Python: Google API - 从消息中获取mime类型

Question

我的目标是使用Google API从指定的邮件中提取数据。当前，我可以找到消息、获取消息数据并将其解码为可读格式。在此之后，我需要查找消息中正确的部分（类型为text/html），然后使用beautiful soup扫描我的链接。不幸的是，我对邮件/ Google API 的结构理解不够，无法扫描邮件中的特定部分。

        try:
            message = gmail_service.users().messages().get(userId='me', id=thread['id'], format='raw').execute()
            print 'Message snippet: %s' % message['snippet']
            msg_str = base64.urlsafe_b64decode(message['raw'].encode('ASCII'))
            mime_msg = email.message_from_string(msg_str)

            print mime_msg        #this line gives the output I quoted
            for parts in mime_msg['payload']:    #this line produces error quoted
                if parts['text/html']:
                    mylink = base64.urlsafe_b64decode(part[0]['body']['data'].encode('UTF-8'))
                    print mylink

这段代码给我报错的信息是：

Traceback (most recent call last):
  File "gmailAPI.py", line 55, in <module>
    for parts in mime_msg['payload']:
TypeError: 'NoneType' object is not iterable

在代码的输出中，我还收到了关于邮件不同部分的信息，这就是我想要的部分。

----boundary_1_81681de2-2c9a-4827-802a-91544e5e6e28
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: base64

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我的代码完整备份链接

编辑: 我修复后的代码

try:
        message = gmail_service.users().messages().get(userId='me', id=thread['id'], format='raw').execute()
        # print 'Message snippet: %s' % message['snippet']
        msg_str = base64.urlsafe_b64decode(message['raw'].encode('ASCII'))
        msg = email.message_from_string(msg_str)

        for part in msg.walk():
            msg.get_payload()
            if part.get_content_type() == 'text/html':
                mytext = base64.urlsafe_b64decode(part.get_payload().encode('UTF-8'))
                # print part.get_payload()
                print mytext

我选择的答案文档链接中的信息对解决我的问题非常有价值。

- LeonH

---

你能提供从API调用返回的消息对象的完整转储吗？ - Brandon Jewett-Hall

好的，我会把它放在外部网站上。这个消息很长，我会稍后编辑我的问题。 - LeonH

@Stormie 但是邮件正文仍然不是人类可读版本吗？我也可以打印出message['snippet']，但仍然无法获得电子邮件消息正文的人类可读版本。 - littletiger

Answer 1

要在Python中迭代多部分消息的各个部分，您应该使用get_payload()方法：https://docs.python.org/2/library/email.message.html#email.message.Message.get_payload 在您的示例中，调用mime_msg['payload']是在查找名为“payload”的消息头，但该消息头并不存在，而且也不是您想要的内容。

一旦您拿到某个部分，您可以使用part['Content-Type']检查其类型，以检查Content-Type头信息。

通常，MIME消息是一个部分的树状结构，因此您可能需要进行递归处理。

- Brandon Jewett-Hall

---

我只是在编辑我的问题，以包括“修复”的代码，因为您的解释并不像您链接的文档那样有帮助。 - LeonH