not printing after "if not" instruction in python

Question

< I have the following code:

print("Options:")
print("Option 1")
print("Option 2")
print("Option 3")
print("Option 4")
choice = int(input("What would you like to choose [1,2,3,4]? "))
while choice not in ['1','2','3','4']:
    choice = int(input("What would you like to choose [1,2,3,4]? "))
    if not choice:
        print: ("Please enter 1, 2, 3 or 4. ")

However, the output when running the module is:

What would you like to do [1,2,3,4]? 5 What would you like to do [1,2,3,4]?

I want this to loop until 1,2,3, or 4 is selected & produce the following output:

What would you like to do [1,2,3,4]? 7 Please enter 1, 2, 3 or 4.

What would you like to do [1,2,3,4]?

Where am I going wrong?

@chepner - you were on the case too quick (impressive) - I was still proofreading! Have amended. — YVB
– YVB, Commented May 14, 2020 at 11:21

chepner · Accepted Answer · 2020-05-14 11:22:45Z

2

if not choice: is equivalent to if choice == 0: in this context, since you will only reach that statement after input has returned and int hasn't raised an exception.

Use the following idiom for a potentially infinite loop that doesn't duplicate the call to input:

...
print("Option 4")
while True:
    choice = int(input("..."))
    if choice in [1, 2, 3, 4]:
        break
    print("Please enter 1, 2, 3, or 4.")

answered May 14, 2020 at 11:22

chepner

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Comments

user12188239 · Accepted Answer · 2020-05-14 11:21:11Z

0

The colon after the line print: ("Please enter 1, 2, 3 or 4. ") might be causing the string to not be printed. I don't know why that doesn't cause a syntax error though.

A pretty rudimentary solution.

print("Options:")
print("Option 1")
print("Option 2")
print("Option 3")
print("Option 4")

choice = "0"
while choice not in ['1','2','3','4']:
    choice = input("What would you like to choose [1,2,3,4]? ")
    if choice not in ['1', '2', '3', '4']:
        print ("Please enter 1, 2, 3 or 4. ")

Try it online!

answered May 14, 2020 at 11:21

user12188239

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