syntax error in python:

Question

Hello,

I'm learning python. It gives syntax error in this script. I'm unable to figure out.

import exceptions
class FOUND(Exception): pass

x = [1,2,3,4,56,73,29,35,12,32,63,12,76,75,89]


while True:

    try:
        test = int(raw_input('Enter integer to be checked in list:'))
        count = -1
        for y in x:
            count += 1
            if y == test:
                raise FOUND
    except ValueError as e:
        print "Not a valid integer (%s)"%(e)
    except FOUND:
        print "Found (%d) at (%d)"%(test,count)
    else:
        print "Not found ,Appending (%d) to list at location (%d)"%(test,count+1)
        x.append(test)
    finally:
        print "The List:"
        print x
        print " "

Invalid syntax & it highlights closing double quote in this line: print "Not a valid integer (%s)"%(e)

Answer 1

You need an empty line between the class ... and x = ...

Answer 2

Try except ValueError as e:, the older syntax you use is invalid in Python 3.

Answer 3

Your code (cut and pasted, no alterations) works fine for me (Python 2.5).

BTW, your test = int... line should be after the try (and indented appropriately) and the %d in "Not a valid integer (%d)" should be a %s.

Syntax for exception handling has been changed for Python 3: make sure any help/tutorials you are following are for the same major version of Python you have installed. There have been signficant changes from 2.x to 3.x.

Answer 4

print without brackets is from python 2, if you are using python 3, you need to use print().

You can't format an exception as %d - %d is for integers.