passing a variable from php to bash

Question

I cannot seem to get a variable passed to my bash script from php. $uaddress and $upassword come up empty no matter what I try.

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Answer 1

You need to pass the variables as arguments to the shell script, and the shell script has to read its arguments.

So in PHP:

$useraddress = escapeshellarg('mytestuser@tpccmedia.com');
$upassword = escapeshellarg('test1234');
$addr = shell_exec("sudo /home/tpccmedia/cgi-bin/member_add_postfixadmin $useraddress $upassword 2>&1");

and in the shell script:

useraddress=$1
upassword=$2