[Tutor] Syntax error help

Evert Rol evert.rol at gmail.com
Fri Mar 30 22:50:36 CEST 2012

> Alright i have been trying to right a (relatively) simple to calculate area and volume below is my current working code
> def areamenu():
>     print 'Square (1)'
>     print 'triangle (2)'
>     print 'rectangle (3)'
>     print 'trapazoid (4)'
>     print 'circle (5)'
> def squareacalc():
>     sidelength = input('enter side length: ')
>     print ' the side length is' sidelength ** 2
> def displaymenu():
>     print 'please make a selection';
>     print 'Area (1)';
>     choice = input(raw_input('enter selection number'):

You're missing a closing parenthesis here.

But see comments below.

>     if (choice == 1):
>         Areamenu():
>     else:
>         print 'choice' , choice, ' is wrong try again'
> def selctiona():
>     Areamenu();
>     choicea = input(raw_input'enter selection');

And here you're missing an openening parenthesis.

>     if (choicea == 1):
>         squareacalc()
>  print 'good bye'
> I keep getting this error
> Traceback (most recent call last):
>   File "<pyshell#3>", line 1, in <module>
>     import Area
>   File "C:\Python27\Area.py", line 10
>     areamenu()
>            ^
> SyntaxError: invalid syntax
> can anyone tell me what im doing wrong i cant see the problem
> help would be appreciated

A syntax error is often a typing error in the code. In this case, forgotten parentheses, but could be forgotten colons or an unclosed string.
The parentheses problem can often be caught by using a good editor: these often lightlight when you close a set of parentheses, so you can spot syntax errors while you are typing.

There are a number of other things wrong here, though.
Style-wise: Python does not need (and prefers not to have) closing semicolons. In addition, there is no need to surround if statements with parentheses: "if choice == 1:" is perfectly fine and much more legible.
Perhaps you think (coming from another language): "but it doesn't hurt, and I like it this way". But then you're still programming in that other language, and just translating to Python; not actually coding in Python.

Also, this is odd, wrong and pretty bad:

   choice = input(raw_input('enter selection number')):

Use either raw_input() (for Python 2.x) or input() (Python 3.x).
It's wrong because you are waiting for input, then use that input as the next prompting string for further input. Like this
>>> choice = input(raw_input('enter selection number'))
enter selection number1
>>> print choice

(the 12 is the 1 I entered before, plus a 2 I just entered as a second entry.)
So just use one function, and the appropriate one for the Python version.
Lastly, choice will be a string, since input() and raw_input() return a string.
In the if-statement, however, you are comparing that string to an integer. Python does not do implicit conversion, so you'll have to convert the string to an integer first, or compare to a string instead.
(Something similar happens for the sidelength, btw.)
Last thing I see glancing over the code: you define areamenu(), but call Areamenu(). Python is case sensitive, so you'd have to call areamenu() instead.

This may be a bit more information than you asked for, but hopefully you don't mind.

Good luck,


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