shell_exec phpfile

Question

Atmoment i use this code:

if(isset($_POST['url']) && isset($_POST['trefwoorden']) )
{
  mysql_connect('localhost', 'crawler', 'whathasbeenseencannotbeunseen');
  mysql_select_db("crawler");
  mysql_query("INSERT INTO jobs (jobID, url, trefwoorden) VALUES ('', '".$_POST['url']."', '".$_POST['trefwoorden']."')");
  $output = shell_exec("./content.php " .mysql_insert_id());
  echo $output;

}

In my content.php the following code:

#!/usr/bin/php
<?php
echo 'HET WERKT';
?>

Now i want to see if my shell_exec actually works by filling in the form and submitting it... but it doesn't echo anything at all. Did i wrote some wrong code?

Answer 1

.. Maybe this will work for you:

shell_exec('php -l content.php'. mysql_insert_id())

Or;

1. PHP is running in safe_mode
2. Apache does not have the permissions to execute the script

Answer 2

Hello,

Try this. This may help you.

shell_exec('usr/local/bin/php -l content.php'. mysql_insert_id())

or

shell_exec('usr/local/bin/php -content.php'. mysql_insert_id())

All the best.

Thanks,

Kanji

Answer 3

Why do you invoke a php file?

Why not including it?

Besides, i agree with Chouchenos because this line of code:

shell_exec('php -l content.php'. mysql_insert_id())

will execute like this "php -l content.php132" if myslq_insert_id() returns the id 132. this might be a problem because content.php132 might not be existant.