PHP: Pass parameter to function, but with variables defined inside function used in parameter

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.

That does not seems to make a lot of sense to me... why would you want to do this? (it's not possible btw, unless you make $test global, which defeats the purpose of defining it inside the function). Variables defined inside function should stay local, not accessible to the outside, to keep the code decoupled. — Felix Kling
– Felix Kling, Commented Oct 18, 2012 at 5:06
The function contains a foreach, with foreach($array as $post), and I'm using the variable $post in the parameter. — Sang Froid
– Sang Froid, Commented Oct 18, 2012 at 5:07
Why are you passing $post as argument? Or do you mean the parameter name is $post? $post will be populated with each element in $array during the loop. I still don't understand and I'm even more confused. — Felix Kling
– Felix Kling, Commented Oct 18, 2012 at 5:09

newfurniturey · Accepted Answer · 2012-10-18 05:07:28Z

2

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.

answered Oct 18, 2012 at 5:07

newfurniturey

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2 Comments

Felix Kling Over a year ago

@Sang: Note that using variable variables is generally a poor design choice. If you would explain which problem you are actually trying to solve, we could help you better.

Mahdi Over a year ago

@FelixKling I believe also it could be better, however variable-variables working also ... I agree with you, if he provides an original part of code, it would be much better for him at least.

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