I need to extract just the text in the body of an eml file but my code keeps giving me some code text and the folders that exist in Outlook. I am working with Python 2.7 and BeautifulSoup. My code is:
import email
from email import message_from_file
import os
import bs4 as bs
import re
# Path to directory where attachments will be stored:
path = "D:\Python"
# To have attachments extracted into memory, change behaviour of 2 following functions:
def file_exists (f):
"""Checks whether extracted file was extracted before."""
return os.path.exists(os.path.join(path, f))
def save_file (fn, cont):
"""Saves cont to a file fn"""
file = open(os.path.join(path, fn), "wb")
file.write(cont)
file.close()
def construct_name (id, fn):
"""Constructs a file name out of messages ID and packed file name"""
id = id.split(".")
id = id[0]+id[1]
return id+"."+fn
def disqo (s):
"""Removes double or single quotations."""
s = s.strip()
if s.startswith("'") and s.endswith("'"): return s[1:-1]
if s.startswith('"') and s.endswith('"'): return s[1:-1]
return s
def disgra (s):
"""Removes < and > from HTML-like tag or e-mail address or e-mail ID."""
s = s.strip()
if s.startswith("<") and s.endswith(">"): return s[1:-1]
return s
def pullout (m, key):
"""Extracts content from an e-mail message.
This works for multipart and nested multipart messages too.
m -- email.Message() or mailbox.Message()
key -- Initial message ID (some string)
Returns tuple(Text, Html, Files, Parts)
Text -- All text from all parts.
Html -- All HTMLs from all parts
Files -- Dictionary mapping extracted file to message ID it belongs to.
Parts -- Number of parts in original message.
"""
Html = ""
Text = ""
Files = {}
Parts = 0
if not m.is_multipart():
if m.get_filename(): # It's an attachment
fn = m.get_filename()
cfn = construct_name(key, fn)
Files[fn] = (cfn, None)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# Not an attachment!
# See where this belongs. Text, Html or some other data:
cp = m.get_content_type()
if cp=="text/plain": Text += m.get_payload(decode=True)
elif cp=="text/html": Html += m.get_payload(decode=True)
else:
# Something else!
# Extract a message ID and a file name if there is one:
# This is some packed file and name is contained in content-type header
# instead of content-disposition header explicitly
cp = m.get("content-type")
try: id = disgra(m.get("content-id"))
except: id = None
# Find file name:
o = cp.find("name=")
if o==-1: return Text, Html, Files, 1
ox = cp.find(";", o)
if ox==-1: ox = None
o += 5; fn = cp[o:ox]
fn = disqo(fn)
cfn = construct_name(key, fn)
Files[fn] = (cfn, id)
if file_exists(cfn): return Text, Html, Files, 1
save_file(cfn, m.get_payload(decode=True))
return Text, Html, Files, 1
# This IS a multipart message.
# So, we iterate over it and call pullout() recursively for each part.
y = 0
while 1:
# If we cannot get the payload, it means we hit the end:
try:
pl = m.get_payload(y)
except: break
# pl is a new Message object which goes back to pullout
t, h, f, p = pullout(pl, key)
Text += t; Html += h; Files.update(f); Parts += p
y += 1
return Text, Html, Files, Parts
def extract (msgfile, key):
"""Extracts all data from e-mail, including From, To, etc., and returns it as a dictionary.
msgfile -- A file-like readable object
key -- Some ID string for that particular Message. Can be a file name or anything.
Returns dict()
Keys: from, to, subject, date, text, html, parts[, files]
Key files will be present only when message contained binary files.
For more see __doc__ for pullout() and caption() functions.
"""
m = message_from_file(msgfile)
From, To, Subject, Date, Body = caption(m)
Text, Html, Files, Parts = pullout(m, key)
Text = Text.strip(); Html = Html.strip()
msg = {"subject": Subject, "from": From, "to": To, "date": Date, "body": Body,
"text": Text, "html": Html, "parts": Parts}
if Files: msg["files"] = Files
return msg
def caption (origin):
"""Extracts: To, From, Subject and Date from email.Message() or mailbox.Message()
origin -- Message() object
Returns tuple(From, To, Subject, Date)
If message doesn't contain one/more of them, the empty strings will be returned.
"""
Date = ""
if origin.has_key("date"): Date = origin["date"].strip()
From = ""
if origin.has_key("from"): From = origin["from"].strip()
To = ""
if origin.has_key("to"): To = origin["to"].strip()
Subject = ""
if origin.has_key("subject"): Subject = origin["subject"].strip()
if origin.has_key("body"): To = origin["body"].strip()
Body = ""
return From, To, Subject, Date, Body
and these are my tries for bs:
"""
# Usage:
f = open("e.eml", "rb")
soup = bs.BeautifulSoup(f,'lxml')
for url in soup.find_all('body'):
print url.get('href')
f.close()
"""
f = open("e.eml", "rb")
soup = bs.BeautifulSoup(f,'lxml')
for body in soup.find_all('p'):
text = soup.get_text()
print text
f.close()
"""def cleanhtml(raw_html):
cleanr = re.compile('<.*?>')
cleantext = re.sub(cleanr, '', raw_html)
return cleantext
f = open("e.eml", "rb")
soup = bs.BeautifulSoup(f,'lxml')
for goal in soup.find_all():
body = soup.find('body')
the_contents_of_body_without_body_tags = body.findChildren()
soup.get_text()
print the_contents_of_body_without_body_tags
f.close()
"""
"""f = open("e.eml", "rb")
message = email.message_from_file(open('e.eml'))
text = Text.text(message)
#print extract(f,f.name)
f.close()
"""
The goal is to have the text written in an eml file printed and then write it on a csv file that will rank it as spam or ham. The only thing remains is to isolate the text body somehow.
Thank you in advance.
