Incorrectly breaking out of a python while loop?

Question

I am trying to build a Twitter filter app that searches your personal timeline of people you follow/RTs from people you follow from keywords of your choosing. (That's not the question, that's the background... I'm not up to that yet, I'm just playing around with the API at the moment!)

I'm just starting to learn Python, I'm previously a Java programmer and I'm wondering how to check user input for valid... well, input!

I have a numbered menu (currently with just 2 items in) and I want the user to type 1 or 2 and if they don't, for it to throw an error message and loop back to the input. I'm currently getting the error message of:

Traceback (most recent call last):
  File "Filtwer.py", line 31, in <module>
    if "1" in menuselect:
TypeError: argument of type 'int' is not iterable

Where line 31 is the start of the if statement in the code block below. I'm not sure if I'm missing something? Not breaking out of the while loop correctly for example? Any help would be much appreciated!

Thanks :)

import twitter
api = twitter.Api(consumer_key='<redacted>',
                  consumer_secret='<redacted>',
                  access_token_key='<redacted>',
                  access_token_secret='<redacted>')

while True:
    menuselect = input("1. Tweet\n2. Get Tweets\n...: ")
    if menuselect == 1 or 2: break
    print "Please enter a valid entry!"

if "1" in menuselect:
    statusinput = raw_input("Tweet: ")
    status = api.PostUpdate(statusinput)
    print "Sucessfully tweeted!"

else:
    timeline5 = api.GetUserTimeline(user_id=<my_twitter_ID>, screen_name='<my_twitter_screenname>', count=5, include_rts='true')
    print [s.text for s in timeline5]

Edit:

Got it to work like this (included comments to show how my answer differed from the answer I chose as correct. Thanks for the help guys! :) )

while True:
#try:
    menuselect = raw_input("1. Tweet\n2. Get Tweets\n...: ")
    if menuselect == "1" or menuselect == "2": break
#catch ValueError:
#   pass
#finally:
    print "Please enter a valid entry!"

if "1" == menuselect:
[...]

Answer 1

To check if menuselect is 1 , you should use == operator NOT in operator. (Because in operator is used to check the member existence)

if 1 == menuselect:

Note: Also, don't use input function in Python 2.x, to get input from the user (because of the security concerns). Use raw_input instead and convert the result to int manually, like this

menuselect = int(raw_input("1. Tweet\n2. Get Tweets\n...: "))

Note 2: You might want to use in operator in the break condition (it is correct here, because you are checking whether the value of menuselect is one of the possible values), like this

if menuselect in (1, 2): break

because

if menuselect == 1 or 2:

will always evaluate to True , as it is evaluated as (menuselect == 1) or 2 , even if menuselect is not 1, the 2 part will make the expression evaluate to Truthy.

Edit: To get around the exception part, when a string is entered instead of integers, you can use try..except like this

while True:
    try:
        menuselect = input("1. Tweet\n2. Get Tweets\n...: ")
        if menuselect int (1, 2): break
    catch ValueError:
        pass
    finally:
        print "Please enter a valid entry!"

Answer 2

Another problem you have is with this line:

if menuselect == 1 or 2: break

Python interprets this as

if (menuselect == 1) or 2: break

Meaning it checks if menuselect is one, and if it is false, it checks if 2 is true. In python, non-zero numbers are interpreted as true, so this condition will always be true.

Change this to either

if menuselect in [1, 2]: break

or the more lengthy

if menuselect == 1 or menuselect == 2: break