Python:Google API - 从消息中获取 mimeTypes

Question

我的目标是使用 Google API 从我指定的电子邮件中获取数据。目前我可以找到消息，获取消息数据并将消息数据解码为可读格式。在此之后，我需要找到消息的正确部分(键入 text/html)，然后使用 beautiful soup 扫描我的链接。 .不幸的是，我对电子邮件/Google API 的结构了解不够，无法扫描邮件的这个特定部分。

        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

这段代码给我的错误是:

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，我还收到了邮件不同部分的信息，这是我想要的部分:

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

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Link to full dump from my code

编辑:我的固定代码

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

在我选择的答案文档链接中找到的信息对于解决我的问题非常宝贵!

Answer 1

要在 Python 中循环访问多部分消息的各个部分，您应该使用 get_payload():https://docs.python.org/2/library/email.message.html#email.message.Message.get_payload

在您的示例中，对 mime_msg['payload'] 的调用正在查找名为“payload”的消息 header ，它不存在，也不是您想要的。

一旦您掌握了零件，就可以使用 part['Content-Type'] 检查其类型以检查 Content-Type header 。

一般来说，MIME 消息是部件树，因此您可能需要递归。