我不知道如何在 python 3.4 中加载 eml 文件。
我想列出所有并在 python 中阅读所有这些。
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4 回答
33
这就是您获取电子邮件内容的方式，即*.eml 文件。这在 Python2.5 - 2.7 上完美运行。在 3 上试一试。它应该也可以。
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写了这个程序，这就是它如此复杂的原因。它从来没有让我失望过。从来没有任何垃圾。如果消息是多部分的，则输出字典将包含一个键“文件”（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() 从消息 id 和多部分消息文件名（如果有的话）构造一个文件名。
在 pullout() 中，Text 和 Html 变量是字符串。对于在线邮件组，可以一次将任何文本或 HTML 打包到不是附件的多部分中。
如果您需要更复杂的内容，请将 Text 和 Html 更改为列表并附加到它们并根据需要添加它们。没什么问题。
也许这里有一些错误，因为它旨在与mailbox.Message()一起使用，而不是与email.Message()一起使用。我在 email.Message() 上试过了，效果很好。
您说，您“希望将它们全部列出”。从哪里？如果你提到 POP3 邮箱或一些不错的开源邮件程序的邮箱，那么你可以使用邮箱模块。如果你想从其他人那里列出它们，那么你就有问题了。例如，要从 MS Outlook 中获取邮件，您必须知道如何阅读 OLE2 复合文件。其他邮件程序很少将它们称为 *.eml 文件，所以我认为这正是您想要做的。然后在 PyPI 上搜索 olefile 或 Compoundfiles 模块，并在 Google 上搜索如何从 MS Outlook 收件箱文件中提取电子邮件。或者让自己一团糟，然后将它们从那里导出到某个目录。当您将它们作为 eml 文件时，请应用此代码。
于 2015-07-13T20:13:55.877 回答
24
我发现这段代码更简单
import email
import os
path = './'
listing = os.listdir(path)
for fle in listing:
if str.lower(fle[-3:])=="eml":
msg = email.message_from_file(open(fle))
attachments=msg.get_payload()
for attachment in attachments:
try:
fnam=attachment.get_filename()
f=open(fnam, 'wb').write(attachment.get_payload(decode=True,))
f.close()
except Exception as detail:
#print detail
pass
于 2017-01-19T17:39:37.763 回答
4
在这里为任何希望从电子邮件中提取文本并获取 .eml 文件列表的人发布此内容 - 让我永远在网上找到一个好的答案。注意：这不会获取电子邮件的附件，只会获取电子邮件中的文本。
import email
from email import policy
from email.parser import BytesParser
import glob
import os
path = '/path/to/data/' # set this to "./" if in current directory
eml_files = glob.glob(path + '*.eml') # get all .eml files in a list
for eml_file in eml_files:
with open(eml_file, 'rb') as fp: # select a specific email file from the list
name = fp.name # Get file name
msg = BytesParser(policy=policy.default).parse(fp)
text = msg.get_body(preferencelist=('plain')).get_content()
fp.close()
text = text.split("\n")
print (name) # Get name of eml file
print (text) # Get list of all text in email
归功于这篇文章中的一些代码：Reading .eml files with Python 3.6 using emaildata 0.3.4
于 2020-07-05T23:13:29.590 回答
2
尝试这个：
#!python3
# -*- coding: utf-8 -*-
import email
import os
SOURCE_DIR = 'email'
DEST_DIR = 'temp'
def extractattachements(fle,suffix=None):
message = email.message_from_file(open(fle))
filenames = []
if message.get_content_maintype() == 'multipart':
for part in message.walk():
if part.get_content_maintype() == 'multipart': continue
#if part.get('Content-Disposition') is None: continue
if part.get('Content-Type').find('application/octet-stream') == -1: continue
filename = part.get_filename()
if suffix:
filename = ''.join( [filename.split('.')[0], '_', suffix, '.', filename.split('.')[1]])
filename = os.path.join(DEST_DIR, filename)
fb = open(filename,'wb')
fb.write(part.get_payload(decode=True))
fb.close()
filenames.append(filename)
return filenames
def main():
onlyfiles = [f for f in os.listdir(SOURCE_DIR) if os.path.isfile(os.path.join(SOURCE_DIR, f))]
for file in onlyfiles:
#print path.join(SOURCE_DIR,file)
extractattachements(os.path.join(SOURCE_DIR,file))
return True
if __name__ == "__main__":
main()
于 2019-06-16T13:48:15.490 回答