Function which produces new variable for use outside of function - PHP - php

Question

How would I alter the function below to produce a new variable for use outside of the function?
PHP Function
function sizeShown ($size)
{
// *** Continental Adult Sizes ***
if (strpos($size, 'continental-')!== false)
{
$size = preg_replace("/\D+/", '', $size);
$searchsize = 'quantity_c_size_' . $size;
}
return $searchsize;
Example
<?php
sizeShown($size);
$searchsize;
?>
This currently produces a null value and Notice: undefined variable.
So the function takes one argument, a variable containing a string relating to size. It checks the variable for the string 'continental-', if found it trims the string of everything except the numbers. A new variable $searchsize is created which appends 'quantity_c_size_' to the value stored in $size.
So the result would be like so ... quantity_c_size_45
I want to be able to call $searchsize outside of the function within the same script.
Can anybody provide a solution?
Thanks.

Answer 1

Try using the global keyword, like so:
function test () {
global $test_var;
$test_var = 'Hello World!';
}
test();
echo $test_var;
However, this is usually not a good coding practice. So I would suggest the following:
function test () {
return 'Hello World!';
}
$test_var = test();
echo $test_var;

Answer 2

In the function 'sizeShown' you are just returning the function. You forgot to echo the function when you call your function.
echo sizeShown($size);
echo $searchsize;
?>
But the way you call $searchsize is not possible.

Answer 3

This is an old question, and I might not be understanding the OP's question properly, but why couldn't you just do this:
<?php
$searchsize = sizeShown($size);
?>
You're already returning $searchsize from the sizeShown method. So if you simply assign the result of the function to the $sizeShown variable, you should have what you want.