在鼠标移动时从画布获取像素颜色

Question

是否可以获取鼠标下方的RGB值像素？有这方面的完整例子吗？这是我到目前为止所拥有的：

 function draw() {
      var ctx = document.getElementById('canvas').getContext('2d');
      var img = new Image();
      img.src = 'Your URL';

      img.onload = function(){
        ctx.drawImage(img,0,0);

      };

      canvas.onmousemove = function(e) {
            var mouseX, mouseY;

            if(e.offsetX) {
                mouseX = e.offsetX;
                mouseY = e.offsetY;
            }
            else if(e.layerX) {
                mouseX = e.layerX;
                mouseY = e.layerY;
            }
            var c = ctx.getImageData(mouseX, mouseY, 1, 1).data;

            $('#ttip').css({'left':mouseX+20, 'top':mouseY+20}).html(c[0]+'-'+c[1]+'-'+c[2]);
      };
    }

原文由 vince83000 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

这是一个完整的、独立的示例。首先，使用以下 HTML：

 <canvas id="example" width="200" height="60"></canvas>
<div id="status"></div>

然后在画布上放置一些具有随机背景颜色的正方形：

 var example = document.getElementById('example');
var context = example.getContext('2d');
context.fillStyle = randomColor();
context.fillRect(0, 0, 50, 50);
context.fillStyle = randomColor();
context.fillRect(55, 0, 50, 50);
context.fillStyle = randomColor();
context.fillRect(110, 0, 50, 50);

并在鼠标悬停时打印每种颜色：

 $('#example').mousemove(function(e) {
    var pos = findPos(this);
    var x = e.pageX - pos.x;
    var y = e.pageY - pos.y;
    var coord = "x=" + x + ", y=" + y;
    var c = this.getContext('2d');
    var p = c.getImageData(x, y, 1, 1).data;
    var hex = "#" + ("000000" + rgbToHex(p[0], p[1], p[2])).slice(-6);
    $('#status').html(coord + "<br>" + hex);
});

上面的代码假定存在 jQuery 和以下实用函数：

 function findPos(obj) {
    var curleft = 0, curtop = 0;
    if (obj.offsetParent) {
        do {
            curleft += obj.offsetLeft;
            curtop += obj.offsetTop;
        } while (obj = obj.offsetParent);
        return { x: curleft, y: curtop };
    }
    return undefined;
}

function rgbToHex(r, g, b) {
    if (r > 255 || g > 255 || b > 255)
        throw "Invalid color component";
    return ((r << 16) | (g << 8) | b).toString(16);
}

function randomInt(max) {
  return Math.floor(Math.random() * max);
}

function randomColor() {
    return `rgb(${randomInt(256)}, ${randomInt(256)}, ${randomInt(256)})`
}

在这里看到它的实际效果：

• https://bl.ocks.org/wayneburkett/ca41a5245a9f48766b7bc881448f9203

 // set up some sample squares with random colors
var example = document.getElementById('example');
var context = example.getContext('2d');
context.fillStyle = randomColor();
context.fillRect(0, 0, 50, 50);
context.fillStyle = randomColor();
context.fillRect(55, 0, 50, 50);
context.fillStyle = randomColor();
context.fillRect(110, 0, 50, 50);

$('#example').mousemove(function(e) {
    var pos = findPos(this);
    var x = e.pageX - pos.x;
    var y = e.pageY - pos.y;
    var coord = "x=" + x + ", y=" + y;
    var c = this.getContext('2d');
    var p = c.getImageData(x, y, 1, 1).data;
    var hex = "#" + ("000000" + rgbToHex(p[0], p[1], p[2])).slice(-6);
    $('#status').html(coord + "<br>" + hex);
});

function findPos(obj) {
    var curleft = 0, curtop = 0;
    if (obj.offsetParent) {
        do {
            curleft += obj.offsetLeft;
            curtop += obj.offsetTop;
        } while (obj = obj.offsetParent);
        return { x: curleft, y: curtop };
    }
    return undefined;
}

function rgbToHex(r, g, b) {
    if (r > 255 || g > 255 || b > 255)
        throw "Invalid color component";
    return ((r << 16) | (g << 8) | b).toString(16);
}

function randomInt(max) {
  return Math.floor(Math.random() * max);
}

function randomColor() {
    return `rgb(${randomInt(256)}, ${randomInt(256)}, ${randomInt(256)})`
}

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<canvas id="example" width="200" height="60"></canvas>
<div id="status"></div>

原文由 Wayne 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

我知道这是一个老问题，但这里有一个替代方案。我将该图像数据存储在一个数组中，然后在画布上发生鼠标移动事件时：

 var index = (Math.floor(y) * canvasWidth + Math.floor(x)) * 4
var r = data[index]
var g = data[index + 1]
var b = data[index + 2]
var a = data[index + 3]

比每次都获取 imageData 容易得多。

原文由 Caio Vertematti 发布，翻译遵循 CC BY-SA 3.0 许可协议