3

原始电子邮件通常看起来像这样

From root@a1.local.tld Thu Jul 25 19:28:59 2013
Received: from a1.local.tld (localhost [127.0.0.1])
    by a1.local.tld (8.14.4/8.14.4) with ESMTP id r6Q2SxeQ003866
    for <ooo@a1.local.tld>; Thu, 25 Jul 2013 19:28:59 -0700
Received: (from root@localhost)
    by a1.local.tld (8.14.4/8.14.4/Submit) id r6Q2Sxbh003865;
    Thu, 25 Jul 2013 19:28:59 -0700
From: root@a1.local.tld
Subject: ooooooooooooooooooooooo
To: ooo@a1.local.tld
Cc: 
X-Originating-IP: 192.168.15.127
X-Mailer: Webmin 1.420
Message-Id: <1374805739.3861@a1>
Date: Thu, 25 Jul 2013 19:28:59 -0700 (PDT)
MIME-Version: 1.0
Content-Type: multipart/mixed; boundary="bound1374805739"

This is a multi-part message in MIME format.

--bound1374805739
Content-Type: text/plain
Content-Transfer-Encoding: 7bit

ooooooooooooooooooooooooooooooo
ooooooooooooooooooooooooooooooo
ooooooooooooooooooooooooooooooo

--bound1374805739--

因此，如果我想编写一个 PYTHON 脚本来获取

From
To
Subject
Body

这是我正在寻找构建的代码还是有更好的方法？

a='<title>aaa</title><title>aaa2</title><title>aaa3</title>'

import re
a1 = re.findall(r'<(title)>(.*?)<(/title)>', a)
4

5 回答 5

17

我真的不明白你的最终代码片段与任何事情有什么关系 - 直到那时你还没有提到任何关于 HTML 的事情，所以我不知道你为什么会突然给出一个解析 HTML 的例子（你应该无论如何都不要使用正则表达式）。

无论如何，为了回答您关于从电子邮件中获取标题的原始问题，Python 在标准库中包含了执行此操作的代码：

import email
msg = email.message_from_string(email_string)
msg['from']  # 'root@a1.local.tld'
msg['to']    # 'ooo@a1.local.tld'
于 2013-07-26T03:03:41.590 回答
14

幸运的是 Python 使这更简单：http ://docs.python.org/2.7/library/email.parser.html#email.parser.Parser

from email.parser import Parser
parser = Parser()

emailText = """PUT THE RAW TEXT OF YOUR EMAIL HERE"""
email = parser.parsestr(emailText)

print email.get('From')
print email.get('To')
print email.get('Subject')

身体比较棘手。打电话email.is_multipart()。如果那是假的，你可以通过调用来获取你的身体email.get_payload()。但是，如果它是真的，email.get_payload()将返回一个消息列表，所以你必须调用get_payload()其中的每一个。

if email.is_multipart():
    for part in email.get_payload():
        print part.get_payload()
else:
    print email.get_payload()
于 2013-07-26T03:09:04.590 回答
2

您的示例电子邮件中不存在“正文”

可以使用电子邮件模块：

import email
    msg = email.message_from_string(email_message_as_text)

然后使用：

print email['To']
print email['From']

... ... ETC

于 2013-07-26T03:04:06.087 回答
1

您可能应该使用email.parser

s = """
From root@a1.local.tld Thu Jul 25 19:28:59 2013
Received: from a1.local.tld (localhost [127.0.0.1])
    by a1.local.tld (8.14.4/8.14.4) with ESMTP id r6Q2SxeQ003866
    for <ooo@a1.local.tld>; Thu, 25 Jul 2013 19:28:59 -0700
Received: (from root@localhost)
    by a1.local.tld (8.14.4/8.14.4/Submit) id r6Q2Sxbh003865;
    Thu, 25 Jul 2013 19:28:59 -0700
From: root@a1.local.tld
Subject: ooooooooooooooooooooooo
To: ooo@a1.local.tld
Cc: 
X-Originating-IP: 192.168.15.127
X-Mailer: Webmin 1.420
Message-Id: <1374805739.3861@a1>
Date: Thu, 25 Jul 2013 19:28:59 -0700 (PDT)
MIME-Version: 1.0
Content-Type: multipart/mixed; boundary="bound1374805739"

This is a multi-part message in MIME format.

--bound1374805739
Content-Type: text/plain
Content-Transfer-Encoding: 7bit

ooooooooooooooooooooooooooooooo
ooooooooooooooooooooooooooooooo
ooooooooooooooooooooooooooooooo

--bound1374805739--
"""

import email.parser

msg = email.parser.Parser().parsestr(s)
help(msg)
于 2013-07-26T03:03:50.317 回答
0

您可以将原始内容写入文件

然后像这样读取文件：

with open('in.txt', 'r') as file:
    raw = file.readlines()

get_list = ['From:','To:','Subject:']
info_list = []

for i in raw:
    for word in get_list:
        if i.startswith(word):
            info_list.append(i)

现在info_list将是：

['From: root@a1.local.tld', 'Subject: ooooooooooooooooooooooo', 'To: ooo@a1.local.tld']

我没有Body:在您的原始内容中看到

于 2013-07-26T03:03:26.990 回答