给定背景颜色,黑色或白色文字？

Question

我正在尝试找到一种方法,用于确定是否使用黑色或白色文本,给定背景颜色(作为十六进制值).以前有人处理过这个吗？有没有有效的方法来做到这一点？

就我而言,我将使用PHP来实现逻辑(尽管欢迎使用其他语言的任何经验).

Answer 1

看一下这个页面:用PHP计算颜色对比度

请记住,如果黑色和白色是您唯一的选择,那么您必然会遇到两种情况都不是特别好的情况.

Answer 2

我认为最好的方法是Luminosity Contrast算法:

function getContrastColor($hexColor) {

        // hexColor RGB
        $R1 = hexdec(substr($hexColor, 1, 2));
        $G1 = hexdec(substr($hexColor, 3, 2));
        $B1 = hexdec(substr($hexColor, 5, 2));

        // Black RGB
        $blackColor = "#000000";
        $R2BlackColor = hexdec(substr($blackColor, 1, 2));
        $G2BlackColor = hexdec(substr($blackColor, 3, 2));
        $B2BlackColor = hexdec(substr($blackColor, 5, 2));

         // Calc contrast ratio
         $L1 = 0.2126 * pow($R1 / 255, 2.2) +
               0.7152 * pow($G1 / 255, 2.2) +
               0.0722 * pow($B1 / 255, 2.2);

        $L2 = 0.2126 * pow($R2BlackColor / 255, 2.2) +
              0.7152 * pow($G2BlackColor / 255, 2.2) +
              0.0722 * pow($B2BlackColor / 255, 2.2);

        $contrastRatio = 0;
        if ($L1 > $L2) {
            $contrastRatio = (int)(($L1 + 0.05) / ($L2 + 0.05));
        } else {
            $contrastRatio = (int)(($L2 + 0.05) / ($L1 + 0.05));
        }

        // If contrast is more than 5, return black color
        if ($contrastRatio > 5) {
            return '#000000';
        } else { 
            // if not, return white color.
            return '#FFFFFF';
        }
}

// Will return '#FFFFFF'
echo getContrastColor('#FF0000');

一些结果:

注意: 字体颜色由上一个功能决定.括号中的数字是对比度.

调用另一种最简单且不太精确的方法YIQ(因为它将RGB颜色空间转换为YIQ):

public function getContrastColor($hexcolor) {               
    $r = hexdec(substr($hexcolor, 1, 2));
    $g = hexdec(substr($hexcolor, 3, 2));
    $b = hexdec(substr($hexcolor, 5, 2));
    $yiq = (($r * 299) + ($g * 587) + ($b * 114)) / 1000;
    return ($yiq >= 128) ? 'black' : 'white';
}                   

Answer 3

  function getTextColour($hex){
    list($red, $green, $blue) = sscanf($hex, "#%02x%02x%02x");
    $luma = ($red + $green + $blue)/3;

    if ($luma < 128){
      $textcolour = "white";
    }else{
      $textcolour = "black";
    }
    return $textcolour;
  }