Question

 # Web scraping

 import urllib
 import smtplib
 from urllib.request import urlopen
 from bs4 import BeautifulSoup

 def make_soup(url):
     thePage= urllib.request.urlopen(url)
     soupdata = BeautifulSoup(thePage, "html.parser")

    return soupdata


 soup = make_soup("http://www.met.gov.kw/?lang=eng")
      for record in soup.select('#newForecast'):
      information = record.text

content = information[48:440]
msg =  content

以下是代码的一部分我遇到问题时，我将内容中存储的信息传输到邮件，然后将其传递给sendmail（）函数，其中正文邮件应该显示为空你能告诉我出了什么问题吗？

server = smtplib.SMTP('smtp.gmail.com', 587)
server.starttls()
server.login("yourEmail", "yourPassword")
server.sendmail("placeholder", "placeholder", msg)
server.quit()

Answer 1

您的变量msg是原始内容字符串。对于电子邮件内容，您必须使用MIME类型（Multipurpose Internet Mail Extensions）。

在Python中，您可以使用对象MIMEText（仅用于文本）或MIMEMultipart（带附件）。它会使用函数as_string()将您的内容转换为等效的MIME格式。 或者您可以使用MIME格式构建自己的String。 ;）

这是我对您的代码的更正。这应该有效：

# import the object MIMEText
from email.mime.text import MIMEText
...
# build a instance of MIMEText from your content string
msg = MIMEText(content)
# Subject, From, To are the information, which the receiver will see
# It's no problem, if you use a fake address here. That's the way, how phishing mail or spam mail works
msg['Subject'] = 'Subject of my email'
msg['From'] = "placeholder"
msg['To'] = "placeholder"

...
# now use as_string() to convert your data to equivalent MIME format
# you can use `print msg.as_string()` to see how it is.
server.sendmail("placeholder", "placeholder", msg.as_string())

更多示例email examples

我正在尝试使用Python发送电子邮件，当我将最后一个字段作为变量传递时，它不会在电子邮件中发送任何内容

1 个答案: