重复@超新星的回答，这里有一个使用ES6类的方法，它在你的回调中保持正确的上下文:

class Mouse { constructor() { this.x = 0; this.y = 0; this.callbacks = { mouseenter: [], mousemove: [], }; } get xPos() { return this.x; } get yPos() { return this.y; } get position() { return `${this.x},${this.y}`; } addListener(type, callback) { document.addEventListener(type, this); // Pass `this` as the second arg to keep the context correct this.callbacks[type].push(callback); } // `handleEvent` is part of the browser's `EventListener` API. // https://developer.mozilla.org/en-US/docs/Web/API/EventListener/handleEvent handleEvent(event) { const isMousemove = event.type === 'mousemove'; const isMouseenter = event.type === 'mouseenter'; if (isMousemove || isMouseenter) { this.x = event.pageX; this.y = event.pageY; } this.callbacks[event.type].forEach((callback) => { callback(); }); } } const mouse = new Mouse(); mouse.addListener('mouseenter', () => console.log('mouseenter', mouse.position)); mouse.addListener('mousemove', () => console.log('mousemove A', mouse.position)); mouse.addListener('mousemove', () => console.log('mousemove B', mouse.position));

真实的答案:不，这是不可能的。

OK, I have just thought of a way. Overlay your page with a div that covers the whole document. Inside that, create (say) 2,000 x 2,000 <a> elements (so that the :hover pseudo-class will work in IE 6, see), each 1 pixel in size. Create a CSS :hover rule for those <a> elements that changes a property (let's say font-family). In your load handler, cycle through each of the 4 million <a> elements, checking currentStyle / getComputedStyle() until you find the one with the hover font. Extrapolate back from this element to get the co-ordinates within the document.

N.b.，别这样。

重复@超新星的回答，这里有一个使用ES6类的方法，它在你的回调中保持正确的上下文:

class Mouse { constructor() { this.x = 0; this.y = 0; this.callbacks = { mouseenter: [], mousemove: [], }; } get xPos() { return this.x; } get yPos() { return this.y; } get position() { return `${this.x},${this.y}`; } addListener(type, callback) { document.addEventListener(type, this); // Pass `this` as the second arg to keep the context correct this.callbacks[type].push(callback); } // `handleEvent` is part of the browser's `EventListener` API. // https://developer.mozilla.org/en-US/docs/Web/API/EventListener/handleEvent handleEvent(event) { const isMousemove = event.type === 'mousemove'; const isMouseenter = event.type === 'mouseenter'; if (isMousemove || isMouseenter) { this.x = event.pageX; this.y = event.pageY; } this.callbacks[event.type].forEach((callback) => { callback(); }); } } const mouse = new Mouse(); mouse.addListener('mouseenter', () => console.log('mouseenter', mouse.position)); mouse.addListener('mousemove', () => console.log('mousemove A', mouse.position)); mouse.addListener('mousemove', () => console.log('mousemove B', mouse.position));

如果你渲染2000 x 2000 <a>个元素，@Tim Down的答案是不性能的:

OK, I have just thought of a way. Overlay your page with a div that covers the whole document. Inside that, create (say) 2,000 x 2,000 elements (so that the :hover pseudo-class will work in IE 6, see), each 1 pixel in size. Create a CSS :hover rule for those elements that changes a property (let's say font-family). In your load handler, cycle through each of the 4 million elements, checking currentStyle / getComputedStyle() until you find the one with the hover font. Extrapolate back from this element to get the co-ordinates within the document. N.B. DON'T DO THIS.

但是你不需要一次呈现400万个元素，而是使用二分搜索。只需使用4 <a>元素即可:

第一步:考虑整个屏幕作为开始搜索区域 步骤2:将搜索区域分割成2 × 2 = 4个矩形<a>个元素 步骤3:使用getComputedStyle()函数确定鼠标悬停在哪个矩形中 步骤4:将搜索区域缩小到该矩形，并重复步骤2。

这样你将需要重复这些步骤最多11次，考虑到你的屏幕不超过2048px。

所以你将生成最多11 x 4 = 44 <a>个元素。

如果你不需要确定鼠标位置精确到一个像素，但说10px的精度是可以的。您将重复这些步骤最多8次，因此您将需要绘制最多8 x 4 = 32 <a>个元素。

生成并销毁<a>元素也不能执行，因为DOM通常很慢。相反，您可以重用初始的4 <a>元素，并在循环执行步骤时调整它们的顶部、左侧、宽度和高度。

现在，创建4 <a>也是多余的。相反，您可以在每个矩形中测试getComputedStyle()时重用相同的<a>元素。因此，不需要将搜索区域分割为2 × 2 <a>元素，只需要重复使用一个<a>元素，使用顶部和左侧样式属性移动它。

所以，所有你需要的是一个<a>元素改变它的宽度和高度最多11次，改变它的顶部和左侧最多44次，你就会有准确的鼠标位置。

这是我的解决方案。它导出窗口。currentMouseX和window。你可以在任何地方使用currentMouseY属性。它首先使用悬停元素的位置(如果有的话)，然后监听鼠标移动来设置正确的值。

(function () {
    window.currentMouseX = 0;
    window.currentMouseY = 0;

    // Guess the initial mouse position approximately if possible:
    var hoveredElement = document.querySelectorAll(':hover');
    hoveredElement = hoveredElement[hoveredElement.length - 1]; // Get the most specific hovered element

    if (hoveredElement != null) {
        var rect = hoveredElement.getBoundingClientRect();
        // Set the values from hovered element's position
        window.currentMouseX = window.scrollX + rect.x;
        window.currentMouseY = window.scrollY + rect.y;
    }

    // Listen for mouse movements to set the correct values
    window.addEventListener('mousemove', function (e) {
        window.currentMouseX = e.pageX;
        window.currentMouseY = e.pageY;
    }, /*useCapture=*/true);
}())

来源:https://github.com/ocproducts/composr/commit/a851c19f925be20bc16bfe016be42924989f262e#diff-b162dc9c35a97618a96748639ff41251R1202

我实现了一个水平/垂直搜索，(首先让一个div充满垂直排列的竖线链接，然后让一个div充满水平排列的竖线链接，然后简单地看看哪个有悬停状态)，就像Tim Down上面的想法，它工作得相当快。遗憾的是，不工作在Chrome 32上的KDE。

jsfiddle.net/5XzeE/4/