Question

我有课程，打算从Git提交中获取用户名和电子邮件：

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

然后用于向用户发送通知（底部是工作版本 - 没有变量 - 显式设置了所有sender和receiver数据，而不是变量 - 他们在这里评论过）：

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 - 如果我使用变量数据（例如to_address = 'user@domain.com'或使用BitbucketData()类to_address = repo.get_user_email() - 我从 Office365 服务器：

... 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：

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' ...

Answer 1

好像您从git命令收到以下内容 -

"'user@domain.com'\n"

您是直接将此传递给smtpconnect，这导致了问题，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()函数。示例 -

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

从get_user_email()函数返回电子邮件时，您需要执行此操作。很可能committer_name也有这个问题，所以你也可能想要这样做。

来自ast.literal_eval() -

的文档

<强> ast.literal_eval（node_or_string）

安全地评估表达式节点或包含Python文字或容器显示的Unicode或Latin-1编码字符串。提供的字符串或节点可能只包含以下Python文字结构：字符串，数字，元组，列表，dicts，布尔值和None。

错误的可变数据类型？

1 个答案: