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From: Tim Chase on 24 Jun 2010 10:25
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(a)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



From: Xianwen Chen on 25 Jun 2010 04:06
On Jun 24, 2:23 pm, Michael Torrie <torr...(a)gmail.com> wrote:
> On 06/24/2010 03:47 AM, Xianwen Chen wrote:
>
> > , but I got error messages. Any hint please?
>
> Why not just use a proxy server:
>
> http://sourceforge.net/projects/imapidproxy/

Hi Michael

Thanks a lot!

Yes, it would be much more convenient. Actually, I have been using a
modified version of Thunderbird [1] to access Yahoo! Mail IMAP server
for months. The reason of using Python to access the mail server is to
have less dependency on other software. Now I don't need Thunderbird
anymore.

Best regards,

Xianwen

[1] http://www.crasseux.com/linux/
From: Xianwen Chen on 25 Jun 2010 04:06
On Jun 24, 2:25 pm, Tim Chase <python.l...(a)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 = 'yourusern...(a)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
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