get exception type in python 1.5.2

Question

How can I get the type of an Exception in python 1.5.2?

doing this:

try:
    raise "ABC"
except Exception as e:
    print str(e)

gives an SyntaxError:

  except Exception as e:
                    ^

SyntaxError: invalid syntax

EDIT: this does not work:

try:
    a = 3
    b = not_existent_variable
except Exception, e:
    print "The error is: " + str(e) + "\n"

a = 3
b = not_existent_variable

as I only get the argument, not the actual error (NameError):

The error is: not_existent_variable

Traceback (innermost last):
  File "C:\Users\jruegg\Desktop\test.py", line 8, in ?
    b = not_existent_variable
NameError: not_existent_variable

Answer 1

It's

except Exception, e:

In Python 1 and 2. (Although as also works in Python 2.6 and 2.7).

(Why on earth are you using 1.5.2!?)

To then get the type of the error you use type(e). To get the type name in Python 2 you use type(e).__name__, I have no idea if that works in 1.5.2, you'll have to check the docs.

Update: It didn't, but e.__class__.__name__ did.