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 $username, if possible with PHP CLI. I guess that won't work from outside function.php since it's private, but how could I do this then?
php -r 'include("/absolute/path/to/function.php"); addUser("some user");'
This should work. Because you are basically executing all that code in between 's. And in that you can include function.php and, should appropriately call addUser().
see phpdoc.
You get your command line argumenst passed in a argv array:
function addUser($username) {
//do some stuff
}
addUser( $argv[1] );
argv -> $argv
php function.php john_doe but now I get an error telling me PHP Parse error: syntax error, unexpected T_STRING, expecting T_FUNCTION. Seems as if the function can't handle the string john_doe, but that's weird. Do you have an idea why?You can use $argv. $argv[0] = the filename, $argv[1] = first thing after filename.
I.e. php function.php "some arg" would be addUser("some arg");
function funcb()
{
echo 'yes';
}
if (php_sapi_name() === 'cli') {
if (count($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!
