Re: quick php for perl coder question

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Thomas Bolioli schreef:
I want to return an array from a function.
I have this:
return array($found, $username, $email, $nickname);
as my code. I have a bug in the code and I am not sure where yet. Should the above statement with this:

array($vars);

not sure what you meant with the line above, it's valid but does nothing.

   $vars = function($type, $abc, $xyz);
   $found = $vars[0];
   $username = $vars[1];
   $email = $vars[2];
   $nickname = $vars[3];

on the other end work? If so, then the bug is somewhere else.

yes this should work assuming you meant this:

<?php

function foo() {
	$found		= true;
	$username	= 'The Joker';
	$email		= 'joker@xxxxxxxxxxx';
	$nickname	= 'joker';

	return array($found, $username, $email, $nickname);
}

$vars 		= foo();
$found 		= $vars[0];
$username 	= $vars[1];
$email 		= $vars[2];
$nickname 	= $vars[3];
?>

although you could do it more perly:

<?php

list($found, $username, $email, $nickname) = foo();

?>

and maybe consider using an assoc array:

<?php

function foo() {
	$found		= true;
	$username	= 'The Joker';
	$email		= 'joker@xxxxxxxxxxx';
	$nickname	= 'joker';

	return array(
		'found' 	=> $found,
		'username' 	=> $username,
		'email'		=> $email,
		'nickname'	=> $nickname,
	);
}

$vars 		= foo();
// using an associative array negates the need to
// define the following variables, you can use
// $vars['found'] (etc) as it is quite descriptive
// in it's own right :-)
$found 		= $vars['found'];
$username 	= $vars['username'];
$email 		= $vars['email'];
$nickname 	= $vars['nickname'];

?>

Thanks in advance,
Tom



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