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[英]I added In-Reply-To and References in my headers while sending Email through SendGrid, however it doesn't work
[英]How correctly set "In-Reply-To" and "Reference" headers in Gmail API
我正在嘗試回復使用 Gmail API 收到的郵件。 我嘗試了以下代碼,它將發送消息附加到我郵箱中的線程,但對於接收器,它作為新消息發送。 聲明 In-Reply-To 和 Reference 標頭的正確方法是什么?
def create_message(origin=None, destination=to, subject=None, msg_txt=None, thr_id=None):
"""Create a message for an email.
Args:
origin: Email address of the sender.
destination: Email address of the receiver.
subject: The subject of the email message.
msg_txt: The text of the email message.
thr_id: the threadId of the message to attach
Returns:
An object containing a base64url encoded email object.
"""
message = MIMEText(msg_txt)
message['to'] = destination
message['from'] = origin
message['subject'] = subject
raw_msg={'raw': (base64.urlsafe_b64encode(message.as_bytes()).decode())}
raw_msg['threadId'] =thr_id
raw_msg['Reference'] = '<CANyAw3CWm33sKL80GMKp-b=8JgXz3MVPkvCVbJ_oK4NuGJcb3w@mail.gmail.com>'
raw_msg['In-Reply-To'] = '<CANyAw3CWm33sKL80GMKp-b=8JgXz3MVPkvCVbJ_oK4NuGJcb3w@mail.gmail.com>'
raw_msg['Message-ID'] = '<CANyAw3CWm33sKL80GMKp-b=8JgXz3MVPkvCVbJ_oK4NuGJcb3w@mail.gmail.com>'
return raw_msg
我的主要方法如下,
def main():
"""Canned reply responder using the Gmail API.
Creates a Gmail API service object and responds to a query with a standard response
whilst giving it a label to ensure only 1 response per thread is sent
"""
# get credentials first and create gmail service object
store = file.Storage('token.json')
creds = store.get()
if not creds or creds.invalid:
flow = client.flow_from_clientsecrets('gmailApiCredentials.json', SCOPES)
creds = tools.run_flow(flow, store)
gmail_service = build('gmail', 'v1', http=creds.authorize(Http()))
# receive email messages
q = 'subject:this is a test message'
messages = list_messages_matching_query(gmail_service, user_id,
query=q,
maxResults=1)
if not messages:
print("No messages to show")
else:
pprint.pprint('Messages to show: {}'.format(messages))
# get thread of first document - so you can label the thread itself if need be
thread_id = messages[0]['threadId']
thread = get_thread(gmail_service, user_id, thread_id)
msg_id = messages[0]['id']
message = get_message(gmail_service, user_id, msg_id)
subject ='Re:this is a test message'
msg = create_message(destination=to, origin=to,
subject=subject,
msg_txt='Hai', thr_id=thread_id)
send_message(gmail_service,"me", msg)
print("Message Sent")
create_message 方法應更改如下以設置“Reference”和“In-Reply-To”標頭。
def create_message(origin=None, destination=TO, subject=None, msg_txt=None, thr_id=None, msgID=None):
"""Create a message for an email.
Args:
origin: Email address of the sender.
destination: Email address of the receiver.
subject: The subject of the email message.
msg_txt: The text of the email message.
thr_id: the threadId of the message to attach
Returns:
An object containing a base64url encoded email object.
"""
message = MIMEText(msg_txt)
message['to'] = destination
message['from'] = origin
message['subject'] = subject
message.add_header('Reference', msgID)
message.add_header('In-Reply-To', msgID)
raw_msg = {'raw': (base64.urlsafe_b64encode(message.as_bytes()).decode())}
raw_msg['threadId'] = thr_id
return raw_msg
正確的msgID可以通過以下方法獲得,
def get_mime_message(service, user_id, msg_id):
try:
message = service.users().messages().get(userId=user_id, id=msg_id,
format='raw').execute()
msg_str = base64.urlsafe_b64decode(message['raw']).decode()
mime_msg = email.message_from_string(msg_str)
return mime_msg
except errors.HttpError as error:
print('An error occurred: %s' % error
可以按如下方式獲取 msgID,
ms = get_mime_message(gmail_service, USER_ID, msg_id)
msgID = format(ms['Message-ID'])
不要將 msg_id 與 msgID 混淆。 msg_id 是特定於 gmail 的,而 msgID 是全局的。 這個 msgID 應該用在“Reference”和“In-Reply-To”標頭中。
threadId
我也遇到了這個問題,並將threadId
到我的“頂級”標題中,如下所示:
return {'raw': base64.urlsafe_b64encode(message.as_string()), 'threadId': thread_id}
似乎成功了!
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