在Python 3中。X，你可以通过导入“imaplib”和“电子邮件”包，以非常简单的方式做到这一点。虽然这是一个老帖子，但也许我的答案可以帮助到这个帖子的新人。

status, data = self.imap.fetch(num, '(RFC822)')
email_msg = email.message_from_bytes(data[0][1]) 
#email.message_from_string(data[0][1])

#If message is multi part we only want the text version of the body, this walks the message and gets the body.

if email_msg.is_multipart():
    for part in email_msg.walk():       
        if part.get_content_type() == "text/plain":
            body = part.get_payload(decode=True) #to control automatic email-style MIME decoding (e.g., Base64, uuencode, quoted-printable)
            body = body.decode()
        elif part.get_content_type() == "text/html":
            continue

现在你可以打印主体变量，它将是明文格式:)如果它对你来说足够好，那么它将很好地选择它作为接受的答案。

下面是xperroni回答的一个版本，它更完整一些。它跳过脚本和样式部分，并翻译charref(例如，')和HTML实体(例如，&)。

它还包括一个简单的纯文本到html的反向转换器。

"""
HTML <-> text conversions.
"""
from HTMLParser import HTMLParser, HTMLParseError
from htmlentitydefs import name2codepoint
import re

class _HTMLToText(HTMLParser):
    def __init__(self):
        HTMLParser.__init__(self)
        self._buf = []
        self.hide_output = False

    def handle_starttag(self, tag, attrs):
        if tag in ('p', 'br') and not self.hide_output:
            self._buf.append('\n')
        elif tag in ('script', 'style'):
            self.hide_output = True

    def handle_startendtag(self, tag, attrs):
        if tag == 'br':
            self._buf.append('\n')

    def handle_endtag(self, tag):
        if tag == 'p':
            self._buf.append('\n')
        elif tag in ('script', 'style'):
            self.hide_output = False

    def handle_data(self, text):
        if text and not self.hide_output:
            self._buf.append(re.sub(r'\s+', ' ', text))

    def handle_entityref(self, name):
        if name in name2codepoint and not self.hide_output:
            c = unichr(name2codepoint[name])
            self._buf.append(c)

    def handle_charref(self, name):
        if not self.hide_output:
            n = int(name[1:], 16) if name.startswith('x') else int(name)
            self._buf.append(unichr(n))

    def get_text(self):
        return re.sub(r' +', ' ', ''.join(self._buf))

def html_to_text(html):
    """
    Given a piece of HTML, return the plain text it contains.
    This handles entities and char refs, but not javascript and stylesheets.
    """
    parser = _HTMLToText()
    try:
        parser.feed(html)
        parser.close()
    except HTMLParseError:
        pass
    return parser.get_text()

def text_to_html(text):
    """
    Convert the given text to html, wrapping what looks like URLs with <a> tags,
    converting newlines to <br> tags and converting confusing chars into html
    entities.
    """
    def f(mo):
        t = mo.group()
        if len(t) == 1:
            return {'&':'&amp;', "'":'&#39;', '"':'&quot;', '<':'&lt;', '>':'&gt;'}.get(t)
        return '<a href="%s">%s</a>' % (t, t)
    return re.sub(r'https?://[^] ()"\';]+|[&\'"<>]', f, text)

我推荐一个名为goose-extractor的Python包 Goose将尝试提取以下信息:

文章的正文 文章主图 任何Youtube/Vimeo电影嵌入文章 元数据描述 元标记

更多:https://pypi.python.org/pypi/goose-extractor/

在Python 3中。X，你可以通过导入“imaplib”和“电子邮件”包，以非常简单的方式做到这一点。虽然这是一个老帖子，但也许我的答案可以帮助到这个帖子的新人。

status, data = self.imap.fetch(num, '(RFC822)')
email_msg = email.message_from_bytes(data[0][1]) 
#email.message_from_string(data[0][1])

#If message is multi part we only want the text version of the body, this walks the message and gets the body.

if email_msg.is_multipart():
    for part in email_msg.walk():       
        if part.get_content_type() == "text/plain":
            body = part.get_payload(decode=True) #to control automatic email-style MIME decoding (e.g., Base64, uuencode, quoted-printable)
            body = body.decode()
        elif part.get_content_type() == "text/html":
            continue

现在你可以打印主体变量，它将是明文格式:)如果它对你来说足够好，那么它将很好地选择它作为接受的答案。

另一个在Python 2.7.9+中使用BeautifulSoup4的例子

包括:

import urllib2
from bs4 import BeautifulSoup

代码:

def read_website_to_text(url):
    page = urllib2.urlopen(url)
    soup = BeautifulSoup(page, 'html.parser')
    for script in soup(["script", "style"]):
        script.extract() 
    text = soup.get_text()
    lines = (line.strip() for line in text.splitlines())
    chunks = (phrase.strip() for line in lines for phrase in line.split("  "))
    text = '\n'.join(chunk for chunk in chunks if chunk)
    return str(text.encode('utf-8'))

解释道:

将url数据读入为html(使用BeautifulSoup)，删除所有脚本和样式元素，并使用.get_text()仅获取文本。分割成行，删除每个标题的开头和结尾空格，然后将多个标题分割成一行，each chunks = (phrase.strip() for line in line for phrase in line。(" "))。然后使用text = '\n'。加入，删除空行，最后返回为批准的utf-8。

注:

一些系统这是运行在https://连接失败，因为SSL问题，你可以关闭验证来解决这个问题。修复示例:http://blog.pengyifan.com/how-to-fix-python-ssl-certificate_verify_failed/ Python < 2.7.9在运行时可能会遇到一些问题 text.encode('utf-8')可能会留下奇怪的编码，可能只需要返回str(text)即可。

我发现的最好的一段代码提取文本没有javascript或不想要的东西:

from urllib.request import urlopen
from bs4 import BeautifulSoup

url = "http://news.bbc.co.uk/2/hi/health/2284783.stm"
html = urlopen(url).read()
soup = BeautifulSoup(html, features="html.parser")

# kill all script and style elements
for script in soup(["script", "style"]):
    script.extract()    # rip it out

# get text
text = soup.get_text()

# break into lines and remove leading and trailing space on each
lines = (line.strip() for line in text.splitlines())
# break multi-headlines into a line each
chunks = (phrase.strip() for line in lines for phrase in line.split("  "))
# drop blank lines
text = '\n'.join(chunk for chunk in chunks if chunk)

print(text)

你只需要安装BeautifulSoup:

pip install beautifulsoup4