How do you ask a question in Python 3?

Question

I'm coding a game, and I don't know how to get the program to let my player type in an answer. Can someone help me, please?

Answer 1

You can search the Internet for various documentations and tutorials on Python.

One of them is on W3School that describes how to handle user input in Python:

https://www.w3schools.com/python/python_user_input.asp

Comments

Why the dislikes?

Answer 2

For Example if you want the user to enter his/age you can prompt the user by using this line of code : 

age = int(input("Enter your age : "))

for accepting the name and other details : 

name = input('Enter your name : ')

Answer 3

To ask a simple question like what school you go to, do this:

school = input("What school do you go to? : ")

print("You go to " , school)

If you want to ask a different question, just replace the variable with what the question is about,  replace the school question with yours then,  print you start of the answer.

Thx!

Answer 4

simple,if you need ask to the player "whats the name of your caracther" for example you can write:

print('Name of your character:')

name = input('')

print("the name of your caracter is {}").name #this line can be wrong, mayble the correct is this:print("the name of your caracter is {name}") i dont rememeber but this you dont use now

Answer 5

In Python 2.7 =  username = raw_input("Enter username:") - This method is for Python2 which is approaching EOL.

In Python3 = username = raw_input("Enter username:")

Interestingly, Python will read the response as a string, so if you want to read a number and treat it as a number, I would recommend age = int(input("Enter your age : "))

Answer 6

player=input('enter anwer')

x=str (player)

Answer 7

Use the "input()" Function

Put the question in the parenthesis

(I recommend using a recent input variable such as "RecentImput" and putting this line of code:

RecentImput = input("Your question here")

Then use an if command to see if the input is any one of a few inputs you will allow)