PHP - 替换图像中的颜色

PHP - Replace colour within image(PHP - 替换图像中的颜色)

问题描述

希望有人能帮忙

我制作了一个屏蔽图像的脚本......但是它依赖于一种颜色来屏蔽(绿屏"样式).问题是，如果我屏蔽的图像包含那种颜色，它就会被破坏.

I have made a script that masks images... however it is reliant on a colour to mask with ( 'green screen' style). The trouble is if the image that I am masking contains that colour it is ruined.

我要做的是在屏蔽图像之前用类似的颜色(例如 0,0,254)替换任何出现的键控颜色 (0,0,255).

What I am looking to do is prior to masking the image replace any occurance of my keying colour (0,0,255) with a similar colour such as 0,0,254.

我找到了一些基于 gif 或 256 色 PNG 的解决方案，因为它们已编入索引..

I have found a few solutions based around gif's or 256 colour PNG as they are indexed..

所以我的问题也是将其转换为 gif 或 256 png 然后查看索引并替换颜色或搜索每个像素并替换颜色是否更有效.

So my question is also will it be more efficient to convert it to a gif or 256 png then look through the index and replace the colour or search through every pixel and replace the colours.

谢谢，

推荐答案

您需要打开输入文件并扫描每个像素以检查色键值.

You need to open the input file and scan each pixel to check for your chromokey value.

像这样:

// Open input and output image
$src = imagecreatefromJPEG('input.jpg') or die('Problem with source');
$out = ImageCreateTrueColor(imagesx($src),imagesy($src)) or die('Problem In Creating image');

// scan image pixels
for ($x = 0; $x < imagesx($src); $x++) {
    for ($y = 0; $y < imagesy($src); $y++) {
        $src_pix = imagecolorat($src,$x,$y);
        $src_pix_array = rgb_to_array($src_pix);

            // check for chromakey color
            if ($src_pix_array[0] == 0 && $src_pix_array[1] == 0 && $src_pix_array[2] == 255) {
                $src_pix_array[2] = 254;
            }


        imagesetpixel($out, $x, $y, imagecolorallocate($out, $src_pix_array[0], $src_pix_array[1], $src_pix_array[2]));
    }
}


// write $out to disc

imagejpeg($out, 'output.jpg',100) or die('Problem saving output image');
imagedestroy($out);

// split rgb to components
function rgb_to_array($rgb) {
    $a[0] = ($rgb >> 16) & 0xFF;
    $a[1] = ($rgb >> 8) & 0xFF;
    $a[2] = $rgb & 0xFF;

    return $a;
}

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