Passing multiple PHP variables to shell_exec()? [duplicate]

Question

I am calling test.sh from PHP using shell_exec method.

$my_url="http://www.somesite.com/";
$my_refer="http://www.somesite.com/";
$page = shell_exec('/tmp/my_script.php $my_url $my_refer');

However, the command line script says it only received 1 argument: the /tmp/my_script.php

When i change the call to:

Code:

$page = shell_exec('/tmp/my_script.php {$my_url} {$my_refer}');

It says it received 3 arguments but the argv[1] and argv[2] are empty.

When i change the call to:

Code:

$page = shell_exec('/tmp/my_script.php "http://www.somesite.com/" "http://www.somesite.com/"');

The script finally receives all 3 arguments as intended.

Do you always have to send just quoted text with the script and are not allowed to send a variable like $var? Or is there some special way you have to send a $var?

Answer 1

Change

$page = shell_exec('/tmp/my_script.php $my_url $my_refer');

to

$page = shell_exec("/tmp/my_script.php $my_url $my_refer");

OR

$page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');

Also make sure to use escapeshellarg on both your values.

Example:

$my_url=escapeshellarg($my_url);
$my_refer=escapeshellarg($my_refer);

Answer 2

There is need to send the arguments with quota so you should use it like:

$page = shell_exec("/tmp/my_script.php '".$my_url."' '".$my_refer."'");

Answer 3

Variables won't interpolate inside of a single quoted string. Also you should make sure the your arguments are properly escaped.

 $page = shell_exec('/tmp/myscript.php '.escapeshellarg($my_url).' '.escapeshellarg($my_refer));

Answer 4

Change

$page = shell_exec('/tmp/my_script.php $my_url $my_refer');

to

$page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');

Then you code will tolerate spaces in filename.