javascript - 在 HTML5 Canvas 中绘制任意形状

Question

我正在尝试使用 HTML5 canvas 元素绘制图形。该图由顶点和连接这些顶点的边组成。我用小圆圈表示顶点，用连​​接圆圈的线表示边。

在我的项目中，我想在按下“下一步”按钮时更改顶点的位置。下面显示的是页面首次加载时 Canvas 的状态:

单击“下一步”按钮时，将根据顶点的新坐标重新绘制图形。但是，渲染似乎有错误，如下所示:

奇怪的是，我没有在任何地方使用 fill() 函数，所以我无法理解为什么要填充部分。以下是用于绘制图形的 draw() 函数的代码:

function draw(position)
{
    var canvas_elem = document.getElementById('drawing');

    // Check the element is in the DOM and the browser supports canvas
    if(canvas_elem.getContext)
    {
        // Initaliase a 2-dimensional drawing context
        var canvas = canvas_elem.getContext('2d');

        // Clear canvas
        canvas.clearRect(0, 0, canvas_elem.width, canvas_elem.height);

        var data = data_array[num_vertices+1+position].split(",");

        // Draw all vertices
        for( var i=0; i<num_vertices; i++)
        {
            canvas.arc(data[2*i], data[2*i+1], radius, 0, 2 * Math.PI, false);
            canvas.fillStyle = 'black';
            canvas.fill();
        }

        // Connect vertices by lines according to adjacency matrix
        for(var i=0;i<num_vertices; i++)
        {
            for(var j=i; j<num_vertices; j++)
            {
                if(adjacency[i][j]==1)
                {
                    canvas.beginPath();
                    canvas.moveTo(data[2*i],data[2*i+1]);
                    canvas.lineTo(data[2*j],data[2*j+1]);
                    canvas.stroke();
                }
            }
        }
    }
    else
    {
        alert("Please use an HTML5 compatible browser.");
    }
}

关于这里可能发生的事情的任何线索？

---

取以下这些变量的值进行调试:

data_array=[
"10",
"0,1,1,0,0,0,0,0,0,0",
"1,0,0,1,0,0,0,0,0,1",
"1,0,0,0,1,0,0,0,0,1",
"0,1,0,0,0,1,0,0,1,1",
"0,0,1,0,0,0,1,0,1,1",
"0,0,0,1,0,0,0,1,1,0",
"0,0,0,0,1,0,0,1,1,0",
"0,0,0,0,0,1,1,0,0,0",
"0,0,0,1,1,1,1,0,0,0",
"0,1,1,1,1,0,0,0,0,0",
"10,2,35.4,23.3,88.9,100,210,350,70,500,412,208,336,112,45,32,89,92,102,23",
"30,4,19.4,35.3,80.9,90,230,310,120,440,400,220,330,105,40,29,80,89,90,18"
]

num_vertices=10

Answer 1

好吧，一个愚蠢的错误。我忘记在绘制每条线后调用 closePath()。