How to assign a variable outside php function [duplicate] - php

Question

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Are PHP Variables passed by value or by reference?
(16 answers)
How to declare a global variable in php?
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Closed 3 years ago.
I would like to be able to assign the name of a variable outside the function so that the function can assign the chosen variable. It does not seem to work. Anyone have any ideas?
Below is my attempt, however the $admin variable is empty after running the function.
function assign_check($variable, $check) {
if (empty($_POST[$check])) {
$variable = "no";
} else {
$variable = $_POST[$check];
}
}
assign_check('$admin', 'admin');
My question is not about the use of global variables.

Answer 1

You can request a reference in the function body.
function assign_check(&$variable, $check) {
$variable = 'hello';
}
And call passing a variable (reference).
assign_check($admin, 'admin');
$admin value is now 'hello';
Fitting that to your code would result in
function assign_check(&$variable, $check) {
$variable = empty($_POST[$check]) ? "no" : $_POST[$check];
}
assign_check($admin', 'admin');
But returning a proper value would be much cleaner code than using references. Using a ternary operator like presented above would it even simplify without need of a function at all.

Answer 2

A normal way to assign the result of a function to a variable name specified outside the function would be to have the function return the result and assign it directly to the variable when you call the function.
function assign_check($check) {
if (empty($_POST[$check])) {
return "no";
} else {
return $_POST[$check];
}
}
$admin = assign_check('admin');
I would do it this way unless there was a compelling reason to do it otherwise.
For the specific type of thing it looks like this function is intended to do, I would suggest looking at filter_input.