Question

我即将使用mandrill和djrill 1.3.0与django 1.7将一些批量电子邮件功能集成到项目中，因为我使用以下方法发送html内容：

from django.core.mail import get_connection

connection = get_connection()
to = ['testaddress1@example.com', 'testaddress1@example.com'] 
for recipient_email in to:
    # I perform some controls and register some info about the user and email address
    subject = u"Test subject for %s" % recipient_email
    text = u"Test text for email body"
    html = u"<p>Test text for email body</p>"
    from_email = settings.DEFAULT_FROM_EMAIL
    msg = EmailMultiAlternatives(
        subject, text, from_email, [recipient_email])
    msg.attach_alternative(html, 'text/html')
    messages.append(msg)
# Bulk send
send_result = connection.send_messages(messages)

此时，send_result是一个int，它等于发送（推送到mandrill）消息的数量。

我需要为每封已发送的邮件获取mandrill响应，以处理mandrill_response [＆＃39; msg＆＃39;] [＆＃39; _id＆＃39;]值和其他一些内容。

djrill提供了＆＃39; send_messages＆＃39; connection方法使用_send调用，它正在向每条消息添加mandrill_response，但如果成功则返回True。

那么，您是否知道在使用djrill发送批量HTML电子邮件时如何获取每条消息的mandrill响应？

Answer 1

Djrill在发送每个EmailMessage对象时附加mandrill_response属性。请参阅Djrill文档中的Mandrill response。

因此，在发送消息后，您可以检查您发送的messages列表中每个对象的该属性。类似的东西：

# Bulk send
send_result = connection.send_messages(messages)

for msg in messages:
   if msg.mandrill_response is None:
       print "error sending to %r" % msg.to
   else:
       # there's one response for each recipient of the msg
       # (because an individual message can have multiple to=[..., ...])
       for response in msg.mandrill_response:
           print "Message _id %s, to %s, status %s" % (
               response['_id'], response['email'], response['status'])

>>> Message _id abc123abc123abc123abc123abc123, to testaddress1@example.com, status sent
>>> ...

使用djrill发送html批量电子邮件时，如何获取每条消息的mandrill响应？

1 个答案: