Re: function I created doesn't work

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Samuel Vogel wrote:
> Explanation of your code:
> 
> $CONTENT = get_content(12, 104, 'merchant');
> echo $CONTENT;
> 
> This does not work, because you don't use a "return" in your function.
> This means that the function does not return a value. Now in the
> function you assign a value to $CONTENT. That works, as you pointed out
> with the second example.
> But after that the line above sets $CONTENT to the empty return value of
> the function. And therefore it is empty!
> 
> so long,
> Samy
> 

Sorry for confusing with 2nd function. Let's take it out, forget about it.

function get_content($client_id, $form_id, $index1)
{
	$query = mysql_query("
		SELECT content
		FROM infos
		WHERE client_id=".$client_id."
			AND form_id=".$form_id."
			AND index1='".$index1."'");
	if (mysql_num_rows($query) > 0)
	{
		$result = mysql_fetch_assoc($query);
		return $result['content'];
	}
	else
	{
		get_content(0, 0, $index1); // get default value
	}
}

$CONTENT = get_content(12, 104, 'merchant');
echo $CONTENT; // empty, nothing

There is "return", right after $result = mysql_fetch_assoc($query);

Some additional info (hopefully will not confuse again :)) I'm pulling
content from table infos for specific client, form and index1. If there
is no record I'm using recursive part (inside "else") to get the default
value (client_id=0, form_id=0).
When echo  the content right before "return" I can see it. But can't see
it in echo after calling the function?!?!

thanks

-afan

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