<script>
function imageChange(img)
{
//document.getElementsByClassName('bgdodge')[0].style.backgroundImage= "url('pictures/gtr.jpg')";
document.getElementsByClassName('bgdodge')[0].style.backgroundImage= "url(img)";
}
</script>
<body>
<div class= "bgdodge">
<div id="clockbox"></div>
<div class="pcenter">
<p onmouseover= "imageChange('pictures/gtr.jpg')" onmouseout= "imageChange('pictures/hellcat.jpeg')"> Dodge Charger SRT Hellcat</p>
</div>
</div>
当我像不带参数的imageChange函数中的注释中那样执行时，它就可以工作。但是，当在onmouseover和mouseout上将图片传递到imageChange并使用“ img”变量时，它不起作用。 Idk为什么
您只是使用字符串“ img”，但实际上是要使用变量：
document.getElementsByClassName('bgdodge')[0].style.backgroundImage= "url("+img+")";
尝试此示例
function screen(){
alert('ddd')
const render = node =>
domtoimage.toPng(node)
.then(dataUrl => {
console.log(performance.now()-pf)
const img = new Image();
img.src = dataUrl;
$('body').append(img);
})
.catch(error =>
console.error('oops, something went wrong!', error)
);
const foo = document.getElementById('foo');
var pf=performance.now();
render(foo);
}<script src="https://cdn.rawgit.com/tsayen/dom-to-image/bfc00a6c5bba731027820199acd7b0a6e92149d8/dist/dom-to-image.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.3/jquery.min.js"></script>
<div id="foo">
<div >foo</div>
<button onclick="screen()">shot</button>
</div>