Python If--elif-else Usage And Clarification

''' This program presents a menu to the user and based upon the selection made invokes already existing programs respectively. ''' import sys def get_numbers(): '''get the upper

Solution 1:

It's about the line if choice == 'x' or 'X'.

Correctly, it should be

if choice == 'x' or choice == 'X'

or simpler

if choice in ('X', 'x')

because the or operator expects boolean expressions on both sides.

The current solution is interpreted as follows:

if (choice == 'x') or ('X')

and you can clearly see that 'X' does not return a boolean value.

Another solution would be of course to check whether if the uppercase letter equals 'X' or the lowercase letter equals 'x', which might look like that:

if choice.lower() == 'x':
    ...

Solution 2:

Your problem is with your if choice == 'x' or 'X': part.To fix that change it to this:

if choice.lower() == 'x':

Solution 3:

ifchoice== 'x' or 'X':

is not doing what you think it's doing. What actually get's parsed is the following:

if (choice == 'x') or ('X'):

You probably want the following:

ifchoice== 'x'orchoice== 'X':

which can be written as

if choice in('x', 'X'):

Solution 4:

As the interpreter says, it is an IndentationError. The if statement on line 31 is indent by 4 spaces, while the corresponding else statement is indent by 5 spaces.