本文介绍了Python:Google API - 从消息中获取 mimeTypes的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
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，我需要找到我的消息的正确部分(类型 text/html)，然后使用 美丽的汤.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.
My goal is to use the Google API to take data from an email I have specified. Currently I can find the message, get the message data and decode the message data into a readable format. After this I need to find the correct part of my message (type text/html) and then scan for my link using beautiful soup. Unfortunately I don't understand enough about the structure of the email/Google API to scan for this specific part of the mail.
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:
And the error that this code gives me is:
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，这就是我想要的部分:
In the output for the code I also receive the information on different parts of the mail, this is the part I want:
----boundary_1_81681de2-2c9a-4827-802a-91544e5e6e28
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: base64
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链接到我的代码中的完整转储
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上找到的信息对于解决我的问题非常宝贵！
The information found on my chosen answers documentation link was invaluable in solving my issue!
推荐答案
要在 Python 中遍历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
To iterate through the parts of a multipart message in Python, you should use 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"的消息标头，该标头不存在，也不是您想要的.
In your example, the call to mime_msg['payload'] is looking up a message header named "payload", which doesn't exist and is not what you want anyway.
一旦你手头有一个部件，你可以使用 part['Content-Type'] 检查它的类型来检查 Content-Type 标头.
Once you have a part in hand, you can check its type using part['Content-Type'] to examine the Content-Type header.
一般来说，MIME 消息是由部分组成的树，因此您可能需要递归.
In general, MIME messages are trees of parts, so you may need to recurse.
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