将div旋转到鼠标位置

Question

我想让div旋转到鼠标所在的点，

所以当我的鼠标在左边时，它必须显示<-；当我的老鼠在右边时，它不得不显示->

这需要朝任何方向发展！

有人有解决方案吗？

编辑：

<style>
    body {
        height: 100vh;
        display: flex;
        align-items: center;
        justify-content: center;
        overflow: hidden;
    }
</style>
<div id="arrow">&gt;</div>
<script type="text/javascript">
    var arrow = document.querySelector("#arrow");
    var arrowRects = arrow.getBoundingClientRect();
    var arrowX = arrowRects.left + arrowRects.width / 2;
    var arrowY = arrowRects.top + arrowRects.height / 2;
    addEventListener("mousemove", function(event) {
        arrow.style.transform = "rotate(" + Math.atan2(event.y - arrowY, event.x - arrowX) + "rad)";
    });
</script>

Answer 1

您可以监听mousemove事件来获取光标的位置。然后，使用旋转变换使箭头旋转。

function getCenter(element) {
    const {left, top, width, height} = element.getBoundingClientRect();
    return {x: left + width / 2, y: top + height / 2}
}
const arrow = document.querySelector("#arrow");
const arrowCenter = getCenter(arrow);
addEventListener("mousemove", ({clientX, clientY}) => {
    const angle = Math.atan2(clientY - arrowCenter.y, clientX - arrowCenter.x);
    arrow.style.transform = `rotate(${angle}rad)`;
});

html  {
    height: 100vh;
    display: flex;
    align-items: center;
    justify-content: center;
}

<div id="arrow">&gt;</div>

Answer 2

元掩码答案的jQuery版本。

var arrow = $('#arrow');
var arrowX = arrow.offset().left + arrow.outerWidth(true) / 2;
var arrowY = arrow.offset().top + arrow.outerHeight(true) / 2;
$(document).on('mousemove', function(event) {
  var rad = Math.atan2(event.pageY - arrowY, event.pageX - arrowX);
  arrow.css('transform', 'rotate('+rad+ 'rad)');
});

body {
  height: 100vh;
  display: flex;
  align-items: center;
  justify-content: center;
  overflow: hidden;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="arrow">&gt;</div>