python - Twisted SMTP 服务器 - 获取完整的收件人列表和 header 数据？

Question

因此，经过 @Jean-PaulCalderone 的巨大耐心和感激之后，我现在拥有了一个支持(尚未测试)TLS 的扭曲邮件服务器!我可以远程登录到它，发出我的 EHLO、MAIL FROM 等，并将消息转储到文件，按照下面的代码片段(下面的 2 个类是从 http://twistedmatrix.com/documents/current/_downloads/emailserver.tac 中删除和修改的:

class ConsoleMessageDelivery:
    implements(smtp.IMessageDelivery)

    def receivedHeader(self, helo, origin, recipients):
        by = helo[1]
        from_ = origin
        for_ = "%s" % (' '.join(map(str, recipients)))
        return "Client: %s\nFrom: %s\nTo: %s\n\n" % (by, from_, for_)

    def validateFrom(self, helo, origin):
        return origin

    def validateTo(self, user):
        return lambda: ConsoleMessage()


class ConsoleMessage:
    implements(smtp.IMessage)

    def __init__(self):
        self.lines = []

    def lineReceived(self, line):
        self.lines.append(line)

    def eomReceived(self):
        print "New message received:"
        print "\n".join(self.lines)
        msgFile = open("./test.msg","w")
        msgFile.writelines(self.lines)
        msgFile.close()
        self.lines = None
        return defer.succeed(None)

    def connectionLost(self):
        # There was an error, throw away the stored lines
        self.lines = None

但是，当我指定多个 RCPT TO 地址时，现在出现了问题 - 只有最后一个位于我的 receiveheader 方法中的“收件人”中？我在“receivedHeader”中处理不当吗？我开始认为这是我的 validateTo 函数中的 return 语句。当我向 2 个收件人发送一条消息时，服务器日志如下所示:

2014-10-21 08:05:10+0000 [ESMTP,2,127.0.0.1] Receiving message for delivery: from=me@home.com to=['you@work1.com', 'you@work2.com']
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] New message received:
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] Client: 127.0.0.1
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] From: me@home.com
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] To: you@work1.com
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] Today is the 21st Oct
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] (still)
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] New message received:
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] Client: 127.0.0.1
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] From: me@home.com
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] To: you@work2.com
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1]
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] Today is the 21st Oct
2014-10-21 08:05:17+0000 [ESMTP,2,127.0.0.1] (still)

此外，如果能够访问“完整”SMTP header (消息 ID 等)就好了。我当然查看了 smtp.IMessage/smtp.IMessageDelivery 的 API 文档 - 但看不到任何相关内容？最终，最好创建一个文件(可能是 2 个，拆分 header /正文)，其名称以某种方式基于 msg ID...

有人可以解释一下我缺少什么吗？非常感谢!

Answer 1

您的代码和输出看起来绝对没问题。收件人不分组的行为是由于lambda语句触发了每个收件人/邮件发出的self.lines.append()。

但是，您可以考虑更改 lineReceived 的行为，以便可以对收到的所有线路进行分组。

祝你好运，希望这会有所帮助。