!Global

dies_irae

Hi folks,
is it possible to somehow use a var that you have made insida a function outside it?
Like global but the other way 🙂

Example:
<?
funcion myfunc() {
$var = "Hello there";
}
myfunc();
echo $var;
?>

Now i want $var to be "Hello there",

note / I dont want to use return,

thanks.

rulian

sure, set the variable that you want to take out of the function as a global variable as well. and just to be safe, return the variable at the end

return $variable;

dies_irae

heh, thanks i saw that in the manuall just after i posted it 🙂 nice..

reagent

That is horrible code design, whoever has to look at your code will not know how the variable is getting set.

I would recommend using pass-by-reference parameters to your function

example:
function myfunct(&$myvar){
$myvar = "two";
}
$myvar = "one";
print $myvar;

myfunct($myvar);
print $myvar;

will print "onetwo" because the function is changing the value of the passed parameter, not making a copy.

dies_irae

horrible code design?

function myfunc() {
global $whatsoever:
$whatsoever = "Hello";
}
myfunc();
echo $whatsoever; // will output "Hello"

it is not a horrible design, if you know what the perpose is..

Anon

my point was that it is not obvious what the purpose is.

say that your function is in an include file and you may have multiple include files in one script, what you would have is this:

require_once("include_file_one.php");
require_once("include_file_two.php");

myfunc();
echo $myvar;

where did $myvar come from? I would think it would be more obvious to use

myfunc($myvar);
echo $myvar;

while global is useful, I feel it is a bad way to design code. I have used it extensively in my code, and now I am in the process of changing it back. I was just trying to save you some hassle in the long run..