Python imaplib抓取电子邮件gmail [英] Python imaplib fetch body emails gmail
问题描述
import email,getpass,imaplib,os
detach_dir =F:\PYTHONPROJECTS#将保存附件的
user = raw_input(输入您的GMail用户名 - >)
pwd = getpass.getpass(输入您的密码 - >)
#连接到gmail imap服务器
m = imaplib.IMAP4_SSL(imap .gmail.com)
m.login(user,pwd)
m.select(PETROLEUM)#这里你可以选择一个像INBOX这样的邮箱,而不是
#使用m.list()获取所有邮箱
resp,items = m.search(无,'(FROMEIA_eLists@eia.gov)')
items = items [0] .split()#获取邮件id
my_msg = []#存储相关msgs在这里请
msg_cnt = 0
break_ = False
项目[:: - 1]中的emailid:
resp,data = m.fetch(emailid,(RFC822))
if(break_):
break
for response_part in data:
if isinstance(response_part,tuple):
msg = email.message_from_string(response_part [1])$ b $ b varSubject = msg ['subject']
如果varSubject [0] =='$':
msg_cnt + = 1
my_msg.append(msg)
打印msg_cnt
打印email.message_from_string(response_part [1])
if(msg_cnt == 5):
break_ = True
如果我打印 email.message_from_string(response_part [1])
,我可以看到它包含第一个信息(标题,从,到,日期...),全文本体。但是,我不能抓住身体本身。 email.message_from_string(response_part [0])
打印邮件IDS和 email.message_from_string(response_part [2])
超出范围。 email.message_from_string(response_part [1] [0])
既没有这样做。
谢谢, / p>
更新
现在,我几乎可以拥有正文。然而,它仍然受到首先发布的信息声明的破坏。我得到一个结果
从nobody Tue Dec 25 11:42:58 2012
US = 3D $ 4.030
EastCst = 3D $ 4.036
NewEng = 3D $ 4.205
CenAtl = 3D $ 4.149
LwrAtl = 3D $ 3.921
Midwst = 3D $ 3.984
GulfCst = 3D $ 3.945
RkyMt = 3D $ 4.195
WCst = 3D $ 4.187
CA = 3D $ 4.268
我想摆脱从nobody Tue Dec 25 11:42:58 2012
哪个是信息。我知道我可以解析第一个相关行的文本寻找...我知道。
实现这个(插入我的第一个样本)的代码是
如果varSubject [0] =='$':
你有一个更好的方法(没有信息字符串)?/ pre>
r,d = m.fetch(emailid,(UID BODY [TEXT ])
msg_cnt + = 1
my_msg.append(msg)
print email.message_from_string(d [0] [1])$ b $ b
更多:现在获取日期的命令是什么?我知道我可以在上面适用的varDate = msg ['date']
,但是如何只获取日月 - 年?感谢解决方案您可以通过执行以下任何一项获取身体的内容:
msg.as_string()
str(msg)
repr(msg)
http://docs.python.org/2.7/library/email.message.html#email.message.Message
I read this already and wrote this script to fetch body for emails in some mail box which title begins with '$' and is sent by some sender.
import email, getpass, imaplib, os detach_dir = "F:\PYTHONPROJECTS" # where you will save attachments user = raw_input("Enter your GMail username --> ") pwd = getpass.getpass("Enter your password --> ") # connect to the gmail imap server m = imaplib.IMAP4_SSL("imap.gmail.com") m.login(user, pwd) m.select("PETROLEUM") # here you a can choose a mail box like INBOX instead # use m.list() to get all the mailboxes resp, items = m.search(None, '(FROM "EIA_eLists@eia.gov")') items = items[0].split() # getting the mails id my_msg = [] # store relevant msgs here in please msg_cnt = 0 break_ = False for emailid in items[::-1]: resp, data = m.fetch(emailid, "(RFC822)") if ( break_ ): break for response_part in data: if isinstance(response_part, tuple): msg = email.message_from_string(response_part[1]) varSubject = msg['subject'] if varSubject[0] == '$': msg_cnt += 1 my_msg.append(msg) print msg_cnt print email.message_from_string(response_part[1]) if ( msg_cnt == 5 ): break_ = True
if I print
email.message_from_string(response_part[1])
, I can see it contains first information (header, from, to, date...), the the full text body. But, I cannot fetch the body itself.email.message_from_string(response_part[0])
prints mails IDS, andemail.message_from_string(response_part[2])
is out of range.email.message_from_string(response_part[1][0])
neither is doing it.Thanks and regards.
UPDATE
Now, I can almost have body text. However, it is still spoilt by an information statement coming first. I get as a result
From nobody Tue Dec 25 11:42:58 2012 US=3D$4.030 EastCst=3D$4.036 NewEng=3D$4.205 CenAtl=3D$4.149 LwrAtl=3D$3.921 Midwst=3D$3.984 GulfCst=3D$3.945 RkyMt=3D$4.195 WCst=3D$4.187 CA=3D$4.268
and I would like to get rid of
From nobody Tue Dec 25 11:42:58 2012
which is information. I know I could parse text look for first relevant line... i know.The code for achieving so (to plug in my first sample) is
if varSubject[0] == '$': r, d = m.fetch(emailid, "(UID BODY[TEXT])") msg_cnt += 1 my_msg.append(msg) print email.message_from_string(d[0][1])
Do you have a better way (no info string) ??? More: what is the command to now fetch the date ? I know that I can do
varDate = msg['date']
where suited above, but how to just fetch day-month-year ? THANKS解决方案You can get the contents of the body by doing any of the following
msg.as_string() str(msg) repr(msg)
http://docs.python.org/2.7/library/email.message.html#email.message.Message
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