Question

I would like to pass a parameter to a function. However, inside this parameter, I use variables that are defined inside the function.
For example:

<?php

function foo($var){
    $test = "test";
    echo $var;
}

foo($test);

?>

In this example, I would like for the function to print out “test”. Of course, this returns an error. However, this is what I am trying to do.

Answer 1

You can accomplish this with using variable-variables, though I don’t understand why you’d want/need to:

function foo($var){
    $test = "test";
    echo $$var;
}

foo('test');

Notice in the echo statement inside foo(), the variable has two $ leading. This is a variable-variable and $var‘s value will be treated as a variable-name. So, by calling foo('test'); (the parameter being a string in this case), $$var will evaluate to $test.