Andrej Mitrovic wrote:
On Apr 20, 1:06 pm, MRAB <pyt...@mrabarnett.plus.com> wrote:
Dodo wrote:
Hello,
I don't understand why this won't execute
import urllib.request as u
import socket
socket.setdefaulttimeout(10)
l = "http://img144.imageshack.us/my.php?image=koumakandg8.jpg" #
supposed to timeout
try:
h = u.urlretrieve(l)
except u.URLError, e: # I tried u.e too, no effect.
print(e)
except:
print("other error")
The error :
...\Python>err.py
File "...\err.py", line 8
except u.URLError, e: # I tried u.e too, no effect.
^
SyntaxError: invalid syntax
In Python 3 it's:
except u.URLError as e:
This a because in Python 2 people sometimes write:
except OSError, IOError:
thinking that it will catch both OSError and IOError.
except (OSError, IOError), e: # Python 2.x
If you put them in a tuple, it will catch them, right?
In Python 2.x:
except (OSError, IOError), e:
In Python 3.x (and also Python 2.6):
except (OSError, IOError) as e:
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