4

我正在编写一个程序来绘制sine curve画布。
HTML:

<canvas id="mycanvas" width="1000" height="100">
    Your browser is not supported.
</canvas>

JavaScript:

var canvas = document.getElementById("mycanvas");
if (canvas.getContext) {
    var ctx = canvas.getContext("2d");
    ctx.lineWidth = 3;
    var x = 0,
        y = 0;
    var timeout = setInterval(function() {
        ctx.beginPath();
        ctx.moveTo(x, y);
        x += 1;
        y = 50 * Math.sin(0.1 * x) + 50;
        ctx.lineTo(x, y);
        ctx.stroke();
        if (x > 1000) {
            clearInterval(timeout);
        }
    }, 10);
}

这真的很好用:http: //jsfiddle.net/HhGnb/

但是,现在我只能为画布宽度提供 100 像素,因此只能看到曲线最左边的 100 像素。http://jsfiddle.net/veEyM/1/
我想存档这个效果:当曲线的右边点大于画布的宽度时,整个曲线可以向左移动,所以我可以看到最右边的点曲线,有点像曲线向左流动。我可以这样做吗?

4

1 回答 1

11

元素的基本思想之一<canvas>是计算机“忘记”绘图命令,只保存像素,就像位图一样。因此,要将所有内容向左移动,您需要清除画布并重新绘制所有内容。

还有一件事我想告诉你——你总是从 x = 0 和 y = 0 开始,但显然在 x = 0 时,y 也不一定等于 0。编辑:实现了这个。

无论如何,我最终得到了这段代码:http: //jsfiddle.net/veEyM/5/

var canvas = document.getElementById("mycanvas");
var points = {}; // Keep track of the points in an object with key = x, value = y
var counter = 0; // Keep track when the moving code should start

function f(x) {
    return 50 * Math.sin(0.1 * x) + 50;
}

if (canvas.getContext) {
    var ctx = canvas.getContext("2d");
    ctx.lineWidth = 3;
    var x = 0,
        y = f(0);
    var timeout = setInterval(function() {
        if(counter < 100) { // If it doesn't need to move, draw like you already do
            ctx.beginPath();
            ctx.moveTo(x, y);
            points[x] = y;
            x += 1;
            y = f(x);
            ctx.lineTo(x, y);
            ctx.stroke();
            if (x > 1000) {
                clearInterval(timeout);
            }
        } else { // The moving part...
            ctx.clearRect(0, 0, 100, 100); // Clear the canvas
            ctx.beginPath();
            points[x] = y;
            x += 1;
            y = f(x);
            for(var i = 0; i < 100; i++) {
                // Draw all lines through points, starting at x = i + ( counter - 100 )
                // to x = counter. Note that the x in the canvas is just i here, ranging
                // from 0 to 100
                ctx.lineTo(i, points[i + counter - 100]);
            }
            ctx.stroke();
        }
        counter++;
    }, 10);
}
于 2011-05-17T15:57:21.920 回答