PHP:如何表达“接近白色"作为颜色? [英] PHP: how to express "near white" as a color?

问题描述

我具有修剪图像周围白色区域的功能，其功能类似于

I have a function for trimming white areas around images that works something like

if(imagecolorat($img, $x, $top) != 0xFFFFFF) {
    //sets where the top part of the image is trimmed
}

问题是某些图像偶尔会出现杂散像素，其接近白色，因此不明显，但由于不是精确的0xFFFFFF而是0xFFFEFF或类似的东西，所以会增加裁剪速度.我该如何重写上面的语句，以使它对于接近白色的图像(例如，低至0xFDFDFD)的评估正确，显然无需测试任何可能的值.

The problem is some images have an occasional stray pixel that is so near white that it's unnoticeable but screws up the cropping because it's not exactly 0xFFFFFF but 0xFFFEFF or something. How could I rewrite the above statement so that it evaluates true for those near white images, say down to 0xFDFDFD, obviously without testing for ever possible value.

推荐答案

$color = imagecolorat($img, $x, $top);
$color = array(
    'red'   => ($color >> 16) & 0xFF,
    'green' => ($color >>  8) & 0xFF,
    'blue'  => ($color >>  0) & 0xFF,
);
if ($color['red']   >= 0xFD
 && $color['green'] >= 0xFD
 && $color['blue']  >= 0xFD) {
    //sets where the top part of the image is trimmed
}

有关所用运算符的说明，请阅读:

For a description of the operators used, please read:

• PHP-两个异常运算符一起用于获取图像像素颜色，请说明

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