【发布时间】:2015-06-24 14:10:58
【问题描述】:
我正在使用以下代码
import smtplib
import mimetypes
from email.mime.multipart import MIMEMultipart
from email import encoders
from email.message import Message
from email.mime.audio import MIMEAudio
from email.mime.base import MIMEBase
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
emailfrom = "sender@example.com"
emailto = "destination@example.com"
fileToSend = "hi.csv"
username = "user"
password = "password"
msg = MIMEMultipart()
msg["From"] = emailfrom
msg["To"] = emailto
msg["Subject"] = "help I cannot send an attachment to save my life"
msg.preamble = "help I cannot send an attachment to save my life"
ctype, encoding = mimetypes.guess_type(fileToSend)
if ctype is None or encoding is not None:
ctype = "application/octet-stream"
maintype, subtype = ctype.split("/", 1)
if maintype == "text":
fp = open(fileToSend)
# Note: we should handle calculating the charset
attachment = MIMEText(fp.read(), _subtype=subtype)
fp.close()
elif maintype == "image":
fp = open(fileToSend, "rb")
attachment = MIMEImage(fp.read(), _subtype=subtype)
fp.close()
elif maintype == "audio":
fp = open(fileToSend, "rb")
attachment = MIMEAudio(fp.read(), _subtype=subtype)
fp.close()
else:
fp = open(fileToSend, "rb")
attachment = MIMEBase(maintype, subtype)
attachment.set_payload(fp.read())
fp.close()
encoders.encode_base64(attachment)
attachment.add_header("Content-Disposition", "attachment", filename=fileToSend)
msg.attach(attachment)
server = smtplib.SMTP("smtp.gmail.com:587")
server.starttls()
server.login(username,password)
server.sendmail(emailfrom, emailto, msg.as_string())
server.quit()
我收到错误 “不接受用户名和密码。在\n5.7.8 https://support.google.com/mail/answer/14257 了解更多信息” 按照那里写的,我已经改变了允许不太安全的应用程序:开
但得到同样的错误! 有什么帮助吗??
【问题讨论】:
-
是您使用的用户名的完整地址。例如:name@gmail.com
-
是的!执行时正确插入用户名密码!