CODEWITHSUNDEEP
phpclinux
Orca
Orca

Reputation: 73

PHP Shell_Exec not working?

I have the following code:

$output = shell_exec("./Program $var1 $var2");
echo "<pre>$output</pre>";

It doesn't work but 

$output = shell_exec("ls");
echo "<pre>$output</pre>";

does work.

$output = shell_exec("top");
echo "<pre>$output</pre>";

also doesn't work for example. Why?

Upvotes: 0

Views: 151

Answers (2)

Orangepill
Orangepill

Reputation: 24645

Depending on the content of $var1 and $var2 you may need to do an escapeshellarg call around it.

 $output = shell_exec("./Program ".escapeshellarg($var1)." ".escapeshellarg($var2));

even if it doesn't work it might be a good idea. Also confirm that your Path is correct. Maybe with a file_exits('./Program'); check

Upvotes: 0

Eric Wich
Eric Wich

Reputation: 1544

This is most certainly a permissions issue. Make sure that the file you're trying to execute with the ./ command from your script has +x perms. Here's a previous thread about giving files executable permissions: Creating executable files in Linux.

If the file already has +x rights, it could be a permissions issue with your script running the commands. Either way, if you can run ls but not ./ and top, has to be permissions.

Edit: The link I gave, I realize has a lot of info about Perl and bash scripts. The important part is that the command to make a file executable is 

chmod +x ProgramName

Upvotes: 1

Related Questions