I'm trying to kill a program in python 3 when the user says 'no' to start a maths quiz, : here is the code I'm using
import sys
while True:
new_item = input("> ")
if new_item == "Yes" or "yes":
break
elif new_item == "no":
sys.exit()
that doesn't work any pointers?
Your problem is here:
if new_item == "Yes" or "yes":
You need to use either:
if new_item in ["Yes", "yes"]:
or:
if new_item == "Yes" or new_item == "yes"
Your original code is parsed as:
if (new_item == "Yes") or "yes":
and this always evaluates to True since "yes" is a true value.
if new_item == "Yes" or "yes":
This conditional is always True. It may be stated as:
(new_item == "Yes") or ("yes")
Non-empty string 'yes' is always evaluated to True.
Change conditional to:
if new_item in ['Yes', 'yes']:
You need to change your if statement, it isn't evaluating properly. You need to use this code to fix your problem:
import sys
while True:
new_item = input("> ")
if new_item == "Yes" or new_item == "yes":
break
elif new_item == "no":
sys.exit()
What about
import sys
while True:
new_item = input("> ")
new_item = new_item.lower()
#everything you wrote in input will be lowercase, no more "or" problems
if new_item.lower() == "yes":
break
elif new_item.lower() == "no":
sys.exit()
Related
If my_input == "n" I want to my program to loop again, which works fine.
But if my else statement is True I dont want it to run the whole program again and just "start" at the my_input variable.
How can I achieve this?
def name_user_validation():
while True:
full_name = input("What is your name? ")
print(f"Hello {full_name}, nice to meet you.")
full_name.split()
print(f"If I understood correctly, your first name is {full_name[0]} and your last name is {full_name[-1]}.")
my_input = input("Is that right? (y/n) ")
if (my_input == "y"):
print("Great!")
break
elif my_input == "n":
print("Oh no :(")
else:
print("Invalid input, try again.")
name_user_validation()
I misunderstood your question, I would probably restructure your code a bit, so you get rid of your while loops and use recursive function calling to go back when you need to,
something like the below
def name_user_validation():
full_name = input("What is your name? ")
print(f"Hello {full_name}, nice to meet you.")
full_name.split() # This line actually doesn't do anything
print(f"If I understood correctly, your first name is {full_name[0]} and your last name is {full_name[-1]}.")
if not accept_input():
name_user_validation()
def accept_input():
my_input = input("Is that right? (y/n) ")
if my_input == "y":
print("Great!")
return True
elif my_input == "n":
print("Oh no :(")
return False
else:
print("Invalid input, try again.")
accept_input()
name_user_validation()
Add another loop that doesn't terminate until user enters acceptable input.
def name_user_validation():
while True:
full_name = input("What is your name? ")
print(f"Hello {full_name}, nice to meet you.")
full_name.split()
print(f"If I understood correctly, your first name is {full_name[0]} and your last name is {full_name[-1]}.")
while True:
my_input = input("Is that right? (y/n) ")
if (my_input == "y"):
print("Great!")
break
elif my_input == "n":
print("Oh no :(")
break
else:
print("Invalid input, try again.")
if my_input == 'y':
break
name_user_validation()
Edit: The program terminates only when my_input = y.
This question already has answers here:
How to test multiple variables for equality against a single value?
(31 answers)
Closed 2 years ago.
ans = input("Enter yes or no: ")
if ans != "Yes" or "yes" or "no" or "No":
print("Can't do that")
if ans == "yes" or "Yes":
print("Great!")
if ans == "no" or "No":
print("Okay, then")
If I type let's say "Okay" it outputs this:
Can't do that!
Great!
Okay, then
instead of "Can't do that". I don't know what's wrong, and I couldn't find questions like this.
do instead something more pythonic:
if ans.lower() in ['no', 'yes']:
and use elif instead of doing another if verification.
Use in:
ans = input("Enter yes or no: ")
if ans in ["Yes", "yes"]:
print("Great!")
elif ans in ["No", "no"]:
print("Okay, then")
else:
print("Can't do that")
Test whether the answer is in a list or a set (using sets in the example below). Otherwise, your first condition evaluates to True. This is because of the operator precedence, Python considers it equivalent to (ans != "Yes") or ("yes") or ("no") or ("No"). And "yes" is True because it is not an empty string (docs), which makes the whole expression evaluate to True as well.
ans = input("Enter yes or no: ")
if ans not in {"Yes", "yes", "no", "No"}:
print("Can't do that")
if ans in {"yes", "Yes"}:
print("Great!")
if ans in {"no" or "No"}:
print("Okay, then")
Better still, make it shorter like so:
ans = input('Enter yes or no: ').lower()
if ans == 'yes':
print('Great!')
elif ans == 'no':
print('Okay, then')
else:
print("Can't do that")
That's not the way to use logical operators in a programming language, it should be:
if (ans != "Yes") or (ans != "yes") or (ans != "no") or (ans != "No")
As you see, you should always repeat your variable.
the or operator assignment is not right please try this
ans = input("Enter yes or no: ")
if ans != "Yes" or "yes" or "no" or "No":
print("Can't do that")
if ans == "yes" or "Yes":
print("Great!")
if ans == "no" or "No":
print("Okay, then")
you can either write
if ans in ('stringone', 'stringtwo'):
dosomething()
Or you can write separate equality tests,
if var == 'stringone' or var == 'stringtwo':
dosomething()
How can I go about creating a input and output with Y\N (yes or no) function in a question?
My example; if my question is Would you like some food? (Y \ N):, how can I do this and have the answers show Yes, please. or No, thank you. for either choice and then proceed to the next question with the same function?
