User input-based if statement gives unexpected output

Question

I am making a little text based game to practise my python skills. I am struggling to get the if statement to show the correct result based on what my user inputs.

weapon_choice = str(input("You can choose between three weapons to defeat the beast!"
                   " Press 1 for Axe, 2 for Crossbow, 3 for Sword."))

if input(1):
    print("You chose the Axe of Might")
elif input(2):
    print("You chose the Sacred Crossbow")
else:
    print("You chose the Elven Sword")

I would expect the output to ask me for a number (1, 2, or 3) and then print the string that is associated with that number. Instead, when I input 1, it prints 1, then 2, and then the string associated with number 3 (the 'else' option), regardless of what number I type. I don't understand why?

Greetings, weary wanderer.
Welcome to Freyjaberg. Choose your weapon.
You can choose between three weapons to defeat the beast! Press 1 for Axe, 2 for Crossbow, 3 for Sword.1
1
2
You chose the Elven Sword

Process finished with exit code 0

What did you expect if input(1) to do other than prompt the user with a 1? — Robin Zigmond
– Robin Zigmond, Commented Sep 21, 2019 at 8:24

Arkistarvh Kltzuonstev · Accepted Answer · 2019-09-21 08:32:29Z

1

Try this:

weapon_choice = input("You can choose between three weapons to defeat the beast!\nPress 1 for Axe, 2 for Crossbow, 3 for Sword.\n")

if weapon_choice=='1':
    print("You chose the Axe of Might")
elif weapon_choice=='2':
    print("You chose the Sacred Crossbow")
else:
    print("You chose the Elven Sword")

Note :

weapon_choice don't need to be cast into str format as input() will be already in string formatting.
Whenever you do input(1) or input(2) it basically prompts user to give another input in stead of checking with the condition.

Output :

michael@arkistarvh:/$ python text_game.py 
You can choose between three weapons to defeat the beast! Press 1 for Axe, 2 for Crossbow, 3 for Sword.
3
You chose the Elven Sword

edited Sep 21, 2019 at 8:32

answered Sep 21, 2019 at 8:25

Arkistarvh Kltzuonstev

6,9837 gold badges32 silver badges62 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

nephele Over a year ago

I see! Thank you so much for explaining this! I was trying to refer to the 'input' the same way you do a list if that makes sense (thinking if the user input 1, then show this). I see the what I did wrong now.

Stoner · Accepted Answer · 2019-09-21 08:25:15Z

1

You should have instead:

weapon_choice = input("You can choose between three weapons to defeat the beast! \n " +  " Press 1 for Axe, 2 for Crossbow, 3 for Sword.")

if weapon_choice == '1':
    print("You chose the Axe of Might")
elif weapon_choice == '2':
    print("You chose the Sacred Crossbow")
elif weapon_choice == '3':
    print("You chose the Elven Sword")
else:
    print("Invalid input selected")

The reason for this is that input(..) results in a string being parsed, so there is no need to have a str(..) around the input(..). Furthermore, you should have a condition where an invalid input is passed, so that the user will be notified more clearly of the root cause of the error.

(Note the \n which signifies the start of a new line)

answered Sep 21, 2019 at 8:25

Stoner

9361 gold badge10 silver badges33 bronze badges

5 Comments

nephele Over a year ago

Thank you! I will add the 4th option for invalid input, that's a great idea!

nephele Over a year ago

Hi @Stoner! I added the 'invalid input' option, but I want it to go back through the loop. I have tried

while True:     weapon_choice <= '4'     if weapon_choice=='1':         print("You chose the Axe of Might")         break     elif weapon_choice=='2':         print("You chose the Sacred Crossbow")         break     elif weapon_choice=='3':         print("You chose the Elven Sword")         break     else:         weapon_choice >= '4'         print("Wanderer, there is no such weapon. Please choose from the ones available.")         continue

but it doesn't work :(

nephele Over a year ago

I get

Traceback (most recent call last):   File "/Users/elissavetvelli/PycharmProjects/Giraffe/2dlist.py", line 10, in <module>     weapon_choice <= '4' NameError: name 'weapon_choice' is not defined

. I have tried lots of things, and at best I got it to show ```Wanderer, there is no such weapon. Please choose from the ones available`` infinitely!

Stoner Over a year ago

@nephele you should have 'weapon_choice' outside of the loop. If you have it inside the loop, it will get "refreshed" during each iteration, hence the infinite loop. In any case, it is sort of a convention here in the community to have 1 question per problem, so it'll be best if you post further questions not related to the original problem as a new, separate question. People here may not check back to closed questions so your new question may get missed out. :')

nephele Over a year ago

Thank you! I will do that from now on - and thanks for taking the time to explain.

bruno0 · Accepted Answer · 2019-09-21 08:30:02Z

1

That's just not how input works in Python.

You correctly assumed that input("some text") would print you that text on the first line (and store the result in a variable weapon_choice), so why do you think that input(number) would return a boolean value telling you if the input was that number?

Instead, what it does is prints the number again and returns an empty string (because you presumably just pressed Enter), so the first two ifs are False and the program goes into else, printing "Invalid input selected".

The result of your first input will be stored in weapon_choice, so you should do the comparing on that variable.

answered Sep 21, 2019 at 8:30

bruno0

1761 silver badge4 bronze badges

Comments

Robin Zigmond · Accepted Answer · 2019-09-21 08:28:10Z

0

if input(1) calls the input function with the argument 1, before using the response to test whether or not to execute the code in its block. Since executing input simply prints its argument to the terminal, and returns whatever the user inputs in response, your program is behaving exactly as your code implies.

Since you already stored the user's response to the first question in a variable called weapon_choice, you should replace if input(1) with if weapon_choice == '1' - and the same for the others.

answered Sep 21, 2019 at 8:28

Robin Zigmond

18.3k2 gold badges25 silver badges38 bronze badges

Comments

ankit singh · Accepted Answer · 2019-09-21 08:37:50Z

0

you can try this script

weapon_choice = str(input("You can choose between three weapons to defeat the beast!"
                   " Press 1 for Axe, 2 for Crossbow, 3 for Sword."))
if weapon_choice == '1':
    print("You chose the Axe of Might")
elif weapon_choice == '2':
    print("You chose the Sacred Crossbow")
else:
    print("You chose the Elven Sword")

edited Sep 21, 2019 at 8:37

answered Sep 21, 2019 at 8:34

ankit singh

5754 silver badges9 bronze badges

1 Comment

roganjosh Over a year ago

No, you can't. raw_input doesn't even exist in Python 3, and Python 2 is soon to be retired

Collectives™ on Stack Overflow

User input-based if statement gives unexpected output

5 Answers 5

1 Comment

5 Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

5 Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related