CODEWITHSUNDEEP
pythonstringinputinteger
Dark Leviathan
Dark Leviathan

Reputation: 489

How to look for a text in an input

I am having a problem with this very simple calculator script. If a user enters a letter or a word (such as: s, the), the system would crash. My idea is to make it print:

print ("WARNING: That is not a valid equation!")

When that happens. However, I cannot just make it see if y is an integer because otherwise, *, +, -,and / would not work. Also, I need a way for it to skip the 

exec(x)
print(x)

part so it does not give me an error and crash the script. Thank you in advance!

Code (Python v3.3.0 - Mac OSX 10.8):

while True:
    x = "x="
    y = input(" >> ")
    x += y
    exec(x)
    print(x)

Edit: After looking at raser's answer. This is what I changed it to. It is a mix of my previous code and both of his answers.

                valid_chars = "0123456789-+/* \n";
                while True:
                    x = "x="
                    y = input(" >> ")
                    x += y
                    if False in [c in valid_chars for c in y]:
                        print("Errors!");
                        continue;
                    if(y == "end" or y == "End" or y == "exit" or y == "Exit" or y == "cancel" or y == "Cancel"):
                         break
                    exec(x)
                    print(x)

This is capable of doing equations such as: 7+9/3*2-3

Upvotes: 0

Views: 95

Answers (3)

Nick Beeuwsaert
Nick Beeuwsaert

Reputation: 1638

you could catch the exception:

#!/usr/bin/env python
while True:
    equation = raw_input(">> ");
    try:
        exec(equation);
    except SyntaxError:
        print("WARNING! That is not a valid function");

or if you want to detect anything not in 0-9*/-= you could just use a regex (which is probably overkill) or do something like this:

#!/usr/bin/env python
valid_chars = "0123456789-+/* \n";
while True:
    equation = raw_input(">> ").strip();
    if False in [c in valid_chars for c in equation]:
        print("Errors!");
        continue;
    exec(equation);

Upvotes: 2

nicholsonjf
nicholsonjf

Reputation: 1001

Here's a really simple solution. Could be done with a lot less code, but it works (once each time you run the script):

cursor = 'calculator: '
x = raw_input(cursor)
if x.isdigit() == False:
    print 'Please enter a number first'
y = raw_input(cursor)
operators = ['*', '+', '-', '/']
if y not in operators:
    print 'Please enter one of the following operators: '.format(','.join(operators))
z = raw_input(cursor)
if z.isdigit() == False:
    print 'Please enter a number after the operator'
else:
    if y == '*':
        print int(x) * int(z)
    elif y == '+':
        print int(x) + int(z)
    elif y == '-':
        print int(x) - int(z)
    else:
        print int(x) / int(z)

Upvotes: 0

asdf
asdf

Reputation: 3067

If you want to check if an input is an integer place a regex match statement before you run integer operations against it.

import re

while True:
    x = "x="
    y = input(" >> ")
    if re.match("^[0-9]{0,}$", y):
        x += y
        exec(x)
        print(x)
    else:
        raise TypeError("WARNING: That is not a valid equation!")

If you want to learn more about how regex works look here --> http://docs.python.org/2/library/re.html

Upvotes: 0

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