Please note that, $argv and $argc need to be declared global, while trying to access within a class method.
<?php
class A
{
public static function b()
{
var_dump($argv);
var_dump(isset($argv));
}
}
A::b();
?>
will output NULL bool(false) with a notice of "Undefined variable ..."
whereas global $argv fixes that.
$argv
(PHP 4, PHP 5, PHP 7)
$argv — Array of arguments passed to script
Description
Contains an array of all the arguments passed to the script when running from the command line.
Note: The first argument $argv[0] is always the name that was used to run the script.
Note: This variable is not available when register_argc_argv is disabled.
Examples
Example #1 $argv example
<?php
var_dump($argv);
?>
When executing the example with: php script.php arg1 arg2 arg3
The above example will output something similar to:
array(4) {
[0]=>
string(10) "script.php"
[1]=>
string(4) "arg1"
[2]=>
string(4) "arg2"
[3]=>
string(4) "arg3"
}
Notes
Note:
This is also available as $_SERVER['argv'].
add a note
User Contributed Notes 9 notes
tufan dot oezduman at googlemail dot com ¶
8 years ago
hamboy75 at example dot com ¶
5 years ago
To use $_GET so you dont need to support both if it could be used from command line and from web browser.
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)==2)
$_GET[$e[0]]=$e[1];
else
$_GET[$e[0]]=0;
}
php at simoneast dot net ¶
3 years ago
Sometimes $argv can be null, such as when "register-argc-argv" is set to false. In some cases I've found the variable is populated correctly when running "php-cli" instead of just "php" from the command line (or cron).
Steve Schmitt ¶
10 years ago
If you come from a shell scripting background, you might expect to find this topic under the heading "positional parameters".
Andreas ¶
2 years ago
How to check if one parameter is given:
if ($argc < 2 )
{
exit( "Usage: program <parameter1>\n" );
}
process( $argv[1] );
andrew w ¶
2 years ago
An easier way to populate $_GET with $argv values.
<?php
if ( isset( $argv ) ) {
parse_str(
join( "&", array_slice( $argv, 1 )
), $_GET );
}
?>
fabio at naoimporta dot com ¶
3 years ago
When you pass an option to the file that intercept the request, it will be transformed into an array item, and the option name will be lost. Only its content is captured.
<?php
var_dump($argv);
?>
call : "php file.php --test=foo baz"
will print
array(3) {
[0] =>
string(16) "file.php"
[1] =>
string(3) "foo"
[2] =>
string(3) "baz"
}
KRowe ¶
4 years ago
Improves on hamboy75's note by providing better support for positional arguments:
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)==2)
$_GET[$e[0]]=$e[1];
else
$_GET[]=$e[0];
}
var_dump($_GET);
Using this modification, arguments without an = are treated as positional (this is not web standard but generally works).
Jesse ¶
6 years ago
If your script is read from standard input or with the -r option, $argv[0] will be "-".
If you use the "--" option to separate PHP's arguments from your script's arguments, $argv[1] will be "--" if your script is read from a file. But if your script is read from standard input or with the -r option, the "--" will be removed.
