Python traceback issues

Question

So I've been learning the basics of Python and threw this bit of code together but i'm having a few issues. So I basically want it to check if the user input matches a value in the 'con' variable and if it matches to print correct, if it's wrong print not recognised.

 #countries.py con = [ "uk" , "japan" , "us" ] uInput = input("Enter the country: ") if uInput == con: print("Correct") else: print("Not Recognised")

Also, I'd like to add lower() to the users input, so the capitalisation doesn't affect the results but don't really know where to add it.

Like I say, I'm new to coding so go easy!

Cheers

user2555451 · Accepted Answer · 2014-03-11 17:29:17Z

2

con is a list and uInput is a string. Meaning, they will never be equal.

Instead, you want to use in here:

if uInput in con:

The above code will test if the value of uInput can be found in con.

Then, you can add str.lower for case-insensitive searching:

if uInput.lower() in con:

This code will test if a lowercase version of the value of uInput can be found in con.

answered Mar 11, 2014 at 17:29

user2555451

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2 Comments

Jam3sn Over a year ago

Ok, so I have done that but it spits out this error in the terminal: Enter the country: uk Traceback (most recent call last): File "<stdin>", line 4, in <module> File "<string>", line 1, in <module> NameError: name 'uk' is not defined

user2555451 Over a year ago

@user3407221 - You must be using Python 2.x. Use raw_input to get the user input instead of input. The former will return a string object where as the latter will try to evaluate its input.

jonrsharpe · Accepted Answer · 2014-03-11 17:30:04Z

You can chain that method straight onto the string that input() returns:

uInput = input("Enter the country: ").lower()

Note also that the input will never be == con, as con is a list and the input is a string; instead, try:

if uInput in con:

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