Chr*_*s R 36 python email mime rfc822
鉴于Python 2.6中的RFC822消息,我如何获得正确的文本/纯文本内容部分?基本上,我想要的算法是这样的:
message = email.message_from_string(raw_message)
if has_mime_part(message, "text/plain"):
mime_part = get_mime_part(message, "text/plain")
text_content = decode_mime_part(mime_part)
elif has_mime_part(message, "text/html"):
mime_part = get_mime_part(message, "text/html")
html = decode_mime_part(mime_part)
text_content = render_html_to_plaintext(html)
else:
# fallback
text_content = str(message)
return text_content
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这些事,我get_mime_part和has_mime_part拍下来,但我不太清楚如何从MIME部分解码的文本.我可以使用编码文本get_payload(),但如果我尝试使用方法的decode参数get_payload()(请参阅doc),当我在text/plain部分调用它时会出现错误:
Run Code Online (Sandbox Code Playgroud)File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/ email/message.py", line 189, in get_payload raise TypeError('Expected list, got %s' % type(self._payload)) TypeError: Expected list, got <type 'str'>
另外,我不知道如何使用HTML并尽可能地将其呈现为文本.
Jar*_*die 84
在多部分电子邮件中,email.message.Message.get_payload()返回每个部分包含一个项目的列表.最简单的方法是遍历消息并获取每个部分的有效负载:
import email
msg = email.message_from_string(raw_message)
for part in msg.walk():
# each part is a either non-multipart, or another multipart message
# that contains further parts... Message is organized like a tree
if part.get_content_type() == 'text/plain':
print part.get_payload() # prints the raw text
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对于非多部分消息,无需进行所有步行.无论content_type如何,您都可以直接使用get_payload().
msg = email.message_from_string(raw_message)
msg.get_payload()
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如果内容是编码的,则需要None将第一个参数传递给get_payload(),然后是True(解码标志是第二个参数).例如,假设我的电子邮件包含MS Word文档附件:
msg = email.message_from_string(raw_message)
for part in msg.walk():
if part.get_content_type() == 'application/msword':
name = part.get_param('name') or 'MyDoc.doc'
f = open(name, 'wb')
f.write(part.get_payload(None, True)) # You need None as the first param
# because part.is_multipart()
# is False
f.close()
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至于获得HTML部分的合理纯文本近似,我发现html2text非常好用.