我正在使用PHP和GD裁剪并输出带有以下代码的图像.它工作正常,但是当我将透明PNG传递给它时,会生成黑色背景.我该如何阻止呢?
//setup
switch ($source_type) {
case IMAGETYPE_JPEG: $source = imagecreatefromjpeg($img_path); break;
case IMAGETYPE_PNG: $source = imagecreatefrompng($img_path); break;
}
// setup cropped destination
$cropped = imagecreatetruecolor($cropped_width, $cropped_height);
// create cropped image
$x = (($source_width / 100) * IMAGE_X) - ($cropped_width / 2);
$y = (($source_height / 100) * IMAGE_Y) - ($cropped_height / 2);
imagecopy(
$cropped,
$source,
0, 0,
$x, $y,
$cropped_width, $cropped_height
);
// output inc header
header('Content-type: image/jpeg');
imagejpeg($cropped);
解决方法:
应该是以下内容:
switch ($source_type)
{
case IMAGETYPE_PNG:
$background = imagecolorallocate($source, 0, 0, 0);
// remove the black
imagecolortransparent($source, $background);
// turn off alpha blending
imagealphablending($source, false);
imagesavealpha($source, true);
break;
}
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