具有透明度的PHP图像颜色分析

时间:2010-01-22 10:34:00

标签: php image-processing gdlib

我目前正在开发一个应用程序,需要分析一些图像并找出它们最接近的颜色。

因此,我找到了一个完全符合该代码的代码片段:

  function analyzeImageColors($im, $xCount =3, $yCount =3)
  {
    //get dimensions for image
    $imWidth =imagesx($im);
    $imHeight =imagesy($im);
    //find out the dimensions of the blocks we're going to make
    $blockWidth =round($imWidth/$xCount);
    $blockHeight =round($imHeight/$yCount);
    //now get the image colors...
    for($x =0; $x<$xCount; $x++) { //cycle through the x-axis
      for ($y =0; $y<$yCount; $y++) { //cycle through the y-axis
        //this is the start x and y points to make the block from
        $blockStartX =($x*$blockWidth);
        $blockStartY =($y*$blockHeight);
        //create the image we'll use for the block
        $block =imagecreatetruecolor(1, 1);
        //We'll put the section of the image we want to get a color for into the block
        imagecopyresampled($block, $im, 0, 0, $blockStartX, $blockStartY, 1, 1, $blockWidth, $blockHeight );
        //the palette is where I'll get my color from for this block
        imagetruecolortopalette($block, true, 1);
        //I create a variable called eyeDropper to get the color information
        $eyeDropper =imagecolorat($block, 0, 0);
        $palette =imagecolorsforindex($block, $eyeDropper);
        $colorArray[$x][$y]['r'] =$palette['red'];
        $colorArray[$x][$y]['g'] =$palette['green'];
        $colorArray[$x][$y]['b'] =$palette['blue'];
        //get the rgb value too
        $hex =sprintf("%02X%02X%02X", $colorArray[$x][$y]['r'], $colorArray[$x][$y]['g'], $colorArray[$x][$y]['b']);
        $colorArray[$x][$y]['rgbHex'] =$hex;
        //destroy the block
        imagedestroy($block);
      }
    }
    //destroy the source image
    imagedestroy($im);
    return $colorArray;
  }

问题在于,每当我提供具有透明度的图像时,GDLib就会将透明度设为黑色,从而产生错误(更暗)的输出。

例如,此图标中箭头周围的白色区域实际上是透明的:

example http://img651.imageshack.us/img651/995/screenshot20100122at113.png

有谁能告诉我如何解决这个问题?

1 个答案:

答案 0 :(得分:1)

你需要imageColorTransparent()。 http://www.php.net/imagecolortransparent

透明度是图像的属性,而不是颜色。因此,使用类似$transparent = imagecolortransparent($im)的内容来查看图像是否有任何透明度,然后只需忽略$ colorArray中的颜色,或者使用其他方法识别函数返回中的透明颜色。这一切都取决于你如何使用返回的数据。

- 中号