[SOLVED] functions :: Undefined variable

[chocopi]

I've never really used functions but now I find myself needing to. I understand the basics but I can't seem to get the variable from within the function to echo outside of it.

 

<?php
function topic($topic,$cat)
{
if(empty($_GET['topic']))
{
	header ("Location: index.php");
} else
	{
		$topic = $_GET['topic'];
		if(!ereg("^([0-9])+$",$topic))
		{
			die("The Topic '{$topic}' You Have Selected Does Not Exist !");
		} else
			{
				$query = mysql_query("SELECT topic_name FROM fori_topic WHERE topic_id='$topic' AND cat_id='$cat'") or die(mysql_error());
				$num_rows = mysql_num_rows($query) or die(mysql_error());
				$row = mysql_fetch_assoc($query) or die(mysql_error());
				$topic_name = trim(ucwords(strtolower($row['topic_name']))); // This one
			}
		if($num_rows == 0)
		{
			die("The Topic Does Not Exist Within The Chosen Category !");
		}
	}
}
?>

 

so when I call it later on with:

 

<?php
require_once("functions/function_topic.php");
topic($topic,$cat);
echo $topic_name;
?>

 

I get Undefined variable notice on the line i'm echoing. I know there is a way to do this but I can't figure it out.

 

Many Thanks

 

~ Chocopi

[rarebit]

Here's an example (you need to return something!)

function myfunk()
{
  $s = "hi";
   return $s;
}
echo myfunk();

You can return most things...

[chocopi]

Thats what I thought but it still gives the same error even after adding

 

return $topic_name;

 

within the function

 

~ Chocopi

[rarebit]

Who's got the banana?

function myfunk()
{
  $s = "hi";
   return $s;
}
$v myfunk();
echo $v;

[chocopi]

i dont understand what you mean

 

Thats gives a parse error when the variable is infront of yhr function

 

~ Chocopi

[rarebit]

And so it should, it needs an equals sign!

$v = myfunk();

[chocopi]

Cheers Rarebit,

 

I never thought that would work as you are then storing the function within the variable, but it does so I'll just shut up

 

Thanks

 

~ Chocopi

[chocopi]

I was wondering, how could I get it to return 2 values

 

You can't return multiple values from a function' date=' but similar results can be obtained by returning a list.[/quote']

 

Is this the only way to do it, or is there a better option ?

 

~ Chocopi

[rarebit]

That's the way:

...
   return list($x, $y);
}

with either:

list($x, $y) = myfunc();

or

$z = myfunc();

[chocopi]

I get what your doing but I can't get it to return the values

 

<?php
$array = array($post,$post_message,$post_subject);
return list($post,$post_message,$post_subject) = $array;
?>

 

And I am using it with

 

list($post,$post_message,$post_subject) = action($topic,$cat);

 

~Chocopi

[rarebit]

That makes no sense to me? lol

For this:

return list($post,$post_message,$post_subject) = $array;

which I presume is the function returning, you do either or, i.e. return the array you created, or return the list set...

[chocopi]

Well if I do

 

return list($variables);

 

I get a notice saying: Unexpected ; expecting =

 

Thats the only reason i added the = $array;

 

~ Chocopi

[wildteen88]

list can only be used in conjunction with arrays. if you want to return multiple variables from your function you're best of of storing all the variables to be returned into an array, then use list out side of your function:

eg:

function myfunc()
{
   // vars to be returned
   $var1 = 'foo';
   $var2 = 'bar';

   // set up an array to store all variables to be returned.
   $return = array($var1, $var2);

    // return the array of variables
    return $return;

    // ALTERNATIVELY you could do:
    // return array($var1, $var2);
    // instead.
}

// gets vars out of array of variables return from myfunc
list(var1, $var2) = myfunc();

// use variables
echo $var1;
echo $var2;

[chocopi]

Thanks Wildteen

 

Thats sorted it.

 

I couldn't seem to get it to work and was pulling my hair out but it was because I was over-ridding variables for no reason

 

But thanks Rarebit + Wildteens !!!

 

~ Chocopi