Question

以下内容在文件中。我使用shell脚本来调用python脚本。这个python脚本发送mail.But在邮件内容中我看到输出如下所示。它全部在线上。什么是我做错了

 /usr/bin/python $DIR/sm.py "$message"  "`cat /tmp/alert.txt`"

输入：以下是alert.txt

的内容

  Thu Jun 28 14:29:26 IST 2012

  Disk usage limit exceeded -Current disk usage is 167G-Configured disk usage is 200HTTPD connections exceeded configured usage limit -Current HTTPD connections is 21-Configured HTTPD connection is 20


  ========================OTHER INFO==================
  Total fds: 8
  Socket fds: 0
  Other fds: 8
  Free memory :Free Memory:183732
  Buffered memory Buffered Memory:78224
  Cache memory : Cache Memory:579040
  Disk usage is 167G
  DB connections 1
  Network connections 21
  CPU Usage: 0.0

输出：

  Thu Jun 28 14:29:26 IST 2012 Disk usage limit exceeded -Current disk usage is 167G-Configured disk usage is 200HTTPD connections exceeded configured usage limit -Current HTTPD connections is 21-Configured HTTPD connection is 20 ========================OTHER INFO================== Total fds: 8 Socket fds: 0 Other fds: 8 Free memory :Free Memory:183732 Buffered memory Buffered Memory:78224 Cache memory : Cache Memory:579040 Disk usage is 167G DB connections 1 Network connections 21 CPU Usage: 0.0

这是sm.py

import logging
import smtplib
import sys
from email.MIMEText import MIMEText
from email.MIMEMultipart import MIMEMultipart


      try:
         smaid = qs[0].id
         gmailUser = 'no-reply@xxxxxxxxxxx.com'
         gmailPassword = '12345'
         mailServer = smtplib.SMTP('smtp.gmail.com', 587)
         mailServer.ehlo()
         mailServer.starttls()
         mailServer.ehlo()
         mailServer.login(gmailUser, gmailPassword)

         to_addr = "xxxxx@xx.com"
         subject = sys.argv[1]
         body = sys.argv[2]
         try:
             msg = MIMEMultipart()
             msg['From'] = gmailUser
             msg['To'] = to_addr
             msg["Content-type"] = "text/html"
             sub = subject
             msg['Subject'] = sub
             body = body

             msg.attach(MIMEText(body, 'html'))
             mailServer.sendmail(gmailUser, to_addr, msg.as_string())
         except:
             write_exception("send exception")

         mailServer.close()
     except:
          write_exception("send exception1")

Answer 1

如果您坚持发送HTML，则需要HTML换行符：

import cgi
# escape special HTML characters
body = cgi.escape(body)
# use HTML line breaks
body = body.replace("\r\n", "\n").replace("\n", "<br />\n")

但可能您既不需要Multipart也不需要HTML，因此您只需使用smtplib.SMTP.sendmail：

headers = (('From', gmailUser), 
           ('To', to_addr), 
           ('Subject', subject))

# normalize newlines to CR+LF, as required by SMTP
body = body.replace("\r\n", "\n").replace("\n", "\r\n")

msg = '\r\n'.join("%s: %s" % kv for kv in headers) + '\r\n'*2 + body

mailServer.sendmail(gmailUser, [to_addr], msg)

此外，您不应通过命令行提供文件内容，因为命令行是长度限制的。相反，你应该通过STDIN提供文件，就像在python ... < /tmp/alert.txt中一样，并通过

读取

import sys
body = sys.stdin.read()

Answer 2

以text/plain而不是text/html发送。你那里没有任何HTML。这应该解决它，因为它取决于客户端正确显示文本。新行在HTML中的效果不同。

Answer 3

您将邮件的内容类型设置为“text / html”，因此用户的邮件代理显然会将主体解释并呈现为html。如果要保留纯文本格式，请不要将内容类型设置为text / html。

Answer 4

您使用的是哪个shell？

"`cat xxx`"

csh中的

将删除所有换行符并返回空格分隔的字符串

> cat >testfile
line 1
line 2
<ctrl-d>
> echo "`cat testfile`"
line 1 line 2

python脚本忽略文件中的换行符

4 个答案: