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Why does my code make me type "yes" or "no" twice to get the result I want instead of just once?

Why does this code make me type yes or no twice to get the result I want instead of just once?
This is for the python dice roll text game, btw...
import random
min = 1
max = 20
# <!--TWO D-20's, A-LA DUNGEONS AND DRAGAONS--!>
roll_again = "yes"
while roll_again == "yes" or roll_again == "y":
print("Rolling the dice")
print("The values are --- ")
print(random.randint(min, max))
print(random.randint(min, max))
roll_again = input("Would you like to play again?")
answer = input()
if answer == ('yes'):print("OK, here we go!")
elif answer == ("no"):print("Sorry about that, please try again another time.")
I am entering into a python class on Monday, this is one of the more common types of beginners code (granted I spiced it up by changing the dice from 6 sided to 20 sided, but it's more fun that way...lol) that I have read in some of the python books and sites I have visited, so I wanted to kind of get my feet wet a bit before I start my class.
So, any idea why I have to type yes or no twice, hitting enter after each time, to get it to run properly?
For the record, I am on Win10 right now but I also mostly use Parrot Security OS (Linux)...
Thanks for any and all feedback which anyone can provide...I know it's probably a stupid noob mistake or oversight, since I don't really know or understand the basics, but the quicker I can grasp them the better...

Python's function input() asks for users to input and waits the answer. As it can be seen in your code, you're doing it two times:
roll_again = input("Would you like to play again?")
answer = input()
The variable roll_again is being redeclarated to the first user's input, then the variable answer is getting the second user's input. You maybe meant to do something like:
roll_again = input("Would you like to play again?")
answer = roll_again
But first of all, there is no need to create an answer variable, you could simply use the roll_again on the if. Also the if statement is out of the while so your code might not work as you're trying to anyways~ (it will re-roll if user inputs yes but it will not print the message you're trying to; that will only happen when the user inputs no as that will go out of the while and blah blah blah)
This should be alright:
import random
min = 1
max = 20
# <!--TWO D-20's, A-LA DUNGEONS AND DRAGAONS--!>
roll_again = "yes"
while roll_again == "yes" or roll_again == "y":
print("Rolling the dice")
print("The values are --- ")
print(random.randint(min, max))
print(random.randint(min, max))
roll_again = input("Would you like to play again?")
if roll_again == 'yes' or roll_again == 'y': print("OK, here we go!")
else: print("Sorry about that, please try again another time.")

Every time you call the input() function, it prompts for input. Since you call it twice for each iteration of your while loop, the user is prompted twice per iteration. You should instead only call the input() function once in your while loop.
You can also avoid using the answer variable if you just use the roll_again variable for your conditions for your if and elif.

Code always doing the same path regardless of input. (PYTHON) [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 last year.
I'm currently writing a program for a school project.
The purpose is for it to work as an online cash register.
answer = "a"
qty1 = 0
while True:
answer = str(input("\nDo you really wish to buy this? (Y/N) "))
if answer == "Y" or "y":
qty1 = int(input("\nHow much quantity of this item would you like to buy? "))
print("\nDo you really wish to buy", qty1, "pieces? (Y/N) ")
answer = str(input(""))
if answer == "Y" or "y":
print("Confirming order and returning to menu.")
break
else:
qty1=0
print("Cancelling order and returning to menu")
break
elif answer == "N" or "n":
print("Okay, returning to menu.")
break
else:
print("not valid answer")
Here is the code for the part of the program I'm having trouble with.
Whenever I reach this part of the program, the input seems to ignore whatever I put and it always goes through the if path.
Does anyone know why this is?
I'm new to programming, so sorry if this is just an easy fix.

Here's the working code which needed some formatting:
Changed these lines -> answer in ("Y", "y") and answer in ("N", "n"). This is because if answer == 'Y' or 'y' is used then 'y' will always evaluate to True which means the first if statement always executes. Also, the error in the print statement has a colon: print: - this was corrected.
answer = "a"
qty1 = 0
while True:
answer = str(input("\nDo you really wish to buy this? (Y/N) "))
print(answer)
if answer in ("Y", "y"):
qty1 = int(input("\nHow much quantity of this item would you like to buy? "))
print("\nDo you really wish to buy", qty1, "pieces? (Y/N) ")
answer = str(input(""))
print(answer)
if answer in ("Y", "y"):
print("Confirming order and returning to menu.")
break
else:
qty1=0
print("Cancelling order and returning to menu")
break
elif answer in ("N" ,"n"):
print("Okay, returning to menu.")
break
else:
print("not valid answer")

Continue outside of loop, dice rolling script

I'm making a dice rolling game. The following code produces an error, saying "continue outside of loop":
import random
repeat="y" or "yes"
while repeat == "y" or "yes"
print("Rolling the dice")
print(random.randint(1,6))
repeat = input("Do you wanna roll again Y/N?").lower()
if repeat =="y" or "yes":
continue
else:
print("Thanks for rolling")
break
Why am I getting this error?

