Question

我在php中生成的图像如下：

$base = imagecreatefromstring(file_get_contents($_GET['imgSRC']));
imagepalettetotruecolor($base);
$background = imagecolorallocate($base, 0, 0, 0);
imagecolortransparent($base, $background);
imagealphablending($base, false);
imagesavealpha($base, true);

图像有透明背景，我想用白色替换任何透明像素。最干净的方法是什么？

Answer 1

创建实体图像并在其上复制透明图像对我来说很有效：

class Solid {
    static function white() {
        return new self(255, 255, 255);
    }

    function __construct($red, $green, $blue) {
        $this->red = $red;
        $this->green = $green;
        $this->blue = $blue;
    }

    function under($pic) {
        $w = imagesx($pic);
        $h = imagesy($pic);

        $buffer = imagecreatetruecolor($w, $h);
        imagefilledrectangle($buffer, 0, 0, $w, $h,
            imagecolorallocate($buffer, $this->red, $this->green, $this->blue));

        imagecopy($buffer, $pic, 0, 0, 0, 0, $w, $h);

        return $buffer;
    }
}

这是我的用法：

$solid_picture = Solid::white()->under($transparent_picture);

如果您的图片具有Alpha通道，请考虑使用imagecopymerge()而不是imagecopy。

用白色PHP GD替换透明bg

1 个答案: