How to send a non-standard IMAP command?

[Xianwen Chen]

On Thu, Jun 24, 2010 at 2:25 PM, Tim Chase <python.list at tim.thechases.com>wrote:

> On 06/24/2010 04:47 AM, Xianwen Chen wrote:
>
>> Thanks a lot for your reply! I thought it would be simpler if the
>> problem was presented in a brief way. Unfortunately, not for this
>> case.
>>
>> Here is the detail. Free Yahoo! mail accounts can be accsessed via
>> IMAP protocal, however, a non-standard shake hand code is needed
>> before log in [1]:
>>
>> ID ("GUID" "1")
>>
>> . This is what I'm now working for. I tried:
>>
>> IMAP4.xatom('','ID ("GUID" "1")','',)
>>
>> and
>>
>> dest_srv.xatom('ID ("GUID" "1")')
>>
>> , but I got error messages. Any hint please?
>>
>
> In general, it would be helpful to include the error-message(s) you get.
>  However, I tried it with a junk Yahoo account I set up:
>
>  from imaplib import IMAP4
>  i = IMAP4("imap.mail.yahoo.com")
>  USER = 'yourusername at yahoo.com'
>  PASS = 'your secret goes here'
>  # per the Wikipedia page you gave
>  # the ID has to happen before login
>  i.xatom('ID ("GUID" "1")')
>
>  i.login(USER, PASS)
>  i.select()
>  typ, data = i.search(None, 'ALL')
>  for num in data[0].split():
>    typ, data = i.fetch(num, '(RFC822)')
>    message = data[0][1].splitlines()
>    subject = [line
>      for line in message
>      if line.lower().startswith('subject: ')
>      ][0]
>    print num, subject
>  i.close()
>  i.logout()
>
> and it worked.
>
> -tkc
>
>
>
>

Hi Tim,

The problem was the password. I was careless. Thanks for your advice. Next
time I'll have error codes posted.

And thanks a lot for your constructive example!

I have a strange problem that

"M = imaplib.IMAP4_SSL(M_addr)
M.debug = 2"

doesn't work. No verbose output at all. Any hint please?

Best regards,

Xianwen


-- 
Xianwen Chen
University of Tromsø
http://u.nu/9uubc
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