我有问题理解如何使用Python电子邮件附件。我已经成功地通过smtplib电子邮件发送了简单的消息。有人能解释一下如何在电子邮件中发送附件吗?我知道网上还有其他的帖子,但作为一个Python初学者,我发现它们很难理解。
当前回答
试试这个我希望这可能会有帮助
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
fromaddr = "youremailhere"
toaddr = input("Enter The Email Adress You want to send to: ")
# instance of MIMEMultipart
msg = MIMEMultipart()
# storing the senders email address
msg['From'] = fromaddr
# storing the receivers email address
msg['To'] = toaddr
# storing the subject
msg['Subject'] = input("What is the Subject:\t")
# string to store the body of the mail
body = input("What is the body:\t")
# attach the body with the msg instance
msg.attach(MIMEText(body, 'plain'))
# open the file to be sent
filename = input("filename:")
attachment = open(filename, "rb")
# instance of MIMEBase and named as p
p = MIMEBase('application', 'octet-stream')
# To change the payload into encoded form
p.set_payload((attachment).read())
# encode into base64
encoders.encode_base64(p)
p.add_header('Content-Disposition', "attachment; filename= %s" % filename)
# attach the instance 'p' to instance 'msg'
msg.attach(p)
# creates SMTP session
s = smtplib.SMTP('smtp.gmail.com', 587)
# start TLS for security
s.starttls()
# Authentication
s.login(fromaddr, "yourpaswordhere)
# Converts the Multipart msg into a string
text = msg.as_string()
# sending the mail
s.sendmail(fromaddr, toaddr, text)
# terminating the session
s.quit()
其他回答
其他的答案也很好,但我还是想分享一种不同的方法,以防有人正在寻找替代方案。
主要的区别是,使用这种方法,你可以使用HTML/CSS来格式化你的邮件,所以你可以有创意,给你的电子邮件一些样式。虽然没有强制使用HTML,但仍然可以只使用纯文本。
注意,该函数接受将电子邮件发送给多个收件人,也允许附加多个文件。
我只在python2上尝试过,但我认为它在python3上也能正常工作:
import os.path
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.application import MIMEApplication
def send_email(subject, message, from_email, to_email=[], attachment=[]):
"""
:param subject: email subject
:param message: Body content of the email (string), can be HTML/CSS or plain text
:param from_email: Email address from where the email is sent
:param to_email: List of email recipients, example: ["a@a.com", "b@b.com"]
:param attachment: List of attachments, exmaple: ["file1.txt", "file2.txt"]
"""
msg = MIMEMultipart()
msg['Subject'] = subject
msg['From'] = from_email
msg['To'] = ", ".join(to_email)
msg.attach(MIMEText(message, 'html'))
for f in attachment:
with open(f, 'rb') as a_file:
basename = os.path.basename(f)
part = MIMEApplication(a_file.read(), Name=basename)
part['Content-Disposition'] = 'attachment; filename="%s"' % basename
msg.attach(part)
email = smtplib.SMTP('your-smtp-host-name.com')
email.sendmail(from_email, to_email, msg.as_string())
我希望这能有所帮助!: -)
python 3的另一种方法(如果有人正在搜索):
import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
from email.mime.base import MIMEBase
from email import encoders
fromaddr = "sender mail address"
toaddr = "receiver mail address"
msg = MIMEMultipart()
msg['From'] = fromaddr
msg['To'] = toaddr
msg['Subject'] = "SUBJECT OF THE EMAIL"
body = "TEXT YOU WANT TO SEND"
msg.attach(MIMEText(body, 'plain'))
filename = "fileName"
attachment = open("path of file", "rb")
part = MIMEBase('application', 'octet-stream')
part.set_payload((attachment).read())
encoders.encode_base64(part)
part.add_header('Content-Disposition', "attachment; filename= %s" % filename)
msg.attach(part)
server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login(fromaddr, "sender mail password")
text = msg.as_string()
server.sendmail(fromaddr, toaddr, text)
server.quit()
确保在你的Gmail帐户上允许“不太安全的应用程序”
from email.mime.multipart import MIMEMultipart
from email.mime.image import MIMEImage
from email.mime.text import MIMEText
import smtplib
msg = MIMEMultipart()
password = "password"
msg['From'] = "from_address"
msg['To'] = "to_address"
msg['Subject'] = "Attached Photo"
msg.attach(MIMEImage(file("abc.jpg").read()))
file = "file path"
fp = open(file, 'rb')
img = MIMEImage(fp.read())
fp.close()
msg.attach(img)
server = smtplib.SMTP('smtp.gmail.com: 587')
server.starttls()
server.login(msg['From'], password)
server.sendmail(msg['From'], msg['To'], msg.as_string())
server.quit()
我能找到的最简单的代码是:
#for attachment email
from django.core.mail import EmailMessage
def attachment_email(request):
email = EmailMessage(
'Hello', #subject
'Body goes here', #body
'MyEmail@MyEmail.com', #from
['SendTo@SendTo.com'], #to
['bcc@example.com'], #bcc
reply_to=['other@example.com'],
headers={'Message-ID': 'foo'},
)
email.attach_file('/my/path/file')
email.send()
它基于官方的Django文档
下面是Oli为python3编写的修改版本
import smtplib
from pathlib import Path
from email.mime.multipart import MIMEMultipart
from email.mime.base import MIMEBase
from email.mime.text import MIMEText
from email.utils import COMMASPACE, formatdate
from email import encoders
def send_mail(send_from, send_to, subject, message, files=[],
server="localhost", port=587, username='', password='',
use_tls=True):
"""Compose and send email with provided info and attachments.
Args:
send_from (str): from name
send_to (list[str]): to name(s)
subject (str): message title
message (str): message body
files (list[str]): list of file paths to be attached to email
server (str): mail server host name
port (int): port number
username (str): server auth username
password (str): server auth password
use_tls (bool): use TLS mode
"""
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = COMMASPACE.join(send_to)
msg['Date'] = formatdate(localtime=True)
msg['Subject'] = subject
msg.attach(MIMEText(message))
for path in files:
part = MIMEBase('application', "octet-stream")
with open(path, 'rb') as file:
part.set_payload(file.read())
encoders.encode_base64(part)
part.add_header('Content-Disposition',
'attachment; filename={}'.format(Path(path).name))
msg.attach(part)
smtp = smtplib.SMTP(server, port)
if use_tls:
smtp.starttls()
smtp.login(username, password)
smtp.sendmail(send_from, send_to, msg.as_string())
smtp.quit()
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