unable to execute shell_exec() with variable

Question

I'm trying to run shell_exec() with variable passed with AJAX from client.

This code causes error (input file doesn't exist!):

$searched_image = escapeshellarg("/home/XXX/XXX/XXX/XXX/XXX/sp_dom1.jpg");
$old_path = getcwd();
chdir('../elevation/source_code/altitudes_system/');
$altitudes_system_result = shell_exec('./predict_altitude.sh -i "{$searched_image}" -p basic -o 0');
chdir($old_path);

But when I replace "{$searched_image}" in shell_exec(...) with /home/XXX/XXX/XXX/XXX/XXX/sp_dom1.jpg code works well:

$old_path = getcwd();
chdir('../elevation/source_code/altitudes_system/');
$altitudes_system_result = shell_exec('./predict_altitude.sh -i /home/XXX/XXX/XXX/XXX/XXX/sp_dom1.jpg -p basic -o 0');
chdir($old_path);

Don't you have any idea why it works like this?

Get rid of the double quotes around {$searched_image}, as escapeshellarg() is likely already adding single quotes for you. Build a string and feed that to shell_exec() so you can echo your command line and see what it actually is. — WWW
– WWW, Commented Feb 15, 2016 at 19:05
Still doesnt work when I remove double quotes. And when I build the string - $searched_image = escapeshellarg('./predict_altitude.sh -i /home/XXX/XXX/XXX/XXX/XXX/sp_dom1.jpg -p basic -o 0'); like this, it outputs exactly what I want to have in my command line - './predict_altitude.sh -i ....' and then when I call shell_exec($searched_image); it doesn't output result but it doesn't event output error.. — Petr Bečka
– Petr Bečka, Commented Feb 15, 2016 at 19:19
In your example in your comment, you're using escapeshellarg() incorrectly. You use it to escape a single argument, not a whole command line. — WWW
– WWW, Commented Feb 15, 2016 at 19:30

fusion3k · Accepted Answer · 2016-02-15 19:14:31Z

1

You write:

'./predict_altitude.sh -i "{$searched_image}" -p basic -o 0'

Inside single-quoted strings variables are not evaluated.

You can use this, instead:

"./predict_altitude.sh -i '{$searched_image}' -p basic -o 0"

Or - to avoid unpredictable evaluations - this:

$cmd = './predict_altitude.sh -i \''.$searched_image.'\' -p basic -o 0';
shell_exec( $cmd );

answered Feb 15, 2016 at 19:14

fusion3k

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