使用Python获取无效的函数名称警告

Question

我正在使用格式化Python代码pylint并获取一些不需要的警告消息,如下所示.

C:204, 0: Invalid function name "sendPasswordViaEmail" (invalid-name)
C:207, 4: Invalid variable name "to" (invalid-name)
W:213, 4: Statement seems to have no effect (pointless-statement)

我正在解释与以下消息相关的代码.

if request.method == 'POST':
        name = request.POST.get('uname')
        email = request.POST.get('uemail')
        range_start = 10 ** (4 - 1)
        range_end = (10 ** 4) - 1
        password = randint(range_start, range_end)
        passw = User(
            uname=name,
            password=password,
            raw_password=password,
        )
        passw.save()
        sendPasswordViaEmail(str(password), email)
    return render(request, 'bookingservice/login.html')


def sendPasswordViaEmail(password, email):
    """ Send email to user"""

    to = email
    gmail_user = settings.EMAIL
    gmail_pwd = settings.PASSWORD
    smtpserver = smtplib.SMTP("smtp.gmail.com", 587)
    smtpserver.ehlo()
    smtpserver.starttls()
    smtpserver.ehlo
    smtpserver.login(gmail_user, gmail_pwd)
    header = 'To:' + to + '\n' + 'From: ' + \
        gmail_user + '\n' + 'Subject:password \n'
    msg = header + '\n Your password is\n\n' + password
    smtpserver.sendmail(gmail_user, to, msg)
    smtpserver.close()

Answer 1

这些消息是警告(您的代码有效),而不是错误.

sendPasswordViaEmail是骆驼案的名字.功能应该是"蛇案".重命名它send_password_via_email.

to太短了.尝试recipient或其他有意义的事情.

包含多个smtpserver.ehlo括号的行什么都不做,因此没用.