我正在使用以下代码通过电子邮件发送 gerrit.txt 的内容，这是 HTML 代码，但它不起作用？它没有显示任何错误，但没有按照预期的方式工作。关于如何解决这个问题的任何输入？
from email.mime.text import MIMEText
from subprocess import check_call,Popen,PIPE
def email (body,subject,to=None):
msg = MIMEText("%s" % body)
msg['Content-Type'] = "text/html;"
msg["From"] = "userid@company.com"
if to!=None:
to=to.strip()
msg["To"] = "userid@company.com"
else:
msg["To"] = "userid@company.com"
msg["Subject"] = '%s' % subject
p = Popen(["/usr/sbin/sendmail", "-t"], stdin=PIPE)
def main ():
Subject ="test email"
email('gerrit.txt',Subject,'userid')
if __name__ == '__main__':
main()
请您参考如下方法:
什么都没有发送的原因是一旦你打开 sendmail 进程，邮件就永远不会写入它。此外，您需要将文本文件的内容读入一个变量以包含在消息中。
这是一个基于您的代码构建的简单示例。我没有对所有内容都使用 MIMEText 对象，因此请修改它以满足您的需要。
from email.mime.text import MIMEText
from subprocess import check_call,Popen,PIPE
def email (body,subject,to=None):
msg = MIMEText("%s" % body)
msg['Content-Type'] = "text/html;"
msg["From"] = "you@yoursite.com"
if to!=None:
to=to.strip()
msg["To"] = to
else:
msg["To"] = "user@domain.com"
msg["Subject"] = '%s' % subject
p = Popen(["/usr/sbin/sendmail", "-t", "-f" + msg["From"]], stdin=PIPE)
(stddata, errdata) = p.communicate(input="To: " + msg["To"] + "\r\nFrom: " + msg["From"] + "\r\nSubject: " + subject + "\r\nImportance: Normal\r\n\r\n" + body)
print stddata, errdata
print "Done"
def main ():
# open gerrit.txt and read the content into body
with open('gerrit.txt', 'r') as f:
body = f.read()
Subject ="test email"
email(body, Subject, "from@domain.com")
if __name__ == '__main__':
main()
