在坐标

处缩放

如果给定具有缩放内容的屏幕坐标，则需要以与缩放匹配的量朝向或远离该点移动原点。

如果放大原点，则朝向鼠标位置移动。并缩小。

因此，使用鼠标获取x，y位置并根据方向取决于放大或缩小的数量

// e is the mouse wheel event
const x = e.offsetX;
const y = e.offsetY;
const amount = e.wheelDelta > 0 ?  1.1 : 1 / 1.1;

然后将其应用于当前比例并移动原点以匹配

scale *= amount;  // the new scale
// move the origin
origin.x = x - (x - origin.x) * amount;
origin.y = y - (y - origin.y) * amount;

然后，您可以使用

设置2D位置和比例

ctx.setTransform(scale, 0, 0, scale, origin.x, origin.y);

实施例

下面是一个简单的视图示例，其中使用了您的小提琴代码。

对象view跟踪当前视图并维护可应用于画布的矩阵。

const ctx = canvas.getContext("2d");
canvas.width = 500;
canvas.height = 500;
const randI = (min, max = min + (min = 0)) => (Math.random() * (max - min) + min) | 0;
const rand = (min, max = min + (min = 0)) => Math.random() * (max - min) + min;

const objects = [];
for (let i = 0; i < 100; i++) {
  objects.push({
    x: rand(canvas.width),
    y: rand(canvas.height),
    w: rand(40),
    h: rand(40),
    col: `rgb(${randI(255)},${randI(255)},${randI(255)})`,
  });
}
canvas.addEventListener("mousewheel", onmousewheel, false);
canvas.addEventListener("DOMMouseScroll", onmousewheel, false);
requestAnimationFrame(drawCanvas); // this will call drawcanvas after all other code has run

const view = (() => {
  const matrix = [1, 0, 0, 1, 0, 0]; // current view transform
  var m = matrix; // alias for clear code
  var scale = 1; // current scale
  var ctx; // reference to the 2D context
  const pos = { x: 0, y: 0 }; // current position of origin
  var dirty = true;
  const API = {
    setContext(_ctx) { ctx = _ctx; dirty = true },
    apply() {
      if (dirty) { this.update() }
      ctx.setTransform(m[0], m[1], m[2], m[3], m[4], m[5])
    },
    getScale() { return scale },
    getPosition() { return pos },
    isDirty() { return dirty },
    update() {
      dirty = false;
      m[3] = m[0] = scale;
      m[2] = m[1] = 0;
      m[4] = pos.x;
      m[5] = pos.y;
    },
    scaleAt(at, amount) { // at in screen coords
      if (dirty) { this.update() }
      scale *= amount;
      pos.x = at.x - (at.x - pos.x) * amount;
      pos.y = at.y - (at.y - pos.y) * amount;
      dirty = true;
    },
  };
  return API;
})();
view.setContext(ctx);

function drawCanvas() {
  if (view.isDirty()) { // has the view changed, then draw all
    ctx.setTransform(1, 0, 0, 1, 0, 0); // default transform for clear
    ctx.clearRect(0, 0, canvas.width, canvas.height);
    view.apply(); // set the 2D context transform to the view
    for (i = 0; i < objects.length; i++) {
      var obj = objects[i];
      ctx.fillStyle = obj.col;
      ctx.fillRect(obj.x, obj.y, obj.h, obj.h);
    }
  }

  requestAnimationFrame(drawCanvas);
}

function onmousewheel(event) {
  var e = window.event || event;
  var x = e.offsetX;
  var y = e.offsetY;
  const delta = e.type === "mousewheel" ? e.wheelDelta : -e.detail;
  if (delta > 0) { view.scaleAt({x, y}, 1.1) }
  else { view.scaleAt({x, y}, 1 / 1.1) }
  e.preventDefault();
}

#canvas {
  border: 2px solid;
}

<canvas id="canvas"></canvas>

javascript / canvas，地图样式点缩放

1 个答案:

在坐标

实施例