cors 污染图像上的映射颜色变换

Question

我需要转换从不允许 crossOrigin 的 s3 服务器获取的图像中的颜色。

这是我需要的功能：

  const img = new Image();
  img.src = src;
  ctx.drawImage(img, 0, 0);
  const imageData = ctx.getImageData(0, 0, canvas.width, canvas.height);
  const colors = [25,50,100,255];
  const data = imageData.data;
  for (let i = 0; i < data.length; i += 4) {
    if(data[i] == 1){data[i] = colors[1]}
    if(data[i] == 2){data[i] = colors[2]}
    if(data[i] == 3){data[i] = colors[3]}
  }

但被 crossOrigin 错误污染：

我知道在这种情况下我不能使用getImageData。我不想读取图像数据。

但也许可以通过其他一些 webGL / canvas 操作来完成我的任务？

无法在服务器上呈现或代理。

Answer 1

WebGL 根本不适用于跨源纹理。草案中有an extension，但尚未标准化，只允许在有限的场景下使用then，所以使用此类纹理的采样器仍然不能用于条件表达式。

使用经典canvas，您可以通过巧妙地使用混合模式来实现大量的色彩效果。有一个列表 on MDN, or a more technical description on W3.

特别是，一个非常有用的技巧涉及使用 color-dodge 操作。通过使用非常明亮的恒定颜色作为源，它有效地将背景图像乘以一个大值，这让我们可以进行阈值操作。

在您的情况下，您可以使用以下代码 "select" 具有特定颜色的所有像素：

// 1. Use 'difference' to make all matching pixels black
ctx.globalCompositeOperation = "difference";
ctx.fillStyle = srcColor;
ctx.fillRect(0, 0, w, h);
// 2. Use 'color-dodge' trick to turn all non-black pixels into white
ctx.globalCompositeOperation = "color-dodge";
ctx.fillStyle = "#fefefe";
ctx.fillRect(0, 0, w, h);

// Steps 3 and 4 are only necessary if full RGB matching is required
// Without these steps, matching will be done on per-channel basis 
// 3*. Desaturate the image, ensuring all three channels have the same value
ctx.globalCompositeOperation = "saturation";
ctx.fillStyle = "#ffffff";
ctx.fillRect(0, 0, w, h);
// 4*. Use color-dodge again, to mask pixels where all 3 components matched
ctx.globalCompositeOperation = "color-dodge";
ctx.fillStyle = "#fefefe";
ctx.fillRect(0, 0, w, h);

// 5. Invert the image to make matching pixels white and the rest black
ctx.globalCompositeOperation = "difference";
ctx.fillStyle = "#ffffff";
ctx.fillRect(0, 0, w, h);
// 6. Multiply by desired color
ctx.globalCompositeOperation = "multiply";
ctx.fillStyle = dstColor;
ctx.fillRect(0, 0, w, h);

这是一个实现您的代码的实例（输出图像的左半部分看起来是黑色的，因为您看不到像 rgb(1,0,0) 这样的深色）：

const canvas = document.getElementById("canvas");
const ctx = canvas.getContext("2d");

function makeImage(w, h) {
  const c = document.createElement("canvas");
  c.width = w;
  c.height = h;
  return c;
}

function paintSource(c) {
  const ctx = c.getContext("2d");
  const data = ctx.getImageData(0, 0, c.width, c.height);
  for (let x = 0; x < c.width; ++x) {
    for (let y = 0; y < c.height; ++y) {
      const i = (y * c.width + x) * 4;
      const v = Math.round((x + y) / 2);
      // 1, 2 or 3
      data.data[i] = Math.floor(x * 3 / c.width) + 1;
      data.data[i + 3] = 255;
    }
  }
  ctx.putImageData(data, 0, 0);
}

function selectColor(srcCanvas, dstCanvas, srcColor, dstColor) {
  const ctx = dstCanvas.getContext("2d");
  ctx.drawImage(srcCanvas, 0, 0);
  const w = srcCanvas.width, h = srcCanvas.height;
  
  ctx.globalCompositeOperation = "difference";
  ctx.fillStyle = srcColor;
  ctx.fillRect(0, 0, w, h);
  ctx.globalCompositeOperation = "color-dodge";
  ctx.fillStyle = "#fefefe";
  ctx.fillRect(0, 0, w, h);
  ctx.globalCompositeOperation = "difference";
  ctx.fillStyle = "#ffffff";
  ctx.fillRect(0, 0, w, h);
  ctx.globalCompositeOperation = "multiply";
  ctx.fillStyle = dstColor;
  ctx.fillRect(0, 0, w, h);
}

const source = makeImage(256, 256);
paintSource(source);

const c1 = makeImage(256, 256);
selectColor(source, c1, "rgb(1,0,0)", "rgb(50,0,0)");
const c2 = makeImage(256, 256);
selectColor(source, c2, "rgb(2,0,0)", "rgb(100,0,0)");
const c3 = makeImage(256, 256);
selectColor(source, c3, "rgb(3,0,0)", "rgb(255,0,0)");

ctx.drawImage(source, 0, 0);
ctx.globalCompositeOperation = "lighter";
ctx.drawImage(c1, 256, 0);
ctx.drawImage(c2, 256, 0);
ctx.drawImage(c3, 256, 0);

<canvas id="canvas" width="512" height="256"></canvas>

这里有一个稍微复杂一点的例子来演示阈值：https://jsfiddle.net/Rivvy/kq2ga90z/35/

cors 污染图像上的映射颜色变换

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