How to get params with shell_exec() in php

Question

I have two php files.

first one execute the second one I want to send a params with the exec

I have no idea how to get the params i sent in the second file

first.php:

$params="hello";
shell_exec('php file.php $params > /dev/null 2>/dev/null &')

file.php:

echo $params;

Thanks!

See this question: http://stackoverflow.com/questions/6826718/pass-variable-to-php-is-running-from-command-line — nickhar, Oct 25 '12 at 15:03

score 2 · Accepted Answer · answered Oct 25 '12 at 15:03

2

Use the $argv array

In your case

echo $argv[1];

answered Oct 25 '12 at 15:03

air4x

score 0 · Answer 2 · answered Oct 25 '12 at 15:04

0

$argv contains any arguments passed to the script.

answered Oct 25 '12 at 15:04

Edd

score 0 · Answer 3 · answered Oct 25 '12 at 15:05

0

Try echo $argv[1]

answered Oct 25 '12 at 15:05

durron597

3 Answers3