Question

I am new to facebook app development and working over my first app.
I am trying to resize an image using imagecopyresampled() function but getting this error for it:

Warning: imagecopyresampled(): supplied argument is not a valid Image resource in …

$url = "http://graph.facebook.com/{$userId}/picture?type=large";
$img = "udp.gif";
file_put_contents($img, file_get_contents($url));
if (exif_imagetype($img) != IMAGETYPE_GIF) {
    $imgFile = 'temp/' . $userId . '_fdp_' . rand() . '.gif';
    if (exif_imagetype($img) == IMAGETYPE_JPEG)
        $imageObject = imagecreatefromjpeg($img);
        if (exif_imagetype($img) == IMAGETYPE_PNG)
            $imageObject = imagecreatefrompng($img);
            $imagegif($imageObject, $imgFile);
            $imgFile1='temp/' . $userId . '_fdp_.gif';
            $imagecopyresampled($imgFile1, $imgFile, 0, 0, 0, 0, 200, 280, 180, 252);
            $img = $imgFile1;
         }
    }
}

Answer 1

You’re misunderstanding the argument types. resource is a PHP type, like string or float. Resources need to be created with the appropriate related function. In your case, you need one resource of the original image, created with a function like imagecreatefromjpeg, and another similar resource to copy the image to, most likely a blank image created with imagecreatetruecolor.

To quote the example from the manual:

$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

What you have instead is a non-existent variable and a string, which is not the required input.