1

我有课，打算从 Git 提交中获取用户名和他的电子邮件：

class BitbucketData(object):

    def get_user_name(self):

        proc = subprocess.Popen("git --no-pager show -s --format='%an'", stdout=subprocess.PIPE)
        committer_name = proc.stdout.read()

        if committer_name:
            return committer_name

    def get_user_email(self):

        proc = subprocess.Popen("git --no-pager show -s --format='%aE'", stdout=subprocess.PIPE)
        committer_email = proc.stdout.read()

        if committer_email:
            return committer_email

然后它用于向用户发送通知（底部是工作版本 - 没有变量 - 全部sender和receiver数据集明确，而不是变量 - 他们在这里评论）：

class Services(object):


    def sendmail(self, event):

        repo = BitbucketData()

        #to_address = repo.get_user_email()
        #to_address = 'user@domain.com'
        #to_name = repo.get_user_name()
        #to_name = 'Test user'

        subject = 'Bamboo build and deploy ready'
        sender = 'Bamboo agent <user@domain.com>'

        text_subtype = 'plain'
        message = """
        Hello, {}.

        Your build ready.
        Link to scenario: URL
        Link to build and deploy results: {})

        """.format('Test user', os.environ['bamboo_resultsUrl'])

        msg = MIMEText(message, text_subtype)
        msg['Subject'] = subject
        msg['From'] = sender
        msg['To'] = 'Test user <user@domain.com>'

        smtpconnect = smtplib.SMTP('outlook.office365.com', 587)
        smtpconnect.set_debuglevel(1)
        smtpconnect.starttls()
        smtpconnect.login('user@domain.com', 'password')
        smtpconnect.sendmail('user@domain.com', 'user@domain.com', msg.as_string())
        smtpconnect.quit()

        print('Mail sent')

        print(repo.get_user_email())

问题是 - 如果我使用来自变量的数据（例如to_address = 'user@domain.com'或使用BitbucketData()类- 我从Office365服务器to_address = repo.get_user_email()收到错误：

...
reply: '250 2.1.0 Sender OK\r\n'
reply: retcode (250); Msg: 2.1.0 Sender OK
send: "rcpt TO:<'user@domain.com'>\r\n"
reply: '501 5.1.3 Invalid address\r\n'
reply: retcode (501); Msg: 5.1.3 Invalid address
...
  File "C:\Python27\lib\smtplib.py", line 742, in sendmail
    raise SMTPRecipientsRefused(senderrs)
smtplib.SMTPRecipientsRefused: {"'user@domain.com'\n": (501, '5.1.3 Invalid address')}

使用变量时，代码如下所示：

class Services(object):


    def sendmail(self, event):

        repo = BitbucketData()

        to_address = repo.get_user_email()
        #to_address = 'user@domain.com'
        to_name = repo.get_user_name()
        #to_name = 'Test user'
        from_address = 'user@domain.com'

        subject = 'Bamboo build and deploy ready'
        sender = 'Bamboo agent <user@domain.com>'

        text_subtype = 'plain'
        message = """
        Hello, {}.

        Your build ready.
        Link to scenario: URL
        Link to build and deploy results: {})

        """.format(to_name, os.environ['bamboo_resultsUrl'])

        msg = MIMEText(message, text_subtype)
        msg['Subject'] = subject
        msg['From'] = sender
        msg['To'] = to_name

        smtpconnect = smtplib.SMTP('outlook.office365.com', 587)
        smtpconnect.set_debuglevel(1)
        smtpconnect.starttls()
        smtpconnect.login('user@domain.com', 'password')
        smtpconnect.sendmail(from_address, to_address, msg.as_string())
        smtpconnect.quit()

        print('Mail sent')

        print(repo.get_user_email())

我（或 Microsoft SMTP ......）在这里做错了什么？

UPD

def sendmail(self, event):

    repo = BitbucketData()
    print(repr(repo.get_user_email()))
    ...

给我：

...
Creating new AutoEnv config
"'user@domain.com'\n"
send: 'ehlo pc-user.kyiv.domain.net\r\n'
...
4

1 回答 1

2

似乎您从 git 命令收到以下输出 -

"'user@domain.com'\n"

您直接将其传递给 smtpconnect，这会导致问题，因为这不是有效的电子邮件地址。您可能需要从中获取实际的字符串。获得它的一种方法是使用ast.literal_eval()function for that 。例子 -

>>> import ast
>>> e = ast.literal_eval("'user@domain.com'\n")
>>> print(e)
user@domain.com

get_user_email()从函数返回电子邮件时，您需要这样做。很可能committer_name也有这个问题，所以你可能也想这样做。

从文档ast.literal_eval()-

ast.literal_eval(node_or_string)

安全地评估包含 Python 文字或容器显示的表达式节点或 Unicode 或 Latin-1 编码字符串。提供的字符串或节点只能由以下 Python 文字结构组成：字符串、数字、元组、列表、字典、布尔值和无。

于 2015-07-24T09:55:02.727 回答