PHP 图像颜色分析与透明度

发布于 2024-08-18 18:25:40 字数 2121 浏览 7 评论 0原文

我目前正在开发一个应用程序,需要分析许多图像并找出它们最接近的颜色。

因此,我找到了一个代码片段,它正是这样做的:

  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;
  }

问题是,每当我提供具有透明度的图像时,GDLib 都会将透明度视为黑色,从而产生比实际情况错误(暗得多)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:

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

谁能告诉我如何解决这个问题?

I am currently working on an application that needs to analyse a number of images and figure out what color they're closest to.

Therefore I found a code snippet that does exactly that:

  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;
  }

Problem is that whenever I provide an image with transparency, GDLib consinders the transparency to be black, thus producing a wrong (much darker) output than is really the case.

For example this icon where the white area around the arrow is actually transparent:

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

Can anyone tell me how to work around this?

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少女的英雄梦 2024-08-25 18:25:40

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

透明度是图像的属性,而不是图像的属性颜色。因此,使用类似 $transparent = imagecolortransparent($im) 来查看图像上是否有任何透明度,然后忽略 $colorArray 中的该颜色,或者使用其他方法来识别 $colorArray 中的透明颜色从你的函数返回。这完全取决于您如何使用返回的数据。

--M

You need imageColorTransparent(). http://www.php.net/imagecolortransparent

Transparency is a property of the image, not of a color. So use something like $transparent = imagecolortransparent($im) to see if there is any transparency on your image, then just ignore that color in your $colorArray or have some other way to identify the transparent color in the return from your function. That all depends on how you're using the returned data.

--M

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