将 subprocess.Popen 命令的输出获取到网页中？ （也许是阿帕奇的问题）
这是我的程序 >
print "Content-type:text/html\r\n\r\n"
print "File starting to execute"
print "<br>"
proc = subprocess.Popen(["sudo", "python", "test3.py"], stdout=subprocess.PIPE)
output = proc.stdout.read()
print "output is %s" %output
print "<br>"
print "File Executed Awesomely"
因此，当我从命令行运行它时，它工作得很好，如下所示 -
[root@localhost html]# python test2.py
Content-type:text/html
File starting to execute
<br>
output is .
Sent 1 packets.
<br>
File Executed Awesomely
[root@localhost html]#
这是完美的“。发送了 1 个数据包。”就是我想要的。但是当我从网页运行它时，网页只是有
File starting to execute
output is
File Executed Awesomely
，所以我最初认为这是因为我在抓取输出时做了一些错误的事情，但我用wireshark（我的另一个程序，它调用发送数据包）监听端口，它看起来像通过网页调用没有显示任何数据包，但当我在命令行上调用它时会显示数据包（以相同的方式）。看看我的 apache error_log->
[Wed Jan 18 18:15:11 2012] [notice] suEXEC mechanism enabled (wrapper: /usr/sbin/suexec)
[Wed Jan 18 18:15:11 2012] [notice] Digest: generating secret for digest authentication ...
[Wed Jan 18 18:15:11 2012] [notice] Digest: done
[Wed Jan 18 18:15:11 2012] [warn] ./mod_dnssd.c: No services found to register
[Wed Jan 18 18:15:11 2012] [notice] Apache/2.2.17 (Unix) DAV/2 configured -- resuming normal operations
关于如何修复它以便我的 apache cgi-bin 脚本以与命令行相同的方式运行的任何建议？
编辑：在几次调用后查看日志，它会重复执行此操作
[Wed Jan 18 18:22:37 2012] [error] [client 10.117.153.89] :
[Wed Jan 18 18:22:37 2012] [error] [client 10.117.153.89] sorry, you must have a tty to run sudo
Here is my program
import subprocess
print "Content-type:text/html\r\n\r\n"
print "File starting to execute"
print "<br>"
proc = subprocess.Popen(["sudo", "python", "test3.py"], stdout=subprocess.PIPE)
output = proc.stdout.read()
print "output is %s" %output
print "<br>"
print "File Executed Awesomely"
So when I run it from the command line it works great, like follows->
[root@localhost html]# python test2.py
Content-type:text/html
File starting to execute
<br>
output is .
Sent 1 packets.
<br>
File Executed Awesomely
[root@localhost html]#
that is perfect the ". Sent 1 packets." is what I want. But when I run it from the webpage, the webpage just has
File starting to execute
output is
File Executed Awesomely
so I originally thought this was because I was doing something wrong grabbing the output but I listened on the port with wireshark (my other program it calls send a packet) and it looks like no packets shows up via the webpage call, but it does when I call it on the command line (the same way). Looking at my apache error_log->
[Wed Jan 18 18:15:11 2012] [notice] suEXEC mechanism enabled (wrapper: /usr/sbin/suexec)
[Wed Jan 18 18:15:11 2012] [notice] Digest: generating secret for digest authentication ...
[Wed Jan 18 18:15:11 2012] [notice] Digest: done
[Wed Jan 18 18:15:11 2012] [warn] ./mod_dnssd.c: No services found to register
[Wed Jan 18 18:15:11 2012] [notice] Apache/2.2.17 (Unix) DAV/2 configured -- resuming normal operations
any suggestions on how I can fix it so my apache cgi-bin script runs the same way as the command line?
EDIT: looking at the log after a few calls it does this repeatedly
[Wed Jan 18 18:22:37 2012] [error] [client 10.117.153.89] :
[Wed Jan 18 18:22:37 2012] [error] [client 10.117.153.89] sorry, you must have a tty to run sudo
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消息
抱歉，您必须有一个 tty 才能运行 sudo是关键。首先，让你的 apache 运行 sudo 至少可以说是危险的，但如果你真的想这样做......有一种方法，编辑/etc/sudoers(visudo< /code>）并找到Defaults requiretty部分（man sudoers）。注意：永远不要让 apache 使用 sudo 运行任何东西，准确指定它需要做什么，仅此而已！
顺便说一句：如果您启用了 SELinux 或其他 LSM 模块，它可能仍然无法工作。
The message
sorry, you must have a tty to run sudois the key. First of all, letting your apache run sudo is dangerous to say the least, but if you really really want to do it... there's a way, edit/etc/sudoers(visudo) and locate theDefaults requirettypart (man sudoers).NB: Never let apache run anything using sudo, specify exactly what it needs to do and nothing more!
BTW: It may still not work if you have SELinux enabled or other LSM module.