Python program decision control issue

Question

I am trying to create a code library where I can store code snippets I find over time and retrieve them. Here is my code so far.

# Python Code Library User Interface

import sys

class CodeLib:
    codes = []
    option = ['help (-h)', 'list (-l)', 'import (-i)', 'quit (-q)']

#What does the user want to do?
    print "Welcome to the Code Library"
    print "What would you like to do? \n %s" % option
    choice = raw_input()
    print choice
    if choice == 'quit' or '-q':
        sys.exit()
    length = len(choice)
# Getting name of file to import if user chooses import before
# viewing the code list
    if choice[0] == 'i':
        _file_ = choice[6:length]
    elif choice[:2] == '-i':
        _file_ = choice[2:length]
# imports file if user chose import before viewing code list
    if choice[0:5] == 'import' or choice [0:2] == '-i':
        import _file_
# If user chose option other than import at beginning
# then perform option chosen then give ability to do something else
    while choice != 'quit' or '-q':
# provides basic help menu
        if choice == 'help' or '-h':
            print
            print '''(list or -l) brings up a list of code snippet names
(import or -i)/file_name brings up the code snippet you chose \n'''
# provides list of code files
        elif choice == 'list' or '-l':
            print codes
            print option[2], "file name"
            _file2_ = raw_input()
            import _file2_
# imports code file
        else:
            _file2_ = raw_input()
            import _file2_
# Next user option                  
        print "What would you like to do?"
        choice = raw_input

Now I know this is very incomplete so far. But the issue I am running into is that no matter what I enter as choice. The program does the first if statement, which exits the program. I have commented out the first if statement and tried again. But then no matter what I enter as choice. It performs this first if statement in the while loop and I just get stuck in a loop. So I'm not sure what is wrong but can someone help me out here?

This gets asked so very many times. I wonder if it'd be asked less if "if" and "or" were easier to search for. — user2357112, Aug 13 '13 at 23:30
Also [if x or y or z == blah](http://stackoverflow.com/questions/15112125/if-x-or-y-or-z-blah) and about 100 other questions. — abarnert, Aug 13 '13 at 23:30
Why do so many people coming from C++ or Java expect this to work? If you write `while (choice != "quit" || "-q")` in C++, it fails for exactly the same reason `while choice != "quit" or "-q"` does in Python… — abarnert, Aug 13 '13 at 23:32
It doesn't exactly "fail". It just doesn't do what OP expects it to do. — Hyperboreus, Aug 13 '13 at 23:34
@abarnert Because we say things like "if she's short or fat, I'm not interested." — 2rs2ts, Aug 13 '13 at 23:37
Sorry for the dumb question. However I may not have been clear enough when. I am basically relearning to program altogether because it has been a long time since I have done anything in c or java — Wickie Lee, Aug 13 '13 at 23:47
@2rs2ts That is the exact reason why I thought it should have worked the way I wrote it. But, as I said. I'm basically relearning everything. With just basic recollection of syntax and method names and a general idea of what they do — Wickie Lee, Aug 14 '13 at 00:24
While everyone else has pointed out the way that the boolean operator works, I should also note that your last line should be `choice = raw_input()`. You have to call `raw_input` or else you won't actually prompt for input. The difference here is that you must use the parentheses (`()`) to call or invoke that function. You don't have them, so, without getting too technical, you're assigning the function itself to `choice`. If your code is working now, then you didn't paste it correctly and you do have those parentheses there. — 2rs2ts, Aug 14 '13 at 01:03
Oh now i see what you are saying. My apologies on that. I must have left that out when pasting my code. the parentheses are there in my code. Sorry about the confusion and thanks for your input. — Wickie Lee, Aug 14 '13 at 02:41

score 7 · Accepted Answer · answered Aug 13 '13 at 23:29

7

The "or '-q'" part checks if the string '-q' is not false, which is true for any nonempty string. Change this line to:

if choice == 'quit' or choice == '-q':

answered Aug 13 '13 at 23:29

BartoszKP

34,786
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3

OP should also check out the argparse module for this kind of thing, it eliminates a lot of needles if-thenning http://docs.python.org/2/library/argparse.html#module-argparse – theodox Aug 13 '13 at 23:31
I am going to look into that argparse module in a bit. Thank you forb the link! I realize all the if statments are a really...elementary way of doing things and I am trying to become more efficient – Wickie Lee Aug 14 '13 at 00:26

score 3 · Answer 2 · answered Aug 13 '13 at 23:29

3

The condition choice == 'quit' or '-q' (and similar) always returns True, because '-q' evaulates to True and (True or Something) is always True.

You most probably wanted something like

choice == 'quit' or choice == '-q'

or

choice in ['quit', '-q']

answered Aug 13 '13 at 23:29

Hyperboreus

31,997
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Ah, now that makes a lit of sense. Thanks for the help! I am going to mess with it some more now and see where it gets me. – Wickie Lee Aug 13 '13 at 23:49
Okay so now everything is working correctly! Thanks a lot for the help! – Wickie Lee Aug 13 '13 at 23:59

score 0 · Answer 3 · answered Aug 13 '13 at 23:33

0

First of all, you are creating an object in a way that I don't believe to be standard. Typically, you define methods for a class with def func(self,arg): blocks inside the class foo: block. The initializer method is indicated by the name __init__(self,args). Note that there are two '_' characters before and after init.

If you just want to run some code in a simple python script, you don't need it to be inside any block. You can just do something like:

print "Welcome to the Code Library"
print "What would you like to do? \n %s" % option
choice = raw_input()
print choice
if choice == 'quit' or '-q':
    sys.exit()
length = len(choice)
#More code goes here

answered Aug 13 '13 at 23:33

cordoro

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This isn't really "non-standard". It is uncommon, and silly, and almost certainly not what the OP actually wants to do… but it work, and there are non-silly use cases for putting code (other than method definitions and class attribute assignments) directly in a class definition… – abarnert Aug 13 '13 at 23:49
1

Also, this isn't related to the OP's question or the problem he's trying to solve. – abarnert Aug 13 '13 at 23:49
Okay. thanks for the info on that. Ill get rid of the class for now until I understand more about methods and functions – Wickie Lee Aug 13 '13 at 23:50

Python program decision control issue

3 Answers3