AttributeError: 'tuple' 对象没有属性 'encode' python

Question

我编写了一个程序，使用 python smtp 库发送带有结果的电子邮件：

def send_mail(sender_mail, receiver_mail):
    msg = MIMEMultipart()
    msg['From'] = sender_mail
    msg['To'] = receiver_mail
    msg['Subject'] = "Test results"
    file = open(RESULTS_FILE, "r")
#   body = file.read()
    ##attachment = MIMEText(file.read())
    attachment = MIMEBase('application', "octet-stream")
    attachment.set_payload(file.read())
    file.close()
    attachment.add_header('Content-Disposition', 'attachment', 
    filename=RESULTS_FILE)
    msg.attach(attachment)
    body = "Attached the test results"
    msg.attach(MIMEText(body, 'plain'))

    server = smtplib.SMTP("172.16.100.9", 25)

    #server.starttls()
    #server.login(sender_mail, password)
    text = msg.as_string()
    server.sendmail(sender_mail, receiver_mail, text)
    server.quit()

但是当我尝试运行它时，似乎该行： text = msg.as_string()导致错误：

    Traceback (most recent call last):
  File "\\filer2\incam\test_scripts\run_only_ci.py", line 268, in <module>
    testresults.analyze_tests(TESTS_FOLDER, RESULTS_FOLDER, sender_mail, receiver_mail)
  File "\\filer2\incam\test_scripts\testresults.py", line 224, in analyze_tests
    send_mail(sender_mail, receiver_mail)
  File "\\filer2\incam\test_scripts\testresults.py", line 119, in send_mail
    text = msg.as_string()
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\message.py", line 158, in as_string
    g.flatten(self, unixfrom=unixfrom)
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\generator.py", line 116, in flatten
    self._write(msg)
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\generator.py", line 195, in _write
    self._write_headers(msg)
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\generator.py", line 222, in _write_headers
    self.write(self.policy.fold(h, v))
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\_policybase.py", line 326, in fold
    return self._fold(name, value, sanitize=True)
  File "C:\Users\fl_ncc\AppData\Local\Programs\Python\Python36\lib\email\_policybase.py", line 369, in _fold
    parts.append(h.encode(linesep=self.linesep, maxlinelen=maxlinelen))
AttributeError: 'tuple' object has no attribute 'encode'

为什么会发生？ 它以前有效，所以我不明白为什么会发生。 谢谢你的帮助。

Answer 1

注意你在send_mail函数中的代码“msg['To'] = receiver_mail”，这是没有必要的，因为sendmail有一个像flow一样的receiver_mail的参数。换句话说，你不必在msg['To'中强调receiver_mail ], msg['To'] 只显示邮件中的收件人信息。

server.sendmail(sender_mail, receiver_mail, text)

当 msg['To'] 被分配给像 ['XXX@XXX','XXX@XXX'] 这样的列表时会发生错误。如果你用像 "msg['To']='XXX@XXX' 这样的单个字符串来设置它“， 一切都会安好的。 然后，当接收者邮件的元素数 >1 时，您可以使用“for 循环”语句。 您还可以注释相关代码，如流动，以避免“for 循环”语句：

# msg['To'] = receiver_mail