Question

我已经汇总了一个代码，可以向多个收件人发送电子邮件。但是，每个收件人都会收到所有邮件，而不是他自己的邮件。

数据框：

email content mark@gmail.com Hi Mark, bla bla eve@gmail.com Hi Eve, bla bla john@gmail.com Hi, John bla bla

for content in df['content']:
    for email in df['email']:
       message = MIMEMultipart()
       message['Subject'] = "Subject"
       message['From'] = 'my email'
       message['Reply-to'] = 'my email'
       message['To'] = '{}'.format(email)

       text = MIMEText(mail)
       message.attach(text)

       server = smtplib.SMTP ('smtp.gmail.com',587)
       server.ehlo()
       server.starttls()
       server.login(login, password) 
       server.sendmail(message['From'], message['To'], message.as_string())
       server.quit()

Answer 1

试试这个：

df = {
      'content': ['test1', 'test2', 'test3'], 
        'email': ['mail1@mail.com', 'mail2@mail.com', 'mail3@mail.com']
     }

x = 0
for email in df['email']:
        print(email)
        print(df["content"][x]+"\n")
        x+=1

Answer 2

只需zip个列并省略一个循环：

for email, content in zip(df['email'], df['content']):
       message = MIMEMultipart()
       message['Subject'] = "Subject"
       message['From'] = 'my email'
       message['Reply-to'] = 'my email'
       message['To'] = '{}'.format(email)

       text = MIMEText(mail)
       message.attach(text)

       server = smtplib.SMTP ('smtp.gmail.com',587)
       server.ehlo()
       server.starttls()
       server.login(login, password) 
       server.sendmail(message['From'], message['To'], message.as_string())
       server.quit()

Answer 3

我正在迭代它两次。而不是使用2 for循环，迭代DataFrame的正确方法是使用

而不是：

for content in df['content']:
    for email in df['email']:

使用此：

for c in df.itertuples():
   message = MIMEMultipart()
   message['To'] = '{}'.format(c[0])

   text = MIMEText(c[1])
   message.attach(text)

将个人电子邮件发送给多个收件人

3 个答案: