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BRIEF: I am trying to write a simple program that checks if a given code is valid. The code is valid if all the three characters entered are digits between 0 and 9. If the code is valid, your progam displays a message Thank you. We will process your order!. The output of your program should be an error message The verification code was not valid. Please check your credit card code again. if the input is not valid.

ATTEMPT: 

lol = raw_input("what's your code?")
rofl=lol.split()
for int in rofl:
 if 9>int(rofl)>0:
   print "Thank you. We will process your order!"
 else:
   print 'The verification code was not valid. Please check your credit card code again.'

PURPOSE:

To learn - I found another way to do this involving just checking the whole thing for being under 1000 and over 0, and it worked perfectly. But I'm just trying to learn more.

REQUEST:

That you give me explicit direction on how to fix my code and provide any additional relevant advice.

CLOSE:

Thank you in advance.

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3
  • 9
    tip: don't name your variables lol or rofl, give them meaning, such as userInput, inputAsArray,... Commented Sep 5, 2011 at 14:16
  • 4
    Another problem: you first rebind "int" in the "for int in rofl:" line, but then you call int(rofl), assuming that it's still the builtin int. Commented Sep 5, 2011 at 14:33
  • 1
    1) regarding what's wrong with my code - some people have already identified it so thank you to them just read on 2) yeh it pretty much is - it's exersizes from a textbook though i'm not studying in an institution or being marked just learning for myself this is one of tasks to try out 3) thanks for the advice - some of the common ones are foo and bar right? 4) I'm a bit confused about what your telling me. I'm sorry - I would really like to know - could you explain slightly differently? -thank you all. Commented Sep 5, 2011 at 15:24

4 Answers 4

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4

What you're trying to do is to check if given input matches a given pattern (three digits in your case), that's basically what regular expressions are made for.

Using a regex, your code would look like this:

import re
code = raw_input("what's your code? ")
if re.match(r'^\d{3}$', code):
    print "Code OK"
else:
    print "Wrong Code!"

In the regex, ^ means "the beginning of the string", \d means "a digit" (it could also be [0-9]), {3} means "repeated 3 times", and $ means "the end of the string".

For more informations about regular expressions, have a look at the re module documentation.

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4 Comments

+1 regex is probably the right approach for this kind of problems. Your answer will be match better if you add some link explaining the basics of regex in python.
this is good, but will actually accept the first 3 digits of a longer number, so you should use re.match(r'^\d{3}$', code)
corrected regex to match only 3 digits (as suggested by @steabert), and added more explanations.
Hi. I'm not familiar with regex - do you suggest I learn this now or wait till I'm up to it?
3

you could:

a = raw_input()

if a.isdigit() and 0<int(a)<1000 and len(a)==3:
    print "yeah!"
else:
    print "Nope."

isdigit() is a cool built-in function that checks strings to see if they are made up entirely of digits! You don't have to wrap the int(a) in a try/except because 'and' acts as a short-circuit operator that will prevent int(a) from being evaluated if isdigit() returns false.

EDIT: added len(a)==3.

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1 Comment

nice alternative, but this will not check if a actually consists of 3 digits, e.g. one might want have people enter zeros explicitly to avoid possible typo's, e.g. 042
2

Your first problem is that .split() doesn't split the string into characters, it splits the string at whitespace.

"12 34 56".split() == ["12", "34", "56"]
  • To go through each character in the input string, don't use .split(), just use for character in string.
  • I find it hard to read code with variable names like lol and rofl, so I've changed this.
  • Your check of 9 > int(digit) > 0 excludes 9 and 0. I assume this is an accident, so I'm using 0 <= int(digit) <= 9 and including them instead.
  • Your current code prints an error message each time it sees an invalid character. Instead I'll break out of the loop after I print it the first time, and print the success message if the loop completes without being broken.
  • If int(digit) fails to convert digit to an int, it will raise an exception. Instead, I'll just compare the character itself to see if it's between "0" and "9", instead of converting it.
code = raw_input("what's your code?")

for digit in code:
    if not ("0" <= digit <= "9"):
        print "The verification code was not valid. Please check your credit card code again."
        break
else:
    print "Thank you. We will process your order!"

A compact alternative to using a loop is to use a generator expression with the any function:

code = raw_input("what's your code?")

if all("0" <= digit <= "9" for digit in code):
    print "Thank you. We will process your order!"
else:
    print "The verification code was not valid. Please check your credit card code again."
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7 Comments

this will give a ValueError when digit is not a digit, I guess one doesn't want that?
@steabert: True, I've modified it to just compare the characters.
also, I think he wanted to check that it is exactly 3 characters long, no?
somehow 'break' has managed to allude me - what does it do in this context? Also I really like your answer of all of them here. May I ask about why when I replace you line: if not ("0" <= digit <= "9"): with if not 0 <= digit <= 9: - and test it, I get out a 'response' to each of the digits individually - and incorrect too!? - thanks for your help.
@rooney: break stops running the loop, and skips the code in the following else:. I'm not sure why that happens, in fact I'm surprised Python doesn't give you an error for comparing an integer to a digit character.
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lol = raw_input("what's your code?")
rofl=lol.split()
# split() doesn't perform as you think.
#         it separates parts of string according to space, tab and line ends.

for int in rofl:
# this is a loop using 'int' as variable name, not a good idea at all 
#   as int is a standard type  
 if 9>int(rofl)>0:
# rofl is a list.
# if you don't redefine int, and use "for d in lol" instead to loop the digits
# then int(d) witll throw an exception if it's not a digit
   print "Thank you. We will process your order!"
 else:
   print 'The verification code was not valid. Please check your credit card code again.'
# You don't test there is exactly 3 digits.
# And you reply to your test in the loop, so once per item in the loop
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