CODEWITHSUNDEEP
phpgd
Ben Holness
Ben Holness

Reputation: 2707

How to make any "extra space" white when using imagecopy in PHP

I am using imagecopy to crop a PNG image to a user specification.

Currently, if the crop area is bigger than the image, any "extra space" becomes black, but I would like it to be white.

Have searched around a bunch, I have discovered that you can use imagefill or imagefilledrectangle to make the background white, however if this is done before imagecopy, then it has no effect and if it is done after imagecopy, it also makes any black parts of the image white.

My code currently looks like this and suffers from black parts of the original image being turned white as well as the extra space:

// Input and output files
$infile = "[Image]";
$outfile = "[Output path for image]"

// Make the image
$orig =imagecreatefromjpeg($infile);
$width = imagesx($orig);
$height = imagesy($orig);
$new = imagecreatetruecolor($width, $height);

// Crop the image
imagecopy($new, $orig,  0, 0, -100, 100, $width, $height);

// Try and make the extra space white
$white = imagecolorallocate($new, 255,255,255);
imagefill($new, 0, 0, $white);

// Save the file
imagepng($new, $outfile);

How can I make that extra space white without affecting the original image? I have no control over what image users might upload, so I can't really pick a transparent color as that color might be part of their original image.

EDIT: This scenario arises when a user chooses a crop size outside of the original image dimensions, something that I do want to be a valid option. The crop is to force a square image, but if the user uploads, say, a landscape rectangle and wants all of their image in the final crop, then the crop will be outside of the image on the top and bottom (which is where I want it to be white instead of black)

Upvotes: 1

Views: 451

Answers (1)

timclutton
timclutton

Reputation: 13004

This happens because you are supplying 'invalid' values to imagecopy() (that is, the crop coordinates are outside the bounds of the source image). GD simply fills in the out of bounds area with black pixels. It would be lovely if it instead used transparent (or any colour) pixels but unfortunately that's not an option.

I don't completely understand what you are trying to do (your source doesn't seem to match your stated goal) but a possible solution involves restricting the crop to the bounds of the image:

$src = imagecreatefromjpeg('JPEG FILE'); // 100x100 image in my test.
$src_w = imagesx($src);
$src_h = imagesy($src);

$user_crop = [
    'x' => -50,
    'y' => -50,
    'width' => 150,
    'height' => 150
];

if ($user_crop['x'] < 0) {
    $user_crop['x'] = 0;
}
if ($user_crop['y'] < 0) {
    $user_crop['y'] = 0;
}
if ($user_crop['x'] + $user_crop['width'] > $src_w) {
    $user_crop['width'] = $src_w - $user_crop['x'];
}
if ($user_crop['y'] + $user_crop['height'] > $src_h) {
    $user_crop['height'] = $src_h - $user_crop['y'];
}

$dest = imagecreatetruecolor($src_w, $src_h);
imagefill($dest, 0, 0, 0x00ffffff); // opaque white.
imagecopy(
    $dest,
    $src,
    $user_crop['x'],
    $user_crop['y'],
    $user_crop['x'],
    $user_crop['y'],
    $user_crop['width'],
    $user_crop['height']
);

header('Content-type: image/png;');
imagepng($dest);
imagedestroy($src);
imagedestroy($dest);
exit;

Note that I've made a few assumptions in this code about placement of the cropped image.

Upvotes: 1

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