基本的Python If语句使用“或”

Question

I am following a beginner program for learning python. I am struggling to find a better way to use an "if" statement with multiple possibilities. I am writing a basic text based game in which the user is able to choose three different difficulty levels: Easy, Medium, Hard. I am just trying to write code that takes in a 1, 2, or 3 to tell the program which level to run. I want to make sure the user input is one of those options, not "4" or "one".

Here is the relevant code I have written:

selected_level = 0
selected_level = raw_input("\n Please select a level by entering 1, 2, or 3 : ")
print selected_level
if selected_level == 1 or selected_level == 2 or selected_level == 3:
    print "Pass"
else:
    print "Fail"
break

All I am trying to do right now is test that the "if" statement works. However, no matter what input I enter it will print the input and then print "Fail" which tells me it is never entering the first part of the "if" statement.

I have tried testing this with inputs 1, 2, 3, 4, one, etc and it always skips to the "else" statement. Is there a better way to write an "if" statement with multiple or values?

Answer 1

Your issue is that raw_input returns a string, but the if-statement is comparing to ints.

"1" does not equal 1.

Try checking if selected_level == "1" ... "2"... etc. and it should work.

I would compare to strings instead of casting the input to an int, as that will crash the program if the user types something unexpected.

Also for a more succinct check:

if selected_level in ("1", "2", "3"):
    print "Pass"

Answer 2

raw_input returns a string. Try int(raw_input(...))

Answer 3

The reason it keeps skipping down to the if statement is because it is comparing the string that you entered to the numbers 1, 2, etc and not to the string value of 1 and 2. If you want to take the numeric value of the input you could do something like:

selected_level = int(raw_input("\\n Please select a level by entering 1, 2, or 3 : "))

This will take the integer value of whatever the user inputs.

Answer 4

raw_input (in Python 2.x) returns a str . so selected_level is a str , not int . You must parse it to int before compare it woth int values

selected_level = 0
selected_level = int(input("\n Please select a level by entering 1, 2, or 3 : "))
print selected_level
if selected_level == 1 or selected_level == 2 or selected_level == 3:
    print "Pass"
else:
    print "Fail"
break