来自python程序的sendmail
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在下面的代码中,从系统发送的邮件将转到用户的垃圾邮件文件夹,下面给出python代码。我怀疑发件人的发件人为root @。
如何纠正这个
def sendmail(to,fr,subject,msg):
sendmail_location = \"/usr/sbin/sendmail\" # sendmail location
p = os.popen(\"%s -t\" % sendmail_location, \"w\")
p.write(\"From: %s\\n\" % fr)
p.write(\"Reply-to: %s\\n\" % fr)
p.write(\"To: %s\\n\" % to)
p.write(\"Content-type: text/html\\n\")
p.write(\"Subject: %s\\n\" % subject)
p.write(\"\\n\") # blank line separating headers from body
p.write(msg)
status = p.close()
The mail format received in the
Delivered-To: harry@8767@gmail.com
Received: by 19.143.162.8 with SMTP id k6gm828f7tfe;
Tue, 19 Apr 2011 22:42:17 -0700 (PDT)
Received: by 10.68.9.168 with SMTP id a5try030f516pbb.481.1303278137028;
Tue, 19 Apr 2011 22:42:17 -0700 (PDT)
Return-Path: <root@.>
Received: from ([174.1.161.204])
by mx.google.com with ESMTPS id v4si18etrt0pbr.108.2011.04.19.22.42.15
(version=TLSv1/SSLv3 cipher=OTHER);
Tue, 19 Apr 2011 22:42:15 -0700 (PDT)
Received-SPF: neutral (google.com: 74.3.161.204 is neither permitted nor denied by best guess record for domain of root@.) client-ip=74.3.161.204;
Authentication-Results: mx.google.com; spf=neutral (google.com: 174.1.161.204 is neither permitted nor denied by best guess record for domain of root@.) smtp.mail=root@.
Received: (qmail 23122 invoked by uid 0); 19 Apr 2011 22:36:27 -0000
Date: 19 Apr 2011 22:36:27 -0000
Message-ID: <2010041954235627.25121.qmail@>
From: admin@xxxxxx.com
Subject: hi
没有找到相关结果
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委婪绷冗诉