问题:

Let's say I have a private function addUser() in function.php that takes $username as an input variable and does some stuff:

function addUser($username) {

//do some stuff

}

Now I want to call this function and pass the value $username, if possible with PHP CLI. I guess that won't work from outside function.php since it's private, but how could I do this then?

回答1:

php -r 'include("/absolute/path/to/function.php"); addUser("some user");'

This should work. Because you are basically executing all that code in between 's. And in that you can include function.php and, should appropriately call addUser().

see phpdoc.

回答2:

You get your command line argumenst passed in a argv array:

function addUser($username) {

//do some stuff

}

addUser( $argv[1] );

回答3:

You can use $argv. $argv[0] = the filename, $argv[1] = first thing after filename.

I.e. php function.php "some arg" would be addUser("some arg");

回答4:

function funcb()
{
    echo 'yes';
}

if (php_sapi_name() === 'cli') {
    if (count($argv) === 1) {
        echo 'See help command'.PHP_EOL;
        exit();
    }

    if (function_exists($argv[1])) {
        $func = $argv[1];
        array_shift($argv);
        array_shift($argv);
        $func(...$argv);
    }
}

This works for me!