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pythonregexemail-parsing

Parsing email headers with regular expressions in python


I'm a python beginner trying to extract data from email headers. I have thousands of email messages in a single text file, and from each message I want to extract the sender's address, recipient(s) address, and the date, and write it to a single, semicolon-delimitted line in a new file.

this is ugly, but it's what I've come up with:

import re

emails = open("demo_text.txt","r") #opens the file to analyze
results = open("results.txt","w") #creates new file for search results

resultsList = []

for line in emails:
    if "From - " in line: #recgonizes the beginning of a email message and adds a linebreak
        newMessage = re.findall(r'\w\w\w\s\w\w\w.*', line)
        if newMessage:
            resultsList.append("\n")        
    if "From: " in line:
        address = re.findall(r'[\w.-]+@[\w.-]+', line)
        if address:
            resultsList.append(address)
            resultsList.append(";")
    if "To: " in line:
        if "Delivered-To:" not in line: #avoids confusion with 'Delivered-To:' tag
            address = re.findall(r'[\w.-]+@[\w.-]+', line)
            if address:
                for person in address:
                    resultsList.append(person)
                    resultsList.append(";")
    if "Date: " in line: 
            date = re.findall(r'\w\w\w\,.*', line)
            resultsList.append(date)
            resultsList.append(";")

for result in resultsList:
    results.writelines(result)


emails.close()
results.close()

and here's my 'demo_text.txt':

From - Sun Jan 06 19:08:49 2013
X-Mozilla-Status: 0001
X-Mozilla-Status2: 00000000
Delivered-To: [email protected]
Received: by 10.48.48.3 with SMTP id v3cs417003nfv;
        Mon, 15 Jan 2007 10:14:19 -0800 (PST)
Received: by 10.65.211.13 with SMTP id n13mr5741660qbq.1168884841872;
        Mon, 15 Jan 2007 10:14:01 -0800 (PST)
Return-Path: <[email protected]>
Received: from bay0-omc3-s21.bay0.hotmail.com (bay0-omc3-s21.bay0.hotmail.com [65.54.246.221])
        by mx.google.com with ESMTP id e13si6347910qbe.2007.01.15.10.13.58;
        Mon, 15 Jan 2007 10:14:01 -0800 (PST)
Received-SPF: pass (google.com: domain of [email protected] designates 65.54.246.221 as permitted sender)
Received: from hotmail.com ([65.54.250.22]) by bay0-omc3-s21.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.2668);
         Mon, 15 Jan 2007 10:13:48 -0800
Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC;
         Mon, 15 Jan 2007 10:13:47 -0800
Message-ID: <[email protected]>
Received: from 65.54.250.200 by by115fd.bay115.hotmail.msn.com with HTTP;
        Mon, 15 Jan 2007 18:13:43 GMT
X-Originating-IP: [200.122.47.165]
X-Originating-Email: [[email protected]]
X-Sender: [email protected]
From: =?iso-8859-1?B?UGF1bGEgTWFy7WEgTGlkaWEgRmxvcmVuemE=?=
 <[email protected]>
To: [email protected], [email protected], [email protected]
Bcc: 
Subject: fotos
Date: Mon, 15 Jan 2007 18:13:43 +0000
Mime-Version: 1.0
Content-Type: multipart/mixed; boundary="----=_NextPart_000_d98_1c4f_3aa9"
X-OriginalArrivalTime: 15 Jan 2007 18:13:47.0572 (UTC) FILETIME=[E68D4740:01C738D0]
Return-Path: [email protected]

The output is:

[email protected];[email protected];[email protected];Mon, 15 Jan 2007 18:13:43 +0000;

This output would be fine except there's a line break in the 'From:' field in my demo_text.txt (line 24), and so I miss '[email protected]'.

I'm not sure how to tell my code to skip line break and still find email address in the From: tag.

More generally, I'm sure there are many more sensible ways to go about this task. If anyone could point me in the right direction, I'd sure appreciate it.


Solution

  • Your demo text is practicallly the mbox format, which can be perfectly processed with the appropriate object in the mailbox module:

    from mailbox import mbox
    import re
    
    PAT_EMAIL = re.compile(r"[0-9A-Za-z._-]+\@[0-9A-Za-z._-]+")
    
    mymbox = mbox("demo.txt")
    for email in mymbox.values():
        from_address = PAT_EMAIL.findall(email["from"])
        to_address = PAT_EMAIL.findall(email["to"])
        date = [ email["date"], ]
        print ";".join(from_address + to_address + date)