duanmeng1862
duanmeng1862
2017-08-12 12:46

从php运行shell脚本作为指定用户

  • shell
  • linux
  • php
  • xampp
  • daemon

I am trying to run a shell script from a PHP script.

PHP code :

<? php
     $sss = escapeshellarg('virtualbox');
     $result = shell_exec("/home/hani/Desktop/launchscript.sh '$sss' 2>&1 ");
     echo "<pre>$result</pre>";
     echo "<br />";
     echo (shell_exec('whoami'));
?>

my shell script :

#!/bin/bash


sss=$1
echo 'the sudo password' |sudo -S service $1 restart

After I run the PHP code in a web server (Xampp), I got this output :

[sudo] password for daemon: Sorry, try again.
[sudo] password for daemon: 
sudo: 1 incorrect password attempt

daemon

Although, I haven't set any password for the daemon user. And after I checked the current user running the PHP code I found it is daemon.
After many researches here and in the net, I found that daemon can't run sudo commands.
I also found that I can fix this by editing the sudoers file and giving permissions to the daemon user to run sudo commands. However this is not a secured solution.
so my question is : How to run that script via the PHP code but not as a daemon?
PS : I tried this in order to change the current user running the PHP file :

$result = shell_exec(" sudo -u hani /home/hani/Desktop/launchscript.sh '$sss' 2>&1 ");  

But I got this output in the browser :

sudo: no tty present and no askpass program specified

and the user remains daemon.

I am using Xampp in Ubuntu 16.04
Another information, I run this command in the terminal to know the owner of the 'httpd' service :

ps -ef | egrep '(httpd)' | grep -v `whoami` | grep -v root | head -n1 | awk '{print $1}'

the output is : daemon

  • 点赞
  • 回答
  • 收藏
  • 复制链接分享

1条回答

为你推荐