Question

是否有一种已知方法可以找到SVG路径与自身的交叉？以＆符号为例，＆amp; ，它是一条在两点相交的线。

我遇到过一个intersection library，但似乎有两种形状相交，而不是一种相交的形状。

我对d3很熟悉所以任何基于javascript的答案都会很棒，但也很高兴听到可以使用的数学方法。

由于

Answer 1

你是对的 - 凯文林赛（@thelonious）的图书馆似乎在这里工作。

您告诉库查找两个相同形状的实例之间的交叉点，去掉Vector2D个对象，剩下的是两组相同的交叉点（Point2D类型的对象库）。

这是主要部分：

var pathEl = path.node();
var intersections = [];

// Kevin Lindsey's library
var shape1 = new Path(pathEl);
var overlays = Intersection.intersectShapes(shape1, shape1);

for (i in overlays.points) {
    if (overlays.points[i].getName() == "Point2D") {
       intersections.push(overlays.points[i]);   
    }
}

以下完整的嵌入式示例：

＆＃13;

var ampersand = "M 72.184621,99.39089 C 68.398038,95.61996 48.405425,73.700329 50.716835,49.985704 C 52.990823,26.655017 76.884556,12.578576 98.97427,11.448114 C 119.34404,10.405671 142.83345,16.156173 152.28457,35.843184 C 162.40413,56.922579 150.99532,81.842705 134.24772,94.48352 C 128.42088,98.881524 127.0609,99.082849 118.46055,102.84216 C 106.06795,108.25911 93.590914,113.54803 80.869078,118.1294 C 54.582831,127.59557 34.539139,149.03858 35.259701,178.23878 C 35.916374,204.84994 59.631137,225.67546 85.210802,229.70364 C 112.43115,233.99018 134.41358,229.54707 153.1347,208.67628 C 161.17912,199.70814 177.58763,184.99294 185.76751,176.14503 C 200.25035,160.47941 207.7442,147.82465 213.06419,126.69158 C 216.66826,112.37483 192.54569,115.67347 196.78314,103.62942 C 222.23036,100.69638 247.81229,99.84462 273.34564,97.96536 C 277.34887,109.81154 263.97786,106.33066 246.61613,121.01207 C 227.0104,137.59107 217.88679,151.73768 201.01195,170.86236 L 189.11358,184.34708 C 181.10521,193.42317 167.95634,207.85044 159.78314,216.77784 C 142.32024,235.85217 126.21297,247.41796 100.27427,252.39343 C 72.543606,257.71262 39.651129,254.69839 20.122962,231.94973 C -0.62641014,207.77846 -4.0848351,167.90434 18.462826,143.66847 C 33.171306,127.85873 41.031184,120.17885 60.724466,112.09432 C 76.147466,105.76283 100.05575,99.431353 112.29677,94.160526 C 139.69178,82.364582 140.40896,53.478721 127.50818,32.380115 C 116.44184,14.281646 83.908653,15.752833 77.904904,37.000557 C 72.689417,55.458561 80.089538,67.982449 91.37226,80.93907 L 187.58994,191.43156 C 199.42503,205.63979 217.24414,228.88851 237.39579,232.51125 C 250.72342,234.90721 267.9319,228.93995 277.27793,220.4821 C 282.25334,229.49138 275.03265,236.84049 269.43939,242.49659 C 251.14471,260.99681 219.58458,257.23653 199.30993,242.48439 C 187.00911,233.53413 178.95611,227.18492 167.95716,215.15746 L 72.184621,99.39089 z ";

var svg = d3.select("body").append("svg")
    .attr("width", 300)
    .attr("height", 300);

var path = svg.append("svg:path")
            .attr("d",ampersand)
            .style("stroke-width", 2)
            .style("stroke", "steelblue")
            .style("fill", "none");

// Taken from https://stackoverflow.com/questions/332422/how-do-i-get-the-name-of-an-objects-type-in-javascript
Object.prototype.getName = function() { 
   var funcNameRegex = /function (.{1,})\(/;
   var results = (funcNameRegex).exec((this).constructor.toString());
   return (results && results.length > 1) ? results[1] : "";
};

// Taken from https://stackoverflow.com/questions/9229645/remove-duplicates-from-javascript-array
function uniq(a) {
    var prims = {"boolean":{}, "number":{}, "string":{}}, objs = [];

    return a.filter(function(item) {
        var type = typeof item;
        if(type in prims)
            return prims[type].hasOwnProperty(item) ? false : (prims[type][item] = true);
        else
            return objs.indexOf(item) >= 0 ? false : objs.push(item);
    });
}

var pathEl = path.node();
var intersections = [];

// Kevin Lindsey's library
var shape1 = new Path(pathEl);
var overlays = Intersection.intersectShapes(shape1, shape1);

for (i in overlays.points) {
   if (overlays.points[i].constructor.name == "Point2D") {
      intersections.push(overlays.points[i]);   
   }
}

// The path will record 2 points for each intersection, so deduping is necessary
var deduped_intersections = uniq(intersections);

var circles = svg.selectAll("circle")
                 .data(deduped_intersections)
                .enter()
                 .append("circle");

var circleAttributes = circles
                       .attr("cx", function (d) { return d.x; })
                       .attr("cy", function (d) { return d.y; })
                       .attr("r", "3")
                       .style("fill", "red");

＆＃13;

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
<script src="http://www.kevlindev.com/gui/2D.js"></script>

＆＃13;

找到SVG路径的交集

1 个答案: