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我有一些词。所有词都在某个“对象”之王中。这个词可以在画布上移动，我需要得到所有交叉点的数组，就像在这个例子中一样，但不需要将文本转换为 SVG。paperjs.org/examples/path-intersections 谢谢。

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1 回答 1

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您可以通过比较两个词的 imageData 来检测两个词之间的所有交点。

无论单词与哪个像素相交，两个单词的 alpha 值都将大于零。

在此处输入图像描述

这是代码和小提琴：http: //jsfiddle.net/m1erickson/ecAvt/

<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>

<style>
    body{ background-color: ivory; padding:20px; }
    #wrapper{
        position:relative;
        width:400px;
        height:400px;
    }
    #canvasTop,#canvasBottom,#canvasDots{
        position:absolute; top:0px; left:0px;
        border:1px solid green;
        width:100%;
        height:100%;
    }
    #canvasTop{
        border:1px solid red;
    }
    #canvasDots{
        border:1px solid blue;
    }
</style>

<script>
$(function(){

    var canvas=document.getElementById("canvasTop");
    var ctx=canvas.getContext("2d");
    ctx.font="192px verdana";
    ctx.strokeText("No",40,220);

    var canvas2=document.getElementById("canvasBottom");
    var ctx2=canvas2.getContext("2d");
    ctx2.font="192px verdana";
    ctx2.strokeStyle="lightgray";
    ctx2.strokeText("Yes",40,300);

    var canvas3=document.getElementById("canvasDots");
    var ctx3=canvas3.getContext("2d");
    ctx3.fillStyle="blue";

    var canvasOffset=$("#canvasTop").offset();
    var offsetX=canvasOffset.left;
    var offsetY=canvasOffset.top;

    var startX;
    var startY;
    var isDown=false;



    var imageData2=ctx2.getImageData(0,0,canvas2.width,canvas2.height);
    var data2=imageData2.data;

    var dotsVisible=false;

    function showIntersections(){

        var imageData=ctx.getImageData(0,0,canvas.width,canvas.height);
        var data=imageData.data;

        ctx3.clearRect(0,0,canvas.width,canvas.height);
        for(var i=0;i<data.length;i+=4){
            if(data[i+3]>120 && data2[i+3]>120){
                var y=parseInt(i/canvas.width/4);
                var x=i/4-y*canvas.width;
                ctx3.beginPath();
                ctx3.arc(x,y,3,0,Math.PI*2);
                ctx3.closePath();
                ctx3.fill();
            }
        }
        dotsVisible=true;
    }

    function handleMouseDown(e){
      mouseX=parseInt(e.clientX-offsetX);
      mouseY=parseInt(e.clientY-offsetY);
      showIntersections();

    }

    function handleMouseUp(e){
      isDown=false;
    }

    function handleMouseOut(e){
      isDown=false;
    }

    function handleMouseMove(e){
      if(dotsVisible){ctx3.clearRect(0,0,canvas.width,canvas.height);}
      mouseX=parseInt(e.clientX-offsetX);
      mouseY=parseInt(e.clientY-offsetY);
      ctx.clearRect(0,0,canvas.width,canvas.height);
      ctx.strokeText("No",mouseX,mouseY+100);

    }

    $("#canvasDots").mousedown(function(e){handleMouseDown(e);});
    $("#canvasDots").mousemove(function(e){handleMouseMove(e);});
    $("#canvasDots").mouseup(function(e){handleMouseUp(e);});
    $("#canvasDots").mouseout(function(e){handleMouseOut(e);});

}); // end $(function(){});
</script>

</head>

<body>
    <h3>Move "NO" with mouse</h3>
    <h3>Click to find intersections</h3>
    <div id="wrapper">
        <canvas id="canvasBottom" width=400 height=400></canvas>
        <canvas id="canvasTop" width=400 height=400></canvas>
        <canvas id="canvasDots" width=400 height=400></canvas>
    </div>
</body>
</html>
于 2013-11-13T16:43:15.920 回答