To be precise, this function replaces the palette in the destination.
(PHP 4 >= 4.0.1, PHP 5, PHP 7, PHP 8)
imagepalettecopy — 将调色板从一个图像复制到另一个
dst
目标图像对象。
src
源图像对象。
没有返回值。
示例 #1 imagepalettecopy() 示例
<?php
// Create two palette images
$palette1 = imagecreate(100, 100);
$palette2 = imagecreate(100, 100);
// Allocate the background to be
// green in the first palette image
$green = imagecolorallocate($palette1, 0, 255, 0);
// Copy the palette from image 1 to image 2
imagepalettecopy($palette2, $palette1);
// Since the palette is now copied we can use the
// green color allocated to image 1 without using
// imagecolorallocate() twice
imagefilledrectangle($palette2, 0, 0, 99, 99, $green);
// Output image to the browser
header('Content-type: image/png');
imagepng($palette2);
imagedestroy($palette1);
imagedestroy($palette2);
?>
actually it doesn't "copy" the palette exactly. It copys the colors from the source palette to the destination image. the palette you end up with in the destination image will be "same colors different order". If you want an EXACT palette copy (at the expense of messing up your image if you aren't careful), then use this code:
<?
// this is a drop-in replacement for imagepalettecopy, except that it make NO attempt to modifiy any of the
// colors in the dest image, just the palette. The result? if you're palette's aren't very similar, the image will look completely different, and likely terrible!
function imagepalettecopy_exact ( $dst_img, $src_img) {
for( $c = 0 ; $c < imagecolorstotal($src_img); $c++) {
$col = imagecolorsforindex($src_img,$c); //get color at index 'c' in the color table
imagecolorset($dst_img,$c,$col[red],$col[green],$col[blue]); //set color at index 'c' to $col in the $dst_image
}
}
?>