How can I get an email message’s text content using Python?

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.

Answers:

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Method 1

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.

Method 2

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 _

Method 3

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"])

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0