背景图像暗或亮？

Question

我正在使用PHP进行ODP-> HTML转换.我有以下问题:

使用style:use-window-font-color属性指定窗口前景色是否应该用作浅色背景色的前景色和白色用于深色背景色.

(OpenDocument规范版本1.0,15.4.4)

如果我有背景图像,如果此图像是亮还是暗,我该如何检查？

你有什么想法？

提前谢谢,Levu

Answer 1

我认为这是一个非常有趣的问题需要解决,所以我编写了一个快速的脚本来完成它.遵循提供的其他建议

<?php

    // TODO supply your own filenames
    $filenames = array(
        'testpics/client-bella-vi.jpg',
        'testpics/istockphoto_8577991-concept-of-business-people-crowd.jpg',
        'testpics/medium-gray.jpg');

    // loop though each file
    foreach ($filenames as $filename) {

        echo "$filename<br/>";

        $luminance = get_avg_luminance($filename,10);
        echo "AVG LUMINANCE: $luminance<br />";

        // assume a medium gray is the threshold, #acacac or RGB(172, 172, 172)
        // this equates to a luminance of 170
        if ($luminance > 170) {
            echo "Black Text<br />";
        } else {
            echo 'White Text<br />';
        }

        echo "<br />";
    }
    exit;

    // get average luminance, by sampling $num_samples times in both x,y directions
    function get_avg_luminance($filename, $num_samples=10) {
        $img = imagecreatefromjpeg($filename);

        $width = imagesx($img);
        $height = imagesy($img);

        $x_step = intval($width/$num_samples);
        $y_step = intval($height/$num_samples);

        $total_lum = 0;

        $sample_no = 1;

        for ($x=0; $x<$width; $x+=$x_step) {
            for ($y=0; $y<$height; $y+=$y_step) {

                $rgb = imagecolorat($img, $x, $y);
                $r = ($rgb >> 16) & 0xFF;
                $g = ($rgb >> 8) & 0xFF;
                $b = $rgb & 0xFF;

                // choose a simple luminance formula from here
                // http://stackoverflow.com/questions/596216/formula-to-determine-brightness-of-rgb-color
                $lum = ($r+$r+$b+$g+$g+$g)/6;

                $total_lum += $lum;

                // debugging code
     //           echo "$sample_no - XY: $x,$y = $r, $g, $b = $lum<br />";
                $sample_no++;
            }
        }

        // work out the average
        $avg_lum  = $total_lum/$sample_no;

        return $avg_lum;
    }