我做了一个完整的演示，可以在每个浏览器中使用这个问题的解决方案的完整源代码:在Javascript中单击Canvas的鼠标坐标。要尝试演示，复制代码并将其粘贴到文本编辑器中。然后将其保存为example.html，最后用浏览器打开该文件。

我在创建一个应用程序，在pdf上有一个画布，这涉及到很多画布的大小调整，比如放大和缩小pdf，然后在每次放大/缩小pdf时，我都必须调整画布的大小以适应pdf的大小，我在stackOverflow中经历了很多答案，并没有找到一个完美的解决方案，最终将解决问题。

我使用的是rxjs和angular 6，没有找到任何针对最新版本的答案。

这里是整个代码片段，这将是有帮助的，任何人利用rxjs在画布上绘制。

  private captureEvents(canvasEl: HTMLCanvasElement) {

    this.drawingSubscription = fromEvent(canvasEl, 'mousedown')
      .pipe(
        switchMap((e: any) => {

          return fromEvent(canvasEl, 'mousemove')
            .pipe(
              takeUntil(fromEvent(canvasEl, 'mouseup').do((event: WheelEvent) => {
                const prevPos = {
                  x: null,
                  y: null
                };
              })),

              takeUntil(fromEvent(canvasEl, 'mouseleave')),
              pairwise()
            )
        })
      )
      .subscribe((res: [MouseEvent, MouseEvent]) => {
        const rect = this.cx.canvas.getBoundingClientRect();
        const prevPos = {
          x: Math.floor( ( res[0].clientX - rect.left ) / ( rect.right - rect.left ) * this.cx.canvas.width ),
          y:  Math.floor( ( res[0].clientY - rect.top ) / ( rect.bottom - rect.top ) * this.cx.canvas.height )
        };
        const currentPos = {
          x: Math.floor( ( res[1].clientX - rect.left ) / ( rect.right - rect.left ) * this.cx.canvas.width ),
          y: Math.floor( ( res[1].clientY - rect.top ) / ( rect.bottom - rect.top ) * this.cx.canvas.height )
        };

        this.coordinatesArray[this.file.current_slide - 1].push(prevPos);
        this.drawOnCanvas(prevPos, currentPos);
      });
  }

这里是修复的代码片段，鼠标坐标相对于画布的大小，不管你如何放大/缩小画布。

const prevPos = {
  x: Math.floor( ( res[0].clientX - rect.left ) / ( rect.right - rect.left ) * this.cx.canvas.width ),
  y:  Math.floor( ( res[0].clientY - rect.top ) / ( rect.bottom - rect.top ) * this.cx.canvas.height )
};
const currentPos = {
  x: Math.floor( ( res[1].clientX - rect.left ) / ( rect.right - rect.left ) * this.cx.canvas.width ),
  y: Math.floor( ( res[1].clientY - rect.top ) / ( rect.bottom - rect.top ) * this.cx.canvas.height )
};

更新(5/5/16):应该使用patriques的答案，因为它既简单又可靠。

Since the canvas isn't always styled relative to the entire page, the canvas.offsetLeft/Top doesn't always return what you need. It will return the number of pixels it is offset relative to its offsetParent element, which can be something like a div element containing the canvas with a position: relative style applied. To account for this you need to loop through the chain of offsetParents, beginning with the canvas element itself. This code works perfectly for me, tested in Firefox and Safari but should work for all.

function relMouseCoords(event){
    var totalOffsetX = 0;
    var totalOffsetY = 0;
    var canvasX = 0;
    var canvasY = 0;
    var currentElement = this;

    do{
        totalOffsetX += currentElement.offsetLeft - currentElement.scrollLeft;
        totalOffsetY += currentElement.offsetTop - currentElement.scrollTop;
    }
    while(currentElement = currentElement.offsetParent)

    canvasX = event.pageX - totalOffsetX;
    canvasY = event.pageY - totalOffsetY;

    return {x:canvasX, y:canvasY}
}
HTMLCanvasElement.prototype.relMouseCoords = relMouseCoords;

最后一行可以方便地获取相对于canvas元素的鼠标坐标。要得到有用的坐标只需要

coords = canvas.relMouseCoords(event);
canvasX = coords.x;
canvasY = coords.y;

我做了一个完整的演示，可以在每个浏览器中使用这个问题的解决方案的完整源代码:在Javascript中单击Canvas的鼠标坐标。要尝试演示，复制代码并将其粘贴到文本编辑器中。然后将其保存为example.html，最后用浏览器打开该文件。

这里是一个简化的解决方案(这并不适用于边框/滚动):

function click(event) {
    const bound = event.target.getBoundingClientRect();

    const xMult = bound.width / can.width;
    const yMult = bound.height / can.height;

    return {
        x: Math.floor(event.offsetX / xMult),
        y: Math.floor(event.offsetY / yMult),
    };
}

三JS r77

var x = event.offsetX == undefined ? event.layerX : event.offsetX;
var y = event.offsetY == undefined ? event.layerY : event.offsetY;

mouse2D.x = ( x / renderer.domElement.width ) * 2 - 1;
mouse2D.y = - ( y / renderer.domElement.height ) * 2 + 1;

在尝试了许多解决方案后。这对我很管用。可能会帮助其他人，因此发帖。从这里开始