邀請碼:872294657
也不是我原創,網上其實有很多。主要是我有個地方一直在多次調用我就給糾正一下僅此而已。
xxyy = [parseInt(bezier_curves(point, i).x), parseInt(bezier_curves(point, i).y)]
貝斯爾曲線方法會被調用2次,浪費手機CPU我給改了一下調用一次就好了。
另外我在某App上測試了這個方法是可以的,代碼公佈如下:
auto.waitFor();
let see_count =300;// rawInput('請輸入滑動次數','1000');
app.launchApp('抖音極速版');
sleep(5000);
console.show(); //開啓日誌(懸浮窗權限)
for (var i = 1; i < see_count; i++) {
toastLog("抖音極速版滑動" + i + '次'+"總計:"+ 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);
}
/**
* 貝塞爾曲線
* @param {座標點} 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;
};
/**
* 曲線滑動
* @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);
};