I thought about using this: valid=("Y": True, "y": True, "N": False, "n": False) but that only shows up as True or False for me, or is there a way to change from True \ False to Yes \ No? or this one:
def user_prompt(yes_no):
while True:
user_input=input(yes_no)
But I'm really not sure how else to proceed with this one, or if there's any other easier solution to this.
I think what you're looking for is a conditional print statement rather than a true/false return statement on the function.
For example:
def user_prompt():
while True:
user_input = input("Would you like some food? (Y \ N)")
print ("Yes, please" if user_input == 'Y' else "No, thank you")
Or, more readable:
def user_prompt():
while True:
user_input = input("Would you like some food? (Y \ N)")
if (user_input == 'Y'):
print("Yes, please")
elif (user_input == 'N'):
print("No, thank you")
I hope I understood your question correctly, you basically check the first letter (incase user enters yes/no) every time the user enters a value and try to verify that if it's Y/N you break the loop if not you keep asking the same question.
def user_prompt(yes_no):
while True:
user_input=input(yes_no)
if user_input[0].lower() == 'y':
print("Yes, please.")
break
elif user_input[0].lower() == 'n':
please("No, thank you.")
break
else:
print("Invalid, try again...")
Not sure if this is the best way, but I have a class based implementation of same
""" Class Questions """
class Questions:
_input = True
# Can add Multiple Questions
_questions = [
'Question 1', 'Question 2'
]
def ask_question(self):
counter = 0
no_of_question = len(self._questions)
while self._input:
if counter >= no_of_question:
return "You have answred all questions"
user_input = input(self._questions[counter])
self._input = True if user_input.lower() == 'y' else False
counter += 1
return "You have opted to leave"
if __name__ == '__main__':
ques = Questions()
print(ques.ask_question())
Firstly there is many ways you can go around this, but I am guessing you found the solution yourself already but here is one that is the best.
def get_yes_no_input(prompt: str) -> bool:
allowed_responses = {'y', 'yes', 'n', 'no'}
user_input = input(prompt).lower()
while user_input not in allowed_responses:
user_input = input(prompt).lower()
return user_input[0] == 'y'
continue = get_yes_no_input('Would you like to proceed? [Y/N]: ')
And there we go.
This question already has answers here:
How to test multiple variables for equality against a single value?
(31 answers)
Closed 6 years ago.
For some reason, no matter what user_input is (Yes, No, "", eserw3) the first if statement will always be triggered. Any insight as to why the elif and the else never get activated? (The below code compiles perfectly without any errors)
Thank you in advance.
def retry():
user_input = raw_input("Would you like to face %s again? (Yes/No)" % (Enemy))
if user_input == "Yes" or "yes":
respawn()
getMove()
elif user_input == "No" or "no":
print "Thanks for playing!"
else:
print "Please enter either Yes or No."
def retry():
user_input = raw_input("Would you like to face %s again? (Yes/No)" % (Enemy))
if user_input == "Yes" or user_input == "yes":
respawn()
getMove()
elif user_input == "No" or user_input == "no":
print "Thanks for playing!"
else:
print "Please enter either Yes or No."
def retry():
user_input = raw_input("Would you like to face %s again? (Yes/No)" % (Enemy)).lower()
if user_input == "yes":
respawn()
getMove()
elif user_input == "no":
print "Thanks for playing!"
else:
print "Please enter either Yes or No."
Change your if condition to
user_input in ["Yes", "yes"]
Reason: When you write user_input == "Yes" or "yes", it evaluates as:
(user_input == "Yes") or "yes"
The second part of OR is a True always(non-zero length string). Hence your problem of if block executing always.
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.
Well my code is working, but when I type No if I want to retry to enter the password it doesn't work; it just goes to the enter password line (line 20). I have tried multiple ways to fix this but I simply cannot.
import time
import os
print ("Hello world.")
time.sleep(1)
print ("Waiting 5 seconds.")
time.sleep(5)
print ("You have waited 10 seconds.")
print ("Executing Chrome.")
time.sleep(1)
print ("Execution failed!")
password = input("Enter the execution password: ")
if password == 'password1234':
os.system ('C:\\Users\\Harry\\AppData\\Local\\Google\\Chrome\\Application\\chrome.exe')
else:
print ("Wrong password!")
time.sleep(1)
passretry = input("Do you want to try again? ")
if passretry == 'yes' or 'Yes':
passretry1 = input("Enter password: ")
if passretry1 == 'password1234':
os.system ('C:\\Users\\Harry\\AppData\\Local\\Google\\Chrome\\Application\\chrome.exe')
elif passretry == 'no' or 'No':
print ("Closing...")
time.sleep(1)
else:
print ("Wrong password.")
time.sleep(.5)
print ("Retry limit exceeded, closing.")
time.sleep(1)
if passretry == 'yes' or 'Yes':
the above if statement is evaluated as: -
if (passretry == 'yes') or 'Yes':
Now, since 'Yes' is evaluated to True, so, your if statement is always True, and hence you always have to enter new password.
You need to change the condition to: -
if passretry in ('yes', 'Yes'):
likewise, the following elif should be changed to: -
elif passretry in ('no', 'No'):
This condition:
if passretry == 'yes' or 'Yes':
means "If passretry == 'yes' is true, or 'Yes' is true". 'Yes' is always true, because a non-empty string counts as true. That's why you're always taking the first code path.
You need to spell things out a little more:
if passretry == 'yes' or passretry == 'Yes':
(Or to make your code a bit more general:
if passretry.lower() == 'yes':
which would allow for people shouting YES.)
You need another complete statement:
passretry == 'yes' or passretry == 'Yes':
The string 'Yes' always evaluates to True.