On 08/07/14 02:41, Pamela Wightley wrote:
Hi All,

I have no programming experience and am trying to  teaching myself
python. Am trying to replicate the code below but I get the error
message below highlighted in yellow:

Unfortunately I can't see anything in yellow, probably due to enmail losing the formatting.

But the code below has several errors and looks like its part
of a bigger program. It also looks completely unrelated to the
errors you show us further down.


choice = input("Choose your option: ")

You seem to be using Python 2 in which case using input() is
frowned upon. Its better to use raw_input() and convert the
resultant string to a number using int() or float() as needed.
[ In Python 3 input() has been rwe oved and raw_input renamed to input()...]


     if choice == 1:

indentation(spacing) is very importanbt in Python.
You should only indent the code inside a code block that follows a structural statement such as if/for/while/def/class etc.

Indenting the if statement after the input() will raise an error.

         add1 = input("Add this: ")


But this should e indented because its part of the if block.


         add2 = input("to this: ")

         print add1, "+", add2, "=", add1 + add2

     elif choice == 2:

         sub2 = input("Subtract this: ")

         sub1 = input("from this: ")

         print sub1, "-", sub2, "=", sub1 - sub2

     elif choice == 3:

         mul1 = input("Multiply this: ")

         mul2 = input("with this: ")

         print mul1, "*", mul2, "=", mul1 * mul2

     elif choice == 4:

         div1 = input("Divide this: ")

         div2 = input("by this: ")

         print div1, "/", div2, "=", div1 / div2

     elif choice == 5:

         loop = 0

Any assistance appreciated.

ERROR MESSAGE

return input (1)

   File "<stdin>", line 1

SyntaxError: 'return' outside function


This seems completely unrelated to the choice code?
You can only use return inside a function (ie a block
starting def....)


HTH
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.flickr.com/photos/alangauldphotos

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