Shell command not working via exec() in php

Question

I'm executing a php function via the PHP exec() function but it seems to only pass through 1 variable at time.

$url = "http://domain.co/test.php?phone=123&msg=testing"
exec('wget '.$url);

Within test.php file I have

$msg = "msg =".$_GET["msg"]." number=".$_GET["phone"];
mail('[email protected]', 'Working?', $msg);

I get an email which returns phone variable only.

But if I change the url as follows

$url = "http://domain.co/test.php?msg=testing&phone=123"

I get msg but not phone? Any ideas on what is causing this strange behavior?

james · Accepted Answer · 2013-07-17 14:07:11Z

4

The & sign is a special character in the Unix shell. You need to escape it:

exec("wget '$url'");

Also, if your URL is based on user input in any way, be very careful to escape it with escapeshellarg. Otherwise your users will be able to run arbitrary Unix commands on your server.

answered Jul 17, 2013 at 14:07

james

13.4k9 gold badges50 silver badges74 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Eun · Accepted Answer · 2013-07-17 14:07:45Z

1

$url = "http://domain.co/test.php?phone=123&msg=testing"
exec('wget "'.$url.'"');

you need to qoute the url

& is the sign for putting a task into background

answered Jul 17, 2013 at 14:07

Eun

4,1885 gold badges32 silver badges51 bronze badges

Collectives™ on Stack Overflow

Shell command not working via exec() in php

2 Answers 2

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Related