Python While Loop with multiple conditions

Question

Can anyone help as to why my while loop isn't working? It's forcing the user to select either 1, 2 or 3, and not letting them proceed, however regardless of whether you put in 1, 2 or 3, it always says that you've entered a different number, and so says "Please choose level 1, 2 or 3"

level = input("Enter your level by typing 1, 2 or 3\n")
int(level)

levelSelect = 1
while levelSelect == 1:
   if level != int(1) or level != 2 or level != 3:
      level = input("Please choose level 1, 2 or 3\n")
      int(level)
   else:
      print("You have selected level", level)
      levelSelect = 0

As another side note, int(1) is always able to be replaced by 1. — mgilson
– mgilson, Commented Jan 9, 2013 at 12:47
...Unless you've done something silly like rebind the name int. — Karl Knechtel
– Karl Knechtel, Commented Jan 9, 2013 at 13:15
You need to change your "or" to "and" in your if statement. It will always fail because level can't be 1 2 and 3. — Rocky Pulley
– Rocky Pulley, Commented Jan 9, 2013 at 15:33

mgilson · Accepted Answer · 2013-01-09 12:40:11Z

4

The line int(level) doesn't do what you think it does. It creates an integer from a string and returns it. It does not operate in place. Because of this, when you get to your if statement, you're comparing a string to integers which is always unequal.

You probably want:

level = int(level)

As a side note, the condition could also be written using the in operator:

if level in (1,2,3):
   print("level is ...")
else:
   print("pick again!")
   #other code ...

answered Jan 9, 2013 at 12:40

mgilson

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Comments

Jdog · Accepted Answer · 2013-01-09 13:27:15Z

1

See mgilson's answer regarding assign level as an integer but your logic in deciding if the 'level' is legitimate is also wrong.

Change:

if level != int(1) or level != 2 or level != 3:

for:

if level not in (1,2,3)

edited Jan 9, 2013 at 13:27

answered Jan 9, 2013 at 12:44

Jdog

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1 Comment

mgilson Over a year ago

Why do you say that logic is wrong? -- The style isn't ideal, but I don't think I see a problem with the logic.

Volatility · Accepted Answer · 2013-01-09 12:47:30Z

0

mgilson has indicated the core problem here, however I will make a suggestion to improve your code.

while True:
    level = input('Enter level: ')
    if level not in ('1','2','3'):
        print('Try again!\n')
    else:
        print('You chose level ', level)
        break

answered Jan 9, 2013 at 12:47

Volatility

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Comments

Rich Tier · Accepted Answer · 2013-01-09 16:50:01Z

0

I think this is what you want (props to with Inbar Rose)

def get_level():
    while True:
        level = int(input("Enter your level by typing 1, 2 or 3\n"))
            if level in [1, 2, 3]:
                return level

edited Jan 9, 2013 at 16:50

answered Jan 9, 2013 at 12:45

Rich Tier

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1 Comment

Inbar Rose Over a year ago

why not put the whole thing into the while loop? pastebin.com/ArVpMNwF :

def choose_level():     while True:         level = int(input("Enter your level by typing 1, 2 or 3\n"))         if level in [1, 2, 3]:             return level

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