使用python和gmail在同一线程中发送多封电子邮件

use*_*813 2 python email

我正在运行一个程序。当该程序得到结果时,它将使用以下功能向我发送电子邮件:

def send_email(message):

    import smtplib

    gmail_user = OMITTED
    gmail_pwd = OMITTED
    FROM = OMITTED
    TO = OMITTED #must be a list

    try:
        #server = smtplib.SMTP(SERVER) 
        server = smtplib.SMTP("smtp.gmail.com", 587) #or port 465 doesn't seem to work!
        server.ehlo()
        server.starttls()
        server.login(gmail_user, gmail_pwd)
        server.sendmail(FROM, TO, message)
        #server.quit()
        server.close()
        print 'successfully sent the mail'
    except:
        print "failed to send mail"  
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免责声明:我在Stack Overflow的某处找到了此代码。不是我的 我删去了某些部分,因为它们似乎没有特殊意义。

有时我的代码会得到很多结果,并且我在不到20秒的时间内收到了150多种不同的电子邮件。

如何修改上面的函数,以便程序在同一线程中将所有结果发送给我?

如果您不明白我的意思,我希望我的收件箱看起来像这样:

sender@gmail.com(150) ...  
... (other emails from other senders)  
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代替:

sender@gmail.com ...  
sender@gmail.com ...  
sender@gmail.com ...  
sender@gmail.com ...  
sender@gmail.com ...  
...  
sender@gmail.com ...  
... (other emails from other senders)
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编辑

为了解决该问题,我要做的就是重新插入我先前删除的代码部分。全部功能是这一功能:

def send_email(TEXT):

    import smtplib

    gmail_user = OMITTED
    gmail_pwd = OMITTED
    FROM = OMITTED
    TO = OMITTED #must be a list
    SUBJECT = "Big brother candidate"
    #TEXT = "Testing sending mail using gmail servers"

    # Prepare actual message
    message = """\From: %s\nTo: %s\nSubject: %s\n\n%s
    """ % (FROM, ", ".join(TO), SUBJECT, TEXT)

    try:
        #server = smtplib.SMTP(SERVER) 
        server = smtplib.SMTP("smtp.gmail.com", 587) #or port 465 doesn't seem to work!
        server.ehlo()
        server.starttls()
        server.login(gmail_user, gmail_pwd)
        server.sendmail(FROM, TO, message)
        #server.quit()
        server.close()
        print 'successfully sent the mail'
    except:
        print "failed to send mail"  
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jmu*_*gz3 5

这是一个古老的问题,但是我不得不回答这个问题,因为有一种方法可以执行OP想要的操作。您可以通过在邮件中添加标题,并在发送另一封电子邮件时引用它们来实现。例如

from email.utils import make_msgid

my_id = make_msgid()

#Build your email as you normally do, and add ID as a message header
message = MIMEMultipart()
message["Message-ID"]  = my_id 
message["Subject"] = "test"
message["From"] = from_email 
# ...etc and send your email using smtp.sendmail

# On the reply (or when sending another email), add the following headers
message["In-Reply-To"] = my_id
message["References"] = my_id

# ...send your email using smtp.sendmail
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当您检查邮件客户端时,您会看到后一封电子邮件将作为对前一封电子邮件的答复而发布,从而创建了通常在常用电子邮件客户端(Gmail,收件箱,Outlook,Yahoo等)中看到的线程。