How to avoid black background when rotating an image 45 degree using PHP?

Question

Hi I have to flip a thumpnail image before merge it with another jpeg file. but when I rotate 45 degree using php. It shows a black background. how can I avoid that. any body can help me.

Answer 1

<?
$image = "130.jpg";
$degrees = 25;
for($i=0;$i<count($data);$i++){
    $ext = "";
    $extarr = "";
    $extarr = explode(".", $data[$i]['name']);
    $ext = array_pop($extarr);
    if($ext == "png"){
        $rotate = imagecreatefrompng("images/".$data[$i]['name']);
        $transColor = imagecolorallocatealpha($rotate, 255, 255, 255, 270);
        $watermark1[$i] = imagerotate($rotate, ((360-$degrees)%360), $transColor);
    }
}

function imagecopymerge_alpha($dst_im, $src_im, $dst_x, $dst_y, $src_x, $src_y, $src_w, $src_h, $opct){
    $w = imagesx($src_im);
    $h = imagesy($src_im);
    $cut = imagecreatetruecolor($src_w, $src_h);
    imagecopy($cut, $dst_im, 0, 0, $dst_x, $dst_y, $src_w, $src_h);
    imagecopy($cut, $src_im, 0, 0, $src_x, $src_y, $src_w, $src_h);
    imagecopymerge($dst_im, $cut, $dst_x, $dst_y, $src_x, $src_y, $src_w, $src_h, (100 - $opct));
}


for($i=0; $i<count($watermark1); $i++){
    if($i == 0) imagecopymerge_alpha($image, $watermark1[$i], $dest_x, $dest_y, 0, 0, $watermark_width, $watermark_height, $opacity);
    else imagecopymerge_alpha($image, $watermark1[$i], ($i*$dest_x)*3, ($i*$dest_y)*15, 0, 0, $watermark_width, $watermark_height, $opacity);
    imagedestroy($watermark1[$i]);
}
header("content-type: image/png");
imagepng($image);
imagedestroy($image);
?>

Also, do your watermark images have alpha channel or are they fully opaque?

Answer 2

Well, if you are generating a jpg, using PHP GD you set the color of the background as the third option of the function imagerotate. In this example I'm gonna assume that you are rotating a jpg image $filename by an arbitrary $angle degrees, and you want a white background, i.e. color code 16777215:

$rotatedImage = imagerotate(imagecreatefromjpeg($filename), ((360-$angle)%360), 16777215);

black is color code 0, which is default, and the rest of the color gamma is in between the two, so you just need to decide which background color you would like

EDIT: for transparent backgrounds, if you are generating a png you would do:

$destimg = imagecreatefromjpeg($filename);
$transColor = imagecolorallocatealpha($destimg, 255, 255, 255, 127);
$rotatedImage = imagerotate($destimg, ((360-$angle)%360), $transColor);

Hope that helps