PHP variables in function argument is not working - php

Question

I've retried solving this, by using a condition and a default attribute as recommended.
User-generated data is declared before to $Variable_1:
<?php
$Variable_1 = 'abc123!' //The user inputs the data
if ($booleanvalue == true) { // User selects if they've put data
name($user_data, $Variable_0 = $Variable_1 );
}
//Then the function will use the user's data from $Variable_1
function name($user_data, $Variable_0 = null) {
//Other code...
}
$Variable_2 = name($user_data);
$data['variable_2'] = $Variable_2;
?>
Is it possible to have $Variable_0 pre-declared and then put as an argument?

Answer 1

you have a few mistakes in your code. and I don't think that you can use a function named name.
you could do it this way for example:
<?php
$Variable_1 = 'abc123!';
function test($data) {
global $Variable_1;
//Other calculations...
return $Variable_1 . $data;
}
$testdata = "huhu";
$Variable_2 = test($testdata);
$data['variable_2'] = $Variable_2;
echo $data['variable_2'];
?>

Answer 2

I agree with the comment by El_Vanja, but you can access a global variable through the magic $GLOBALS array anywhere.
<?php
// what you might actually want
function name($variable = 'abc123!')
{
// if no value is passed into the function the default value 'abc123!' is used
}
$variable = 'abc123!';
// what you could do
function name2($variable)
{
// $variable can be any value
// $globalVariable is 'abc123!';
$globalVariable = $GLOBALS['variable'];
}
I'd also like to point out that currently you have no way of controlling what type of data is passed to the function. You might consider adding types.
<?php
<?php
// string means the variable passed to the function has to be a ... well string
function name(string $variable = 'abc123!'): void
{
// void means the function doesn't return any values
}
name(array()); // this throws a TypeError