介绍 function getOperatedCurves (sourceCurs:曲线[],targetCus:,曲线[])
{才能
,,,let 来源:,Polyline |, Circle =, (sourceCurs [0], instanceof 圆),?,sourceCurs [0], as Circle :, new 折线().Combine (sourceCurs) [0];
,,,let 目标:,Polyline |, Circle =, (targetCus [0], instanceof 圆),?,targetCus [0], as Circle :, new 折线().Combine (targetCus) [0];
,,,试一试
,,,{
,,,,,if (! source.IsClose | |, ! target.IsClose), throw new 错误(“不是封闭曲线“);
,,,}
,,,catch (err)
,,,{
,,,,,console.log(错);
,,,}
,,,let interPts =, source.IntersectWith(目标,IntersectOption.OnBothOperands);
,,,let sourceContainerTarget =, isTargetCurInSourceCur(源,目标);
,,,let targetContainerSource =, isTargetCurInSourceCur(目标,,源);
,,,let isContainer =, sourceContainerTarget | |, targetContainerSource;
,,,let intersectionList:,曲线[],=,[];,//交集
,,,let unionList:,曲线[],=,[];,//并集
,,,let 子表:,曲线[],=,[];,//补集
,,/*
,,,*两封闭区域有交点并且不是包含关系,则通过交点把区域分割
,,*/,,,if (interPts.length ,,, ! isContainer)
,,,{
,,,,,let pars1 =, interPts.map (p =祝辞,source.GetParamAtPoint (p)) .sort ((a, b),=祝辞,a 作用;b);
,,,,,let pars2 =, interPts.map (p =祝辞,target.GetParamAtPoint (p)) .sort ((a, b),=祝辞,a 作用;b);
,,,,,let cus1:, Array,=, source.GetSplitCurves (pars1);
,,,,,cus1.forEach (pl =比;
,,,,,{
,,,,,,,if (isTargetCurInSourceCur(目标,,pl))
,,,,,,,{
,,,,,,,,,intersectionList.push (pl);
,,,,,,,}
,,,,,,,
,,,,,,,{
,,,,,,,,,subList.push (pl);
,,,,,,,,,unionList.push (pl);
,,,,,,,}
,,,,,})
,,,,,let cus2:, Array,=, target.GetSplitCurves (pars2);
,,,,,cus2.forEach (pl =比;
,,,,,{
,,,,,,,if (isTargetCurInSourceCur(来源,,pl