canvas.onmousedown = function(e) {
    pos_left = e.pageX - e.currentTarget.offsetLeft;
    pos_top = e.pageY - e.currentTarget.offsetTop;
    console.log(pos_left, pos_top)
}

HTMLElement.offsetLeft

HTMLElement。offsetLeft read-only属性返回当前元素的左上角在HTMLElement中向左偏移的像素数。offsetParent节点。

对于块级元素，offsetTop, offsetLeft, offsetWidth和offsetHeight描述了一个元素相对于offsetParent的边界框。

However, for inline-level elements (such as span) that can wrap from one line to the next, offsetTop and offsetLeft describe the positions of the first border box (use Element.getClientRects() to get its width and height), while offsetWidth and offsetHeight describe the dimensions of the bounding border box (use Element.getBoundingClientRect() to get its position). Therefore, a box with the left, top, width and height of offsetLeft, offsetTop, offsetWidth and offsetHeight will not be a bounding box for a span with wrapped text.

HTMLElement.offsetTop

HTMLElement。offsetTop只读属性返回当前元素相对于offsetParent节点顶部的距离。

MouseEvent.pageX

pageX只读属性返回事件相对于整个文档的X(水平)坐标(像素)。此属性考虑页面的任何水平滚动。

MouseEvent.pageY

鼠标事件。pageY只读属性返回事件相对于整个文档的Y(垂直)像素坐标。此属性考虑页面的任何垂直滚动。

如需进一步解释，请参阅Mozilla开发者网络:

https://developer.mozilla.org/en-US/docs/Web/API/MouseEvent/pageX https://developer.mozilla.org/en-US/docs/Web/API/MouseEvent/pageY https://developer.mozilla.org/en-US/docs/Web/API/HTMLElement/offsetLeft https://developer.mozilla.org/en-US/docs/Web/API/HTMLElement/offsetTop

基于@Patrick Boos的解决方案，但修复了中间滚动条的潜在问题。

export function getRelativeCoordinates(event: MouseEvent, referenceElement: HTMLElement) {
  const position = {
    x: event.pageX,
    y: event.pageY,
  };

  const offset = {
    left: referenceElement.offsetLeft,
    top: referenceElement.offsetTop,
  };

  let reference = referenceElement.offsetParent as HTMLElement;

  while (reference) {
    offset.left += reference.offsetLeft;
    offset.top += reference.offsetTop;
    reference = reference.offsetParent as HTMLElement;
  }

  const scrolls = {
    left: 0,
    top: 0,
  };

  reference = event.target as HTMLElement;
  while (reference) {
    scrolls.left += reference.scrollLeft;
    scrolls.top += reference.scrollTop;
    reference = reference.parentElement as HTMLElement;
  }

  return {
    x: position.x + scrolls.left - offset.left,
    y: position.y + scrolls.top - offset.top,
  };
}

我必须得到光标的位置在一个非常宽的div与滚动条。目标是将元素拖动到div的任意位置。

将鼠标位置放在滚动画面深处的较远位置。

$('.canvas').on('mousemove', function(e){
    $(dragElement).parent().css('top', e.currentTarget.scrollTop + e.originalEvent.clientY );
    $(dragElement).parent().css('left', e.currentTarget.scrollLeft + e.originalEvent.clientX )
});

下面计算鼠标位置与canvas元素的关系:

const example = document.getElementById('example');

example.onmousemove = function(e) { 
    const x = e.pageX - e.currentTarget.offsetLeft; 
    const y = e.pageY - e.currentTarget.offsetTop; 
}

基于@Spider的解决方案，我的非JQuery版本是这样的:

// Get the container element's bounding box
var sides = document.getElementById("container").getBoundingClientRect();

// Apply the mouse event listener
document.getElementById("canvas").onmousemove = (e) => {
  // Here 'self' is simply the current window's context
  var x = (e.clientX - sides.left) + self.pageXOffset;
  var y = (e.clientY - sides.top) + self.pageYOffset;
}

这适用于滚动和缩放(在这种情况下，有时它返回浮动)。

关于这个问题的难点可以在这里找到:http://www.quirksmode.org/js/events_properties.html#position

使用这里描述的技术，您可以在文档中找到鼠标的位置。然后你只需检查它是否在元素的边界框内，你可以通过调用element. getboundingclientrect()来找到它，它将返回一个具有以下属性的对象:{bottom, height, left, right, top, width}。由此，判断偶数是否发生在元素内部是很简单的。