FUNCTION with $_POST var doesn`t work

Benjamin_Roth

All im tryinto do is a function that gives back a Variable ($x) that has the whole entry on the global $_POST (It´s definitely set) array in it. I tried different ways but all those doesnt work. So this is my code:

function returnpostarray()
	{
	if(isset($_POST))
		{
		while(list($variable,$return)=each($_POST))
			{
			$x=$x.$variable."°";
			}
		}
	return $x;

P.S. The "°" in the code is just a seperator i use to explode the string later in my script.

Any help would be great.
Regards Benjamin

jesus_hairdo

How about trying

function returnpostarray()
{
    if (isset($_POST))
    {
        foreach ($_POST as $value)
        {
            $x .= $value."#";
        }
    }
    return $x;
}

Where # is the seperator.

Is there a need to have it all as a string and to then explode into an array ??

if not just use

$x[]=$value;

This will create an array with a numerical index.

Jamie

Benjamin_Roth

It´s not right that it doesn´t work at all. The function itself gets the variables out of the $_POST -Array and puts them into the variable $x. I tried it. But the variable is not available in the whole script where the function is included. Don´t know what it is?
I tried everything from making that variable global($x) to return($x) it but i isn´t set in the whole script.
Any ideas?

Benjamin

diego25

How are you calling the script? If you're not assigning the return value to any variable then it's not going to be available in other parts of your script.

Diego

Benjamin_Roth

I don´t need the last 3 character so i do a substr().
I´m using the $x variable like this (Outside of the function):

$newvar = substr($x, -3);

I also tried to echo $x and $newvar but this also has no effect.

Benjamin

jamie_kerwick

are you actually calling the function though ??

You would need to do something like

$x = returnpostarray();
$newvar = substr($x, -3);

as a function only 'runs' when it is called in your script

Benjamin_Roth

That was the problem. I forgot to use the function this way. I thought $x would be available outside of the function because of the (return $x) inside of the function. Damn, i was spending hours and all i found out is, that i made such a silly mistake. Ok it solved the problem too, of course.
Thanx for your help.

Benjamin