a problem passing values to a function

[earlparkerii]

I'm having a problem passing values to variables in a function. The first line of the function is:

function PrintImage($image1, $image2, $image3)

The values I'm trying to pass are:

PrintImage("PumpNeedle.gif", "PumpNozzle-Std.gif", "PumpNozzleScrew-HiFlow.gif");

The statements within the function read a filename from a directory and compare the name read with the name supplied (such as PumpNeedle.gif). If there's a match, the graphic is displayed on the page.

The code works flawlessly until I try to turn it into a function. When the function is called all I get is a blank page. The testing I've done seems to indicate that the values in the call line aren't being passed to the variables in the 'function' line.

Doubtless the solution is simple and I'm just not seeing it. Any advice would be much appreciated.

[ayzee01]

are you passing them correctly??

<?

function printImage($image1, $image2, $image3)
{
//DEBUG:
echo "Image 1: ".$image1."<br>";
echo "Image 2: ".$image2."<br>";
echo "Image 3: ".$image3."<br>";
}

call the function by:

<?
printImage("whateva.gif","whateva2.gif","whateva3.gif");
?>

The above is the correct way to call the function with parameters. You should see output, if you dont - then f##k knows ;o)

[Anon]

I don't know if this is your problem but have you global'd everything?

an example:

$a = 1;

function add_number() {

return ($a + 2);

}

$x = add_number();

Now the function wouldn't be able to add the two numbers together because it can't find $a... to solve this problem add this line to the beginning of the function:

global $a;

now the function can get the value of $a...

[ayzee01]

if he's passing the variables as parameters when he calls the function, he wont need to set everything as global.

this is bad programming practice and is best to avoid doing if possible.