PHP passing $_GET in linux command prompt

From this answer on ServerFault:

Use the php-cgi binary instead of just php, and pass the arguments on the command line, like this:

php-cgi -f index.php left=1058 right=1067 class=A language=English

Which puts this in $_GET:

Array
(
    [left] => 1058
    [right] => 1067
    [class] => A
    [language] => English
)

You can also set environment variables that would be set by the web server, like this:

REQUEST_URI='/index.php' SCRIPT_NAME='/index.php' php-cgi -f index.php left=1058 right=1067 class=A language=English

Typically, for passing arguments to a command line script, you will use either argv global variable or getopt:

// bash command:
//   php -e myscript.php hello
echo $argv[1]; // prints hello

// bash command:
//   php -e myscript.php -f=world
$opts = getopt('f:');
echo $opts['f']; // prints world

$_GET refers to the HTTP GET method parameters, which are unavailable in command line, since they require a web server to populate.

If you really want to populate $_GET anyway, you can do this:

// bash command:
//   export QUERY_STRING="var=value&arg=value" ; php -e myscript.php
parse_str($_SERVER['QUERY_STRING'], $_GET);
print_r($_GET);
/* outputs:
     Array(
        [var] => value
        [arg] => value
     )
*/

You can also execute a given script, populate $_GET from the command line, without having to modify said script:

export QUERY_STRING="var=value&arg=value" ; \
php -e -r 'parse_str($_SERVER["QUERY_STRING"], $_GET); include "index.php";'

Note that you can do the same with $_POST and $_COOKIE as well.


I don't have a php-cgi binary on Ubuntu, so I did this:

% alias php-cgi="php -r '"'parse_str(implode("&", array_slice($argv, 2)), $_GET); include($argv[1]);'"' --"
% php-cgi test1.php foo=123
<html>
You set foo to 123.
</html>

%cat test1.php
<html>You set foo to <?php print $_GET['foo']?>.</html>

Tags:

Linux

Php

Recent Posts