PHP Shell_Exec not working with variable

Question

I'm trying to pass a variable into shell_exec, I've looked at other tutorials but they still dont help.

I want to pass $var[1] into the shell_exec so it will just display this specific file or directory's information

$output = shell_exec('ls -l'.$var[1]);

What is inside $var[1]? Can you do print_r($var[1]) and include the output in your question? — gvlasov
– gvlasov, Commented Feb 2, 2015 at 14:51

baao · Accepted Answer · 2015-02-02 14:34:15Z

1

You should add some whitespace:

$output = shell_exec('ls -l '.$var[1]);
                        // ^here

answered Feb 2, 2015 at 14:34

baao

73.5k18 gold badges152 silver badges208 bronze badges

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

7 Comments

Kingsta1993 Over a year ago

LOL, can't believe i missed that..haha thank you though!

baao Over a year ago

You're welcome, please mark the answer as correct if it has helped you

Kingsta1993 Over a year ago

Why does it not display information about a directory? For example if type the name of a folder it doesn't work - will mark as correct once it allows me(10min timer)

baao Over a year ago

I'm not really sure why it doesn't work without seeing what your variable looks like. But if you echo $output, it should work or at least throw an error

Kingsta1993 Over a year ago

$var[1] just gets a file or directory name so like test.php etc

|

Collectives™ on Stack Overflow

PHP Shell_Exec not working with variable

1 Answer 1

7 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

7 Comments

Your Answer

Sign up or log in

Post as a guest

Related