将鼠标移到元素上-获取准确的鼠标位置X Y

Question

我创建表，处理程序将鼠标移动起来。但是在左上角，我得到.offsetX.offsetY等于-5-5。为什么？我需要0 \ 0。

<table cellpadding="0" 
       id="target" 
       cellspacing="0"
       width="602" 
       height="500"  
       style="float:left;
              position:relative;
              background: url(/content/games/kamikaze2/back.jpg) no-repeat 0 0;">
    <tbody>...
</tbody>
</table>

<script type="text/javascript">

    $("#target").mousemove(function (event) {
        var msg = "Handler for .mousemove() called at ";
        msg += event.offsetX + ", " + event.offsetY;
        $("#log").append("<div>" + msg + "</div>");
    });
</script>

Answer 1

获取元素内鼠标坐标的最准确方法是通过计算之间的差

• [MouseEvent.clientX，MouseEvent.clientY和
• Element.getBoundingClientRect()

const getMousePos = (evt) => {
  const pos = evt.currentTarget.getBoundingClientRect();
  return {
    x: evt.clientX - pos.left,
    y: evt.clientY - pos.top
  };
};

document.querySelector("#target").addEventListener('mousemove', (evt) => {
  const mPos = getMousePos(evt);
  evt.currentTarget.textContent = `Mouse position x:${mPos.x}  y:${mPos.y}`;
});

#target {
  position: absolute;
  left: 50px;
  top: 20px;
  width: 200px;
  height: 100px;
  background: #0bf;
  transform: translate(20px, 30px); /* works even with translate */
}

<div id="target"></div>

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

减去这些值，您可以获取精确的鼠标位置，x：0，y：0是元素的左上角。即使元素是CSS转换的，即transform: translate(-50%, -50%)，返回的值仍然正确。