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
下面是我从SoccerPlayer的帖子中找到的内容的组合,下面的链接使我更容易附加一个xlsx文件。在这里找到
file = 'File.xlsx'
username=''
password=''
send_from = ''
send_to = 'recipient1 , recipient2'
Cc = 'recipient'
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = send_to
msg['Cc'] = Cc
msg['Date'] = formatdate(localtime = True)
msg['Subject'] = ''
server = smtplib.SMTP('smtp.gmail.com')
port = '587'
fp = open(file, 'rb')
part = MIMEBase('application','vnd.ms-excel')
part.set_payload(fp.read())
fp.close()
encoders.encode_base64(part)
part.add_header('Content-Disposition', 'attachment', filename='Name File Here')
msg.attach(part)
smtp = smtplib.SMTP('smtp.gmail.com')
smtp.ehlo()
smtp.starttls()
smtp.login(username,password)
smtp.sendmail(send_from, send_to.split(',') + msg['Cc'].split(','), msg.as_string())
smtp.quit()
其他回答
下面是我从SoccerPlayer的帖子中找到的内容的组合,下面的链接使我更容易附加一个xlsx文件。在这里找到
file = 'File.xlsx'
username=''
password=''
send_from = ''
send_to = 'recipient1 , recipient2'
Cc = 'recipient'
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = send_to
msg['Cc'] = Cc
msg['Date'] = formatdate(localtime = True)
msg['Subject'] = ''
server = smtplib.SMTP('smtp.gmail.com')
port = '587'
fp = open(file, 'rb')
part = MIMEBase('application','vnd.ms-excel')
part.set_payload(fp.read())
fp.close()
encoders.encode_base64(part)
part.add_header('Content-Disposition', 'attachment', filename='Name File Here')
msg.attach(part)
smtp = smtplib.SMTP('smtp.gmail.com')
smtp.ehlo()
smtp.starttls()
smtp.login(username,password)
smtp.sendmail(send_from, send_to.split(',') + msg['Cc'].split(','), msg.as_string())
smtp.quit()
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()
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText
from email.MIMEImage import MIMEImage
import smtplib
msg = MIMEMultipart()
msg.attach(MIMEText(file("text.txt").read()))
msg.attach(MIMEImage(file("image.png").read()))
# to send
mailer = smtplib.SMTP()
mailer.connect()
mailer.sendmail(from_, to, msg.as_string())
mailer.close()
从这里改编。
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!: -)
Gmail版本,使用Python 3.6(请注意,您需要更改Gmail设置,以便能够通过smtp从它发送电子邮件:
import smtplib
from email.mime.text import MIMEText
from email.mime.multipart import MIMEMultipart
from email.mime.application import MIMEApplication
from os.path import basename
def send_mail(send_from: str, subject: str, text: str,
send_to: list, files= None):
send_to= default_address if not send_to else send_to
msg = MIMEMultipart()
msg['From'] = send_from
msg['To'] = ', '.join(send_to)
msg['Subject'] = subject
msg.attach(MIMEText(text))
for f in files or []:
with open(f, "rb") as fil:
ext = f.split('.')[-1:]
attachedfile = MIMEApplication(fil.read(), _subtype = ext)
attachedfile.add_header(
'content-disposition', 'attachment', filename=basename(f) )
msg.attach(attachedfile)
smtp = smtplib.SMTP(host="smtp.gmail.com", port= 587)
smtp.starttls()
smtp.login(username,password)
smtp.sendmail(send_from, send_to, msg.as_string())
smtp.close()
用法:
username = 'my-address@gmail.com'
password = 'top-secret'
default_address = ['my-address2@gmail.com']
send_mail(send_from= username,
subject="test",
text="text",
send_to= None,
files= # pass a list with the full filepaths here...
)
要与任何其他电子邮件提供商一起使用,只需更改smtp配置。
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