imagepalettecopy

(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

说明

imagepalettecopy(GdImage $dst, GdImage $src): void

imagepalettecopy() 把调色板从 src 图像复制到 dst 图像。

参数

dst

目标图像对象。

src

源图像对象。

返回值

没有返回值。

更新日志

版本 说明
8.0.0 dstsrc 现在接受 GdImage 实例;之前接受 resource

示例

示例 #1 imagepalettecopy() 示例

<?php
// 创建两个调色板图像
$palette1 = imagecreate(100, 100);
$palette2 = imagecreate(100, 100);

// 在第一个调色板图像中
// 将背景分配为绿色
$green = imagecolorallocate($palette1, 0, 255, 0);

// 将调色板从图像 1 复制到图像 2
imagepalettecopy($palette2, $palette1);

// 由于调色板现已复制,可以使用
// 分配给图像 1 的绿色,而无需
// 两次使用 imagecolorallocate()
imagefilledrectangle($palette2, 0, 0, 99, 99, $green);

// 输出图像到浏览器
header('Content-type: image/png');

imagepng($palette2);
?>

添加备注

用户贡献的备注 2 notes

up
1
buzz at nospam dot oska dot com
19 years ago
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
}
}

?>
up
-1
Los Olvidados
22 years ago
To be precise, this function replaces the palette in the destination.
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