On Nov 12, 1:46 pm, Laszlo Nagy <[EMAIL PROTECTED]> wrote:
> KeefTM wrote:
> > Hello, I am getting an odd error when trying to establish an IMAP
> > connection:
>
> > File "/Library/Frameworks/Python.framework/Versions/2.4//lib/python2.4/
> > imaplib.py&q
Nevermind. It always seems I figure out what I did wrong right after I
post. Turns out I was using the wrong port.
On Nov 12, 12:51 pm, KeefTM <[EMAIL PROTECTED]> wrote:
> Hello, I am getting an odd error when trying to establish an IMAP
> connection:
>
> File &q
Hello, I am getting an odd error when trying to establish an IMAP
connection:
File "/Library/Frameworks/Python.framework/Versions/2.4//lib/python2.4/
imaplib.py", line 904, in _get_response
raise self.abort("unexpected response: '%s'" % resp)
imaplib.abort: unexpected response: '220 libertydis
On Nov 2, 3:36 pm, "Andrew Koenig" <[EMAIL PROTECTED]> wrote:
> <[EMAIL PROTECTED]> wrote in message
>
> news:[EMAIL PROTECTED]
>
> > I would want to sort by name first, then sub sort by location. Any
> > ideas? Thanks!
>
> In Python 2.3 and later, sorting is stable -- so you can sort successively
On Nov 2, 3:36 pm, "Andrew Koenig" <[EMAIL PROTECTED]> wrote:
> <[EMAIL PROTECTED]> wrote in message
>
> news:[EMAIL PROTECTED]
>
> > I would want to sort by name first, then sub sort by location. Any
> > ideas? Thanks!
>
> In Python 2.3 and later, sorting is stable -- so you can sort successively
On Nov 2, 2:45 pm, [EMAIL PROTECTED] wrote:
> Hello, I have been sorting a list of dicts using the following
> function:
>
> result_rs = sorted(unsort_rs, key=itemgetter(orderby))
>
> and this works fine. Now I am looking to perform a subsort as well.
> For example, I have this:
>
> test = [{'name'
Hello, I have been sorting a list of dicts using the following
function:
result_rs = sorted(unsort_rs, key=itemgetter(orderby))
and this works fine. Now I am looking to perform a subsort as well.
For example, I have this:
test = [{'name': 'John Smith', 'location': 'CA',},{'name': 'John
Smith', '