You cannot affect 2 walues as teh same time, chose "y" or "Y" in your expressions.
repeat="y"
while (repeat == "y") or (repeat == "yes"):
You have to check your indentation too,
repeat = input has to be inside the loop, so you get a new input to check each time.
The final print can be put outside the loop at the very end.
Here is a working example, and you can type anything starting with "y"/"Y" to continue rolling:
import random
repeat="y"
while (repeat[0]) == "y":
print("Rolling the dice")
print(random.randint(1,6))
repeat = input("Do you wanna roll again Y/N?").lower()
print("Thanks for rolling")

Ask for input until exact string is input problem

This is my Python code:
choice = (input("Are you signed up? (y/n) "))
print(choice)
while True:
if choice != "y" or choice != "n":
choice = (input("Are you signed up? (y/n) "))
print(choice)
else:
break
I want the program to keep asking the user for input until they input "y" or "n". The problem is it never accepts the input, even if they do input "y" or "n".
When I print what they inputted (that's a real word) it says that it's "y" or "n" but it still won't accept it.
I've been using Python for some time now and I'm just trying something out today but I can't figure this out for some reason, I feel like it's something simple and obvious but I'm stupid to notice it.
Can someone explain this to me, please? Thank you.

Since choice cannot be 'y' and 'n' at the same time, one of choice != 'y' and choice != 'n' is always true.
You need to use and, not or.

Well, suppose the user inputs "y".
Your code will enter the while True loop and encounter the if condition
Is choice != "y" or choice != "n", where choice == 'y', true or false?
"y" != "y" is false, check the next condition
"y" != "n" is true, so the whole condition is true
Ask for input again
Goto (1)
Same thing happens for choice == "n".
However, you want a condition that says not (choice == "y" or choice == "n"). According to De Morgan's laws:
not (a or b) == (not a) and (not b)
So your code translates into:
(not (choice == "y")) and (not (choice == "n"))
=> choice != "y" and choice != "n"

As everyone has already pointed out the problem in your code, I am just adding a cleaned version of your code.
while True:
choice = (input("Are you signed up? (y/n) "))
if choice == "y" or choice == "n":
print(f"Your choice is {choice}")
break

You need and instead of or in this case.
if choice != "y" and choice != "n":
or if you prefer you can do
if not (choice == "y" or choice == "n"):
You can also do:
if choice not in ("y", "n"): # THIS ONE IS PROBABLY THE BEST
or because these are each 1-character you can do:
if choice not in "yn":
You can also do:
if choice not in "Randy!" or choice in "Rad!":

My python code has errors, but just in the try- except loop. [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 4 years ago.
My python code has some errors in the try- except loop. If you enter an input which is not yes or no, it first prints the "yes" reply output, then once you answer the question it shows the output for if you do not enter yes or no. Here's the code:
playAgain = None
while playAgain != "Yes" or "No" or "yes" or "no" or "y" or "n":
try:
playAgain = str(input("Do you want to play again? Enter Yes or No: "))
if playAgain == "Yes" or "y" or "yes":
displayIntro()
elif playAgain == "No" or "no" or "n":
print("Oh, well. The magic 8 ball will see you next time...")
sys.exit()
except:
print("That wasn't yes or no, idiot. The magic 8 ball will not give a fortune to such an imbocile.")
Please help and thank you!

playAgain != "Yes" or "No" or "yes" or "no" or "y" or "n"
Is not a correct way to do this.
When you say playAgain != "Yes" then you need to do the same for the remaining expression. So a valid way to do what you were intended to do is the following:
playAgain != "Yes" or playAgain != "No" or playAgain != "yes" or playAgain != "no" or playAgain != "y" or playAgain != "n"
But this is ugly and too long.
Instead, use
playAgain not in ["Yes", "No", "yes", "no", "y", "n"]
In Python, we have some handy ways to deal with such problems. You can use the in operator to check if the string in question exists (or does not exist) in a list of possible values. It's also very nice to read: "if playAgain (is) not in [this list of values]".
You can even manipulate the input so it's easier for you to work with. That is, you lower all the letters and you don't check for case-sensitive input (if you really don't care about case sensitive input; do you really care if is Yes or yEs?):
playAgain.lower() not in ["yes", "y"]
Something like this should do:
while True:
playAgain = str(input("Do you want to play again? Enter Yes or No: "))
if playAgain.lower() in ["yes", "y"]:
# do something with your yes input. Consider `break` out of the endless loop.
elif playAgain.lower() in ["no", "n"]:
# do something with your no input. Consider `break` out of the endless loop.
else:
print("That wasn't yes or no.")
Note that the above loop is endless. You need to break-out according to your program logic. That is, you need to put a break statement somewhere when you need to break out of the endless loop.