In a function how do I keep calling imagecopy to stack image layers on the newly created image each time?

[3dron]

I have code like this in a function: 

 function stacklayer($folder, $layer) {

       $newlayer = imagecreatefrompng($folder . '/' . $layer . '.png');
        imagecopy($baseimage,$newlayer,0,0,0,0,$w,$h);
        imagedestroy($newlayer); 
}

stacklayer('Layer1', 'Image1')

But $baseimage does not have the new layers added on top of $baseimage created outside the function. 

In this manner:

$baseimage = imagecreatefrompng("Base.png");
$w = imagesx($baseimage); $h = imagesy($baseimage);
imagealphablending($baseimage,true);

 

 

If I run this outside the function it adds the layers as intended.

$folder = 'Layer1', $layer = 'Image1';
$newlayer = imagecreatefrompng($folder . '/' . $layer . '.png');
imagecopy($baseimage,$newlayer,0,0,0,0,$w,$h);
imagedestroy($newlayer); 

 

Any help?

Thanks

3dron

[requinix]

Variables defined outside of functions are not available inside of functions.

If you didn't see PHP give you a warning about that then you probably do not have your environment set up properly for PHP development. Find your php.ini and set

display_errors = on
error_reporting = -1

Then restart PHP and/or your web server and try your script again.

[kicken]

$baseimage, $w, and $h do not exist inside your function.  You need to pass them in as additional parameters.

function stacklayer($baseimage, $w, $h, $folder, $layer) {
    $newlayer = imagecreatefrompng($folder . '/' . $layer . '.png');
    imagecopy($baseimage,$newlayer,0,0,0,0,$w,$h);
    imagedestroy($newlayer); 
} 

Then include them when you call your function.

$baseimage = imagecreatefrompng("Base.png");
$w = imagesx($baseimage); $h = imagesy($baseimage);
stackLayer($baseimage, $w, $h, 'Layer1', 'Image1');

 

Edited March 26, 2022 by kicken

[3dron]

Thanks all works now.