也不是我原创，网上其实有很多。主要是我有个地方一直在多次调用我就给纠正一下仅此而已。
xxyy = [parseInt(bezier_curves(point, i).x), parseInt(bezier_curves(point, i).y)]
贝斯尔曲线方法会被调用2次，浪费手机CPU我给改了一下调用一次就好了。
另外我在某App上测试了这个方法是可以的，代码公布如下：
auto.waitFor(); let see_count =300;// rawInput('请输入滑动次数','1000'); app.launchApp('抖音极速版'); sleep(5000); console.show(); //开启日志（悬浮窗权限） for (var i = 1; i < see_count; i++) { toastLog("抖音极速版滑动" + i + '次'+"总计:"+ see_count + "次") sml_move(device.width / 2, device.height*0.8, device.width / 2, device.height*0.1, 300); let delayTime = random(8000, 12000); sleep(delayTime); } /** * 贝塞尔曲线 * @param {坐标点} ScreenPoint * @param {偏移量} Offset */ function bezier_curves(ScreenPoint, Offset) { cx = 3.0 * (ScreenPoint[1].x - ScreenPoint[0].x); bx = 3.0 * (ScreenPoint[2].x - ScreenPoint[1].x) - cx; ax = ScreenPoint[3].x - ScreenPoint[0].x - cx - bx; cy = 3.0 * (ScreenPoint[1].y - ScreenPoint[0].y); by = 3.0 * (ScreenPoint[2].y - ScreenPoint[1].y) - cy; ay = ScreenPoint[3].y - ScreenPoint[0].y - cy - by; tSquared =Offset * Offset; tCubed = tSquared * Offset; result = { "x": 0, "y": 0 }; result.x = (ax * tCubed) + (bx * tSquared) + (cx * Offset) + ScreenPoint[0].x; result.y = (ay * tCubed) + (by * tSquared) + (cy * Offset) + ScreenPoint[0].y; return result; }; /** * 曲线滑动 * @param {*} qx * @param {*} qy * @param {*} zx * @param {*} zy * @param {*} time */ function sml_move(qx, qy, zx, zy, time) { var xxy = [time]; var point = []; var dx0 = { "x": qx, "y": qy }; var dx1 = { "x": random(qx - 100, qx + 100), "y": random(qy, qy + 50) }; var dx2 = { "x": random(zx - 100, zx + 100), "y": random(zy, zy + 50), }; var dx3 = { "x": zx, "y": zy }; for (var i = 0; i < 4; i++) { eval("point.push(dx" + i + ")"); }; for (let i = 0; i < 1; i += 0.08) { let newPoint=bezier_curves(point, i); xxyy = [parseInt(newPoint.x), parseInt(newPoint.y)] xxy.push(xxyy); } gesture.apply(null, xxy); };
