Function in array executed before calling it - php

Question

I have an array like this:
$cmds = array(
'info' => call_user_func('_info', $cmd1),
'e1' => 'something'
);
The func itself:
function _info($cmd1){
if(strlen($cmd1) < 2){
die('Invalid Username');
}else{
echo 'cmd1 > 2';}
}
The problem is it being executed before i call it giving undefined variable error because $cmd1 not exist yet, it will be exist when i call it in a foreach function.
Been searching for a solution for hours but didn't find.
I tried setting the array as this 'info' => _info($cmd1) and 'info' => '_info' and putted $cmd1 directly in the func like this
function _info(){
if(strlen($cmd1) < 2){die('invalid');
}
But i still get the same error as it being called before i want to be.

Answer 1

You can send parameters to be executed later
$cmds = array(
'info' => [ '_info', '$cmd1' ], // single quotes
'e1' => 'something'
);
In the meantime your cmd1 is defined
$cmd1 = "hello";
foreach ($cmds as $key => $cmdset) {
if (!is_array($cmdset)) {
// do something with e1 => something
} else {
$function = array_shift($cmdset);
foreach ($cmdset as &$param) {
// Only variables are treated below, values are passed through as is
if (stripos($param, '$') === 0) {
// Remove $
$cmd1 = substr($param, 1);
// Set param to the value of cmd1
$param = $$param; // variable varaible $$
}
}
call_user_func_array($function, $cmdset);
}
}
While this works it is really bad practice to do like this. Depending on a variable function that is not clear if it is set or not. The code is hard to understand.
You should think about another solution where you can ensure that $cmd1 is set or can be retrieved and if it is not have some way to deal with that as well.