如何在python中读取eml文件？

Question

我不知道如何在python 3.4中加载eml文件.
我想列出所有并在python中读取所有这些内容.

Answer 1

这是您获取电子邮件内容的方式,即*.eml文件.这在Python2.5 - 2.7上完美运行.尝试3.它也应该工作.

from email import message_from_file
import os

# Path to directory where attachments will be stored:
path = "./msgfiles"

# 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 = caption(m)
    Text, Html, Files, Parts = pullout(m, key)
    Text = Text.strip(); Html = Html.strip()
    msg = {"subject": Subject, "from": From, "to": To, "date": Date,
        "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()
    return From, To, Subject, Date

# Usage:
f = open("message.eml", "rb")
print extract(f, f.name)
f.close()

我使用邮箱为我的邮件组编程了这个,这就是为什么它如此复杂.它永远不会让我失望.从来没有任何垃圾.如果消息是多部分,则输出字典将包含一个键"文件"(子字典),其中包含提取的其他非文本或html文件的所有文件名.这是一种提取附件和其他二进制数据的方法.您可以在pullout()中更改它,或者只是更改file_exists()和save_file()的行为.

construct_name()构造一个消息id和多部分消息文件名的文件名,如果有的话.

在pullout()中,Text和Html变量是字符串.对于在线邮件组,可以将任何文本或HTML打包到多部分中,而不是一次性附件.

如果您需要更复杂的内容,请将Text和Html更改为列表并附加到它们并根据需要添加它们.什么都没有问题.

也许这里有一些错误,因为它打算使用mailbox.Message(),而不是使用email.Message().我在email.Message()上尝试过它并且运行正常.

你说,你"希望列出所有".来自哪里？如果你参考POP3邮箱或一些不错的开源邮件的邮箱,那么你使用邮箱模块.如果您想从其他人列出它们,那么您就遇到了问题.例如,要从MS Outlook获取邮件,您必须知道如何读取OLE2复合文件.其他邮件很少将它们称为*.eml文件,因此我认为这正是您想要做的.然后在PyPI上搜索olefile或compoundfiles模块和Google,了解如何从MS Outlook收件箱文件中提取电子邮件.或者保存自己一团糟,然后将它们从那里导出到某个目录.当您将它们作为eml文件时,请应用此代码.

Answer 2

我发现这个代码更简单

import email
import os

path = './'
listing = os.listdir(path)

for fle in listing:
    if str.lower(fle[-3:])=="eml":
        msg = email.message_from_file(open(fle))
        attachments=msg.get_payload()
        for attachment in attachments:
            try:
                fnam=attachment.get_filename()
                f=open(fnam, 'wb').write(attachment.get_payload(decode=True,))
                f.close()
            except Exception as detail:
                #print detail
                pass

Answer 3

在这里发布此内容是为了任何想要从电子邮件中提取文本并获取 .eml 文件列表的人 - 我花了很长时间才在网上找到了一个很好的答案。注意：这不会获取电子邮件的附件，而只会获取电子邮件中的文本。

import email
from email import policy
from email.parser import BytesParser
import glob
import os

path = '/path/to/data/' # set this to "./" if in current directory

eml_files = glob.glob(path + '*.eml') # get all .eml files in a list
for eml_file in eml_files:
    with open(eml_file, 'rb') as fp:  # select a specific email file from the list
        name = fp.name # Get file name
        msg = BytesParser(policy=policy.default).parse(fp)
    text = msg.get_body(preferencelist=('plain')).get_content()
    fp.close()
 
    text = text.split("\n")
    print (name) # Get name of eml file
    print (text) # Get list of all text in email

归功于这篇文章中的一些代码：Reading .eml files with Python 3.6 using emaildata 0.3.4

Answer 4

Dalen 答案的 Python 3 版本。基本上修复了语法问题。（由于缺乏声誉，无法发表评论，作为答案也更清晰）。

# 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 += str(m.get_payload(decode=True))
        elif cp=="text/html":
            Html += str(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 = email.message_from_file(msgfile)
    From, To, Subject, Date = caption(m)
    Text, Html, Files, Parts = pullout(m, key)
    Text = Text.strip(); Html = Html.strip()
    msg = {"subject": Subject, "from": From, "to": To, "date": Date,
        "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.__contains__("date"): Date = origin["date"].strip()
    From = ""
    if origin.__contains__("from"): From = origin["from"].strip()
    To = ""
    if origin.__contains__("to"): To = origin["to"].strip()
    Subject = ""
    if origin.__contains__("subject"): Subject = origin["subject"].strip()
    return From, To, Subject, Date