shell_exec phpfile

Question

I currently 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 I have 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 write wrong code?

Did you install the CLI (command-line interface) to PHP? Does /usr/bin/php exist? Is content.php executable? Did you try to run it from the shell? — Edgar Bonet, Oct 04 '10 at 09:41
I might tell a mistake but shouldn't it be `$output = shell_exec("./content.php");$output .= mysql_insert_id();` ? 'Cause it doesn't seem to me that `mysql_insert_id()` is a part of the shell cmd... I might be wrong, enlight me! http://www.php.net/manual/en/function.shell-exec.php — Shikiryu, Oct 04 '10 at 10:57

score 1 · Accepted Answer · 2010-10-05T06:21:37.750

1

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

edited Oct 05 '10 at 06:21

answered Oct 04 '10 at 11:47

Also not working for me, got some random tests working but my actual code won't start. Will reply if any of your code work :)! – Oct 04 '10 at 12:47

score 0 · Answer 2 · answered Oct 04 '10 at 10:42

0

.. Maybe this will work for you:

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

Or;

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

answered Oct 04 '10 at 10:42

Bas van Dorst

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Doesn't give my echo back, returns: no errors detected in content.php – Oct 04 '10 at 11:29

score 0 · Answer 3 · edited Mar 31 '12 at 00:08

0

Why do you invoke a PHP file and 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 mysql_insert_id() returns the id 132.

This might be a problem because content.php132 might not be existant.

edited Mar 31 '12 at 00:08

рüффп

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answered Oct 04 '10 at 14:02

ITroubs

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Because i need to get a process on the background to speed up my process of crawling webpage. – Oct 04 '10 at 14:28
hmm try this mysql_query("INSERT INTO jobs (jobID, url, trefwoorden,used) VALUES ('', '".$_POST['url']."', '".$_POST['trefwoorden']."',0)"); and add a line in that takes the lowes jobID that is not used to the content.php and lat it work on that job. – ITroubs Oct 04 '10 at 14:37

shell_exec phpfile

3 Answers3