问题
I am writing a simple sendmail function to myself and I keep getting this error:
NameError: name 'SMTPException' is not defined
What is wrong with my code? Any suggestions?
import smtplib
sender = "user@gmail.com"
receiver = ["user@gmail.com"]
message = "Hello!"
try:
session = smptlib.SMTP('smtp.gmail.com',587)
session.ehlo()
session.starttls()
session.ehlo()
session.login(sender,'password')
session.sendmail(sender,receiver,message)
session.quit()
except SMTPException:
print('Error')
回答1:
In Python, you will need to fully qualify the name by prefixing it with its module:
except smtplib.SMTPException:
This is true unless you specifically import the unqualified name (but I wouldn't recommend doing this for your program, just showing what's possible):
from smtplib import SMTPException
回答2:
That misspelling occurred many times to me as well! One way to circumvent this "problem", is to use yagmail.
Jokes aside, I recently created yagmail to make it easier to send emails.
For example:
import yagmail
yag = yagmail.SMTP('user@gmail.com', 'password')
yag.send(contents = "Hello!")
It uses several shortenings here, for example when To
is not defined, it will send a mail to the same email who registered on the server. Also the port and host are the default, which makes it very concise.
In fact, since it seems you want to close the connection immediately, you can even use this one-liner:
yagmail.SMTP('user@gmail.com', 'password').send(contents = "Hello!")
For security, you can keep your password in the keyring
(see documentation) such that you do not have to keep your personal password in your scripts, very important! It'll even save you more precious screen-estate.
Going all-in with the package (@gmail.com
is default), you can get away with the following:
yagmail.SMTP('user').send('', 'Hello!')
Good luck.
来源:https://stackoverflow.com/questions/13115724/new-to-python-gmail-smtp-error