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
}
}
?>
imagepalettecopy
(PHP 4 >= 4.0.1, PHP 5, PHP 7)
imagepalettecopy — Copy the palette from one image to another
설명
void imagepalettecopy
( resource
$destination
, resource $source
)
imagepalettecopy() copies the palette from the
source image to the
destination image.
인수
-
destination -
The destination image resource.
-
source -
The source image resource.
반환값
값을 반환하지 않습니다.
예제
Example #1 imagepalettecopy() example
<?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);
?>
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User Contributed Notes 2 notes
buzz at nospam dot oska dot com ¶
19 years ago
