Python smtplib：只有其中的第一封电子邮件到达目的地 [英] Python smtplib: only the first email out of several arrives to destination

问题描述

我整理了一个函数 send_email [1]，用于发送支持纯文本和html消息的电子邮件。效果很好，但是我有一个问题，我不太了解如何调试。我的系统将 sendmail 作为其MTA。

I have put together a function send_email [1] to send emails with support for plain-text and html messages. It works well, but I have an issue I don't quite know how to debug. My system has sendmail as its MTA.

该函数在for循环中调用，如下所示：

The function is called in a for loop, as follows:

for data in data_set:
    subject, message = process(data)  # Irrelevant stuff

    send_email(subject, message, "[e-mail]")

在 smtplib [1]我看到对 send_email 的所有调用均已成功完成。奇怪的行为是：

In the debug output of smtplib [1] I see that all calls to send_email completed successfully. The weird behavior is:

• 如果消息是短的（我测试过单行），所有已发送的邮件实际上到达 [e-mail]


• 如果消息是不短的（也就是说，我使用真正的 process_data 函数生成的多行），只有第一 smtplib 的调试输出报告了每封电子邮件的成功，电子邮件确实到达了，而其他电子邮件则没有。


• 如果消息同样不短 但，目标地址为不同每条消息，然后所有消息到达其预期的目的地。

• If the message is "short" (I tested with a single line), all sent messages actually arrive to [e-mail]
• If the message is "not short" (that is, the multiple lines I generate with the real process_data function), only the first email does arrive, while the others don't, even though the debug output of smtplib in [1] reports success for each and all of the emails.
• If the message is equally "not short" but the destination address is different for each message, then all messages arrive to their intended destinations.

对于后一种情况， for 循环如下所示： p>

For the latter case, the for loop would look like:

addresses = ["[e-mail]", "[e-mail]", ...]
for data, addr in zip(data_set, addresses):
    subject, message = process(data)  # Irrelevant stuff

    send_email(subject, message, addr)

当然，预期的行为是针对不同数据的不同地址，但我担心如果不了解为什么会发生这种情况

The intended behavior is of course different addresses for different data, but I'm concerned that not understanding why this happens might bite me in an unexpected way later on.

[1]我的发送邮件功能：

[1] My send mail function:

import smtplib
import socket
import getpass
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def send_email (subject, message, to, reply_to='', cc='', html_message=''):

    COMMASPACE = ", "
    user, host = get_user_and_host_names()
    sender = '%s@%s' % (user, host)

    receivers = make_address_list(to)
    copies = make_address_list(cc)

    msg = MIMEMultipart('alternative')

    msg['Subject'] = subject
    msg['From'] = sender
    msg['To'] = COMMASPACE.join(receivers)
    if reply_to:
        msg.add_header('Reply-to', reply_to)
    if len(copies):
        msg.add_header('CC', COMMASPACE.join(copies))

    # According to RFC 2046, the last part of a multipart message, in this case
    # the HTML message, is best and preferred.
    if message:
        msg.attach( MIMEText(message, 'plain'))
    if html_message:
        msg.attach( MIMEText(html_message, 'html'))

    smtpObj = smtplib.SMTP('localhost')
    smtpObj.set_debuglevel(1)
    smtpObj.sendmail(sender, receivers, msg.as_string())
    smtpObj.quit()
    print "\nSuccessfully sent email to:", COMMASPACE.join(receivers)

def get_user_and_host_names():
    user = getpass.getuser()
    host = socket.gethostname()
    return user, host

def make_address_list (addresses):
    if isinstance(addresses, str):
        receivers = addresses.replace(' ','').split(',')
    elif isinstance(addresses, list):
        receivers = addresses
    return receivers

推荐答案

我的工作解决方案以这种方式格式化列表（经过多个小时的实验）：

My Working solution to format your list in this manner (after many hours of experiment):

[ [e-mail]， secon [e-mail]， [e-mail]]

我使用过：

    to_address = list(str(self.to_address_list).split(","))

将我的QString转换为字符串，然后用，分割成列表

to convert my QString into string then into list with splitting with a ","

在我的情况下，COMMASPACE无法正常工作，因为在拆分时默认已添加空格。

In my case COMMASPACE was not working, because on splitting a space was already added by default.

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