Question

我正在用libGdx做第一场比赛。我必须做像this video

这样的线条绘制动画

这是我的绘制代码

/*-----------------------------------------------------------------------------------*/
/*	RETINA.JS
/*-----------------------------------------------------------------------------------*/
(function () {

    var regex = /(media\/cache\/filter_[A-Z]+)/i //Added this
    function t(e) {
        this.path = e;
        var t = this.path.split("."),
            n = t.slice(0, t.length - 1).join("."),
            r = t[t.length - 1];
        this.at_2x_path = (n + '.' + r).replace(regex, '$1_retina') //Changed that
    }
    function n(e) {
        this.el = e, this.path = new t(this.el.getAttribute("src"));
        var n = this;
        this.path.check_2x_variant(function (e) {
            e && n.swap()
        })
    }
    var e = typeof exports == "undefined" ? window : exports;
    e.RetinaImagePath = t, t.confirmed_paths = [], t.prototype.is_external = function () {
        return !!this.path.match(/^https?\:/i) && !this.path.match("//" + document.domain)
    }, t.prototype.check_2x_variant = function (e) {
        var n, r = this;
        if (this.is_external()) return e(!1);
        if (this.at_2x_path in t.confirmed_paths) return e(!0);
        n = new XMLHttpRequest, n.open("HEAD", this.at_2x_path), n.onreadystatechange = function () {
            return n.readyState != 4 ? e(!1) : n.status >= 200 && n.status <= 399 ? (t.confirmed_paths.push(r.at_2x_path), e(!0)) : e(!1)
        }, n.send()
    }, e.RetinaImage = n, n.prototype.swap = function (e) {
        function n() {
            t.el.complete ? (t.el.setAttribute("width", t.el.offsetWidth), t.el.setAttribute("height", t.el.offsetHeight), t.el.setAttribute("src", e)) : setTimeout(n, 5)
        }
        typeof e == "undefined" && (e = this.path.at_2x_path);
        var t = this;
        n()
    }, e.devicePixelRatio > 1 && (window.onload = function () {
        var e = document.getElementsByTagName("img"),
            t = [],
            r, i;
        for (r = 0; r < e.length; r++) i = e[r], t.push(new n(i))
    })
})();

我可以画2点之间的直线，但需要绘制动画。

任何帮助？

Answer 1

根据您的代码，您只需要使坐标值动态化：

Vector2 startPt = null;
Vector2 endPt = null;
Vector2 movingPt = null;

void myInit()
{
  startPt = new Vector2(10, 10);
  endPt = new Vector2(100, 100);
  movingPt = new Vector2(startPt.x, startPt.y);
}

void update( float delta )
{
  // add your animation here, below is an example
  float PixelsPerSecond = 100f;
  float dt = delta * PixelsPerSecond;
  if ( movingPt.x < endPt.x ) movingPt.x += dt;
  else movingPt.x = endPt.x;
  if ( movingPt.y < endPt.y ) movingPt.y += dt;
  else movingPt.y = endPt.y;
}

void draw()
{

  // add code here (or in a base class) to clear the screen (once) each loop

  ShapeRenderer shapeRenderer=new ShapeRenderer();
  shapeRenderer.setProjectionMatrix(cam.combined);
  shapeRenderer.begin(ShapeRenderer.ShapeType.Line);
  shapeRenderer.setColor(Color.BLACK);
  shapeRenderer.line(startPt, movingPt);
  shapeRenderer.end();
}

在libGDX中绘制线条动画

1 个答案: