it, feel free to edit my answer to include it.
def main()
while True: # while user wants to test variables
spuriousCorrelate(directory) # or whatever your program is doing
while True: # while not received valid answer
choice = input('Would you like to test another variable? (y/n) ')
if choice == 'y':
break # let's test next variable
elif choice == 'n':
return # no more testing, exit whole program
else:
print('This is not a valid answer please try again.')

Exit while loop by user hitting ENTER key

I am a python newbie and have been asked to carry out some exercises using while and for loops. I have been asked to make a program loop until exit is requested by the user hitting <Return> only. So far I have:
User = raw_input('Enter <Carriage return> only to exit: ')
running = 1
while running == 1:
Run my program
if User == # Not sure what to put here
Break
else
running == 1
I have tried: (as instructed in the exercise)
if User == <Carriage return>
and also
if User == <Return>
but this only results in invalid syntax.
Please could you advise me on how to do this in the simplest way possible.
Thanks

I ran into this page while (no pun) looking for something else. Here is what I use:
while True:
i = input("Enter text (or Enter to quit): ")
if not i:
break
print("Your input:", i)
print("While loop has exited")

The exact thing you want ;)
https://stackoverflow.com/a/22391379/3394391
import sys, select, os
i = 0
while True:
os.system('cls' if os.name == 'nt' else 'clear')
print "I'm doing stuff. Press Enter to stop me!"
print i
if sys.stdin in select.select([sys.stdin], [], [], 0)[0]:
line = raw_input()
break
i += 1

Actually, I suppose you are looking for a code that runs a loop until a key is pressed from the keyboard. Of course, the program shouldn't wait for the user all the time to enter it.
If you use raw_input() in python 2.7 or input() in python 3.0, The program waits for the user to press a key.
If you don't want the program to wait for the user to press a key but still want to run the code, then you got to do a little more complex thing where you need to use kbhit() function in msvcrt module.
Actually, there is a recipe in ActiveState where they addressed this issue. Please follow this link
I think the following links would also help you to understand in much better way.
python cross platform listening for keypresses
How do I get a single keypress at a time
Useful routines from the MS VC++ runtime
I hope this helps you to get your job done.

Use a print statement to see what raw_input returns when you hit enter. Then change your test to compare to that.

This works for python 3.5 using parallel threading. You could easily adapt this to be sensitive to only a specific keystroke.
import time
import threading
# set global variable flag
flag = 1
def normal():
global flag
while flag==1:
print('normal stuff')
time.sleep(2)
if flag==False:
print('The while loop is now closing')
def get_input():
global flag
keystrk=input('Press a key \n')
# thread doesn't continue until key is pressed
print('You pressed: ', keystrk)
flag=False
print('flag is now:', flag)
n=threading.Thread(target=normal)
i=threading.Thread(target=get_input)
n.start()
i.start()

You need to find out what the variable User would look like when you just press Enter. I won't give you the full answer, but a tip: Fire an interpreter and try it out. It's not that hard ;) Notice that print's sep is '\n' by default (was that too much :o)

if repr(User) == repr(''):
break

a very simple solution would be, and I see you have said that you
would like to see the simplest solution possible.
A prompt for the user to continue after halting a loop Etc.
raw_input("Press<enter> to continue")

user_input=input("ENTER SOME POSITIVE INTEGER : ")
if((not user_input) or (int(user_input)<=0)):
print("ENTER SOME POSITIVE INTEGER GREATER THAN ZERO") #print some info
import sys #import
sys.exit(0) #exit program
'''
#(not user_input) checks if user has pressed enter key without entering
# number.
#(int(user_input)<=0) checks if user has entered any number less than or
#equal to zero.
'''

Here is the best and simplest answer. Use try and except calls.
x = randint(1,9)
guess = -1
print "Guess the number below 10:"
while guess != x:
try:
guess = int(raw_input("Guess: "))
if guess < x:
print "Guess higher."
elif guess > x:
print "Guess lower."
else:
print "Correct."
except:
print "You did not put any number."

You are nearly there. the easiest way to get this done would be to search for an empty variable, which is what you get when pressing enter at an input request. My code below is 3.5
running = 1
while running == 1:
user = input(str('Enter <Carriage return> only to exit: '))
if user == '':
running = 0
else:
print('You had one job...')

I recommend to use u\000D. It is the CR in unicode.

Here's a solution (resembling the original) that works:
User = raw_input('Enter <Carriage return> only to exit: ')
while True:
#Run my program
print 'In the loop, User=%r' % (User, )
# Check if the user asked to terminate the loop.
if User == '':
break
# Give the user another chance to exit.
User = raw_input('Enter <Carriage return> only to exit: ')
Note that the code in the original question has several issues:
The if/else is outside the while loop, so the loop will run forever.
The else is missing a colon.
In the else clause, there's a double-equal instead of equal. This doesn't perform an assignment, it is a useless comparison expression.
It doesn't need the running variable, since the if clause performs a break.

If you want your user to press enter, then the raw_input() will return "", so compare the User with "":
User = raw_input('Press enter to exit...')
running = 1
while running == 1:
Run your program
if User == "":
break
else
running == 1

The following works from me:
i = '0'
while len(i) != 0:
i = list(map(int, input(),split()))