我正在尝试仅访问电子邮件的正文,并设法使用电子邮件模块和email.getpayload()函数来获取此邮件。唯一的问题是,似乎有“垃圾文本”出现,它似乎采用不同的格式,数量和内容,具体取决于谁发送电子邮件。有没有办法只访问电子邮件的真实主体?谢谢!
这就是我所拥有的:
msg = email.message_from_bytes(data[0][1])
body = ''
if msg.is_multipart():
for part in msg.walk():
if part.is_multipart():
for subpart in part.get_payload():
if subpart.is_multipart():
for subsubpart in subpart.get_payload():
body = body + str(subsubpart.get_payload(decode=True)) + '\n'
else:
body = body + str(subpart.get_payload(decode=True)) + '\n'
else:
body = body + str(part.get_payload(decode=True)) + '\n'
else:
body = body + str(msg.get_payload(decode=True)) + '\n'
答案 0 :(得分:0)
from email import message_from_file
import os
# Path to directory where attachments will be stored:
path = "./msgfiles"
# To have attachments extracted into memory, change behaviour of 2 following functions:
def file_exists (f):
"""Checks whether extracted file was extracted before."""
return os.path.exists(os.path.join(path, f))
def save_file (fn, cont):
"""Saves cont to a file fn"""
file = open(os.path.join(path, fn), "wb")
file.write(cont)
file.close()
def construct_name (id, fn):
"""Constructs a file name out of messages ID and packed file name"""
id = id.split(".")
id = id[0]+id[1]
return id+"."+fn
def disqo (s):
"""Removes double or single quotations."""
s = s.strip()
if s.startswith("'") and s.endswith("'"): return s[1:-1]
if s.startswith('"') and s.endswith('"'): return s[1:-1]
return s
def disgra (s):
"""Removes < and > from HTML-like tag or e-mail address or e-mail ID."""
s = s.strip()
if s.startswith("<") and s.endswith(">"): return s[1:-1]
return s
def pullout (m, key):
"""Extracts content from an e-mail message.
This works for multipart and nested multipart messages too.
m -- email.Message() or mailbox.Message()
key -- Initial message ID (some string)
Returns tuple(Text, Html, Files, Parts)
Text -- All text from all parts.
Html -- All HTMLs from all parts
Files -- Dictionary mapping extracted file to message ID it belongs to.
Parts -- Number of parts in original message.
"""
Html = ""
Text = ""
Files = {}
Parts = 0
if not m.is_multipart():
if m.get_filename(): # It's an attachment
fn = m.get_filename()
cfn = construct_name(key, fn)
Files[fn] = (cfn, None)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# Not an attachment!
# See where this belongs. Text, Html or some other data:
cp = m.get_content_type()
if cp=="text/plain": Text += m.get_payload(decode=True)
elif cp=="text/html": Html += m.get_payload(decode=True)
else:
# Something else!
# Extract a message ID and a file name if there is one:
# This is some packed file and name is contained in content-type header
# instead of content-disposition header explicitly
cp = m.get("content-type")
try: id = disgra(m.get("content-id"))
except: id = None
# Find file name:
o = cp.find("name=")
if o==-1: return Text, Html, Files, 1
ox = cp.find(";", o)
if ox==-1: ox = None
o += 5; fn = cp[o:ox]
fn = disqo(fn)
cfn = construct_name(key, fn)
Files[fn] = (cfn, id)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# This IS a multipart message.
# So, we iterate over it and call pullout() recursively for each part.
y = 0
while 1:
# If we cannot get the payload, it means we hit the end:
try:
pl = m.get_payload(y)
except: break
# pl is a new Message object which goes back to pullout
t, h, f, p = pullout(pl, key)
Text += t; Html += h; Files.update(f); Parts += p
y += 1
return Text, Html, Files, Parts
def extract (msgfile, key):
"""Extracts all data from e-mail, including From, To, etc., and returns it as a dictionary.
msgfile -- A file-like readable object
key -- Some ID string for that particular Message. Can be a file name or anything.
Returns dict()
Keys: from, to, subject, date, text, html, parts[, files]
Key files will be present only when message contained binary files.
For more see __doc__ for pullout() and caption() functions.
"""
m = message_from_file(msgfile)
From, To, Subject, Date = caption(m)
Text, Html, Files, Parts = pullout(m, key)
Text = Text.strip(); Html = Html.strip()
msg = {"subject": Subject, "from": From, "to": To, "date": Date,
"text": Text, "html": Html, "parts": Parts}
if Files: msg["files"] = Files
return msg
def caption (origin):
"""Extracts: To, From, Subject and Date from email.Message() or mailbox.Message()
origin -- Message() object
Returns tuple(From, To, Subject, Date)
If message doesn't contain one/more of them, the empty strings will be returned.
"""
Date = ""
if origin.has_key("date"): Date = origin["date"].strip()
From = ""
if origin.has_key("from"): From = origin["from"].strip()
To = ""
if origin.has_key("to"): To = origin["to"].strip()
Subject = ""
if origin.has_key("subject"): Subject = origin["subject"].strip()
return From, To, Subject, Date
Usage:
f = open("message.eml", "rb")
print extract(f, f.name)
f.close()
我使用邮箱为我的邮件组编程了这个,这就是为什么它如此复杂。它永远不会让我失望。从来没有任何垃圾。如果message是multipart,则输出字典将包含一个键“files”(一个子字典),其中包含提取的其他非文本或html文件的所有文件名。这是一种提取附件和其他二进制数据的方法。您可以在pullout()中更改它,或者只是更改file_exists()和save_file()的行为。 construct_name()构造一个消息id和多部分消息文件名的文件名,如果有的话。也许这里有一些错误,因为它打算使用mailbox.Message,而不是使用email.Message。但这是我目前所拥有的,并且不能告诉你更多,直到你向我们提供更多信息。也许你只是忘记了decode = True并且正在获得一些rfc822 junky crappy end of file。