使用shell_exec（）运行shell脚本时，由putenv（）设置的环境变量不存在。 的PHP

Question

I have a PHP script in which I am setting a environment variable using putenv() and then I am calling a shell script using shell_exec() in this PHP script. In shell script I am trying to use that environment variable.

My PHP script is like:

<?php
...
...
putenv("USERHOME=/home/user/mnt");
...
...
shell_exec("sudo ./script.sh arg1 arg2");
...
...
?>

My shell script is like:

#!/bin/sh
...
...
echo "USER HOME=" $USERHOME
...
<---Another use of $USERHOME----->
...
exit $?

I am using Xampp for my application and PHP version is 5.3.8.

Previously it was working fine on fedora core 3. But recently I moved my application to CentOS 7 and now it is not working. On CentOS 7, I am getting NULL value of USERHOME variable.

Answer 1

In PHP < 5.4 there's a safe_mode directive that limits what environment variables you can write with putenv . Edit your php.ini file and remove USERHOME from the safe_mode_protected_env_vars setting.

http://php.net/manual/en/ini.sect.safe-mode.php#ini.safe-mode-protected-env-vars

Answer 2

After some research and debugging, I am able to get the correct value of environment variable in my shell script, which is set by putenv() in my PHP script.

First of all, I am new to PHP and here also. And I missed some things in question while posting it. Now I have edited my question and provided all the required details and script contents correctly.

Root Cause and Resolution:

In my PHP script I was running the shell script through shell_exec() and was using sudo in the command shell_exec("sudo ./script.sh arg1 arg2"); This sudo is the main problem here.

Removing it and calling the script simply shell_exec("./script.sh arg1 arg2"); did the trick.

The shell_exec() call with sudo was working fine on Fedora Core 3. But when I moved my application to CentOS 7, it stopped working. After removing sudo from the command it is running as expected on CentOS 7.