Given an RFC822 message in Python 2.6, how can I get the right text/plain content part? Basically, the algorithm I want is this:
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
Of these things, I have get_mime_part
and has_mime_part
down pat, but I'm not quite sure how to get the decoded text from the MIME part. I can get the encoded text using get_payload()
, but if I try to use the decode
parameter of the get_payload()
method (see the doc) I get an error when I call it on the text/plain part:
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'>
In addition, I don't know how to take HTML and render it to text as closely as possible.
In a multipart e-mail, email.message.Message.get_payload()
returns a list with one item for each part. The easiest way is to walk the message and get the payload on each part:
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
For a non-multipart message, no need to do all the walking. You can go straight to get_payload(), regardless of content_type.
msg = email.message_from_string(raw_message)
msg.get_payload()
If the content is encoded, you need to pass None
as the first parameter to get_payload()
, followed by True (the decode flag is the second parameter). For example, suppose that my e-mail contains an MS Word document attachment:
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()
As for getting a reasonable plain-text approximation of an HTML part, I've found that html2text works pretty darn well.
Flat is better than nested ;)
from email.mime.multipart import MIMEMultipart
assert isinstance(msg, MIMEMultipart)
for _ in [k.get_payload() for k in msg.walk() if k.get_content_type() == 'text/plain']:
print _
To add on @Jarret Hardie's excellent answer:
I personally like to transform that kind of data structures to a dictionary that I can reuse later, so something like this where the content_type
is the key and the payload
is the value:
import email
[...]
email_message = {
part.get_content_type(): part.get_payload()
for part in email.message_from_bytes(raw_email).walk()
}
print(email_message["text/plain"])
来源:https://stackoverflow.com/questions/1463074/how-can-i-get-an-email-messages-text-content-using-python