$argc

(PHP 4, PHP 5, PHP 7)

$argc — The number of arguments passed to script

설명

Contains the number of arguments passed to the current script when running from the command line.

Note: The script's filename is always passed as an argument to the script, therefore the minimum value of $argc is 1.

Note: This variable is not available when register_argc_argv is disabled.

예제

Example #1 $argc example


<?php
var_dump($argc);
?>

When executing the example with: php script.php arg1 arg2 arg3

위 예제의 출력 예시:

int(4)

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User Contributed Notes 4 notes

down

Tejesember ¶

13 years ago


To find out are you in CLI or not, this is much better in my opinion:
<?php
if (PHP_SAPI != "cli") {
    exit;
}
?>

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-6

karsten at typo3 dot org ¶

15 years ago


Note: when using CLI $argc (as well as $argv) are always available, regardless of register_argc_argv, as explained at http://docs.php.net/manual/en/features.commandline.php

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-14

elm at r3m0ve dot gmx dot ch ¶

10 years ago


To decide whether my script is run from CLI I simply create a PHP script that handles only CLI invocations.

File cron.php:
<?php

// Set environment variables your application depends on
$_SERVER[ 'HTTP_HOST' ] = 'domain.tld';
// $_SERVER[ 'REQUEST_URI' ] = '/some/URI/if/needed';

// Use the environment to read out required values
$task = $_SERVER[ 'argv' ][ 1 ];

// Instanciate the dispatcher or whatever you use
$dispatcher = new Dispatcher();
$dispatcher->handle( $task );

?>

This way my application doesn't have to know about CLI at all.

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-21

anonymous ¶

6 years ago


int main(int argc, char *argv[])
{
  fprintf(stdout,"argumen count : %d\n",argc);
  fprintf(stdout,"argumen vector : %s\n",argv);
  return 0;
}

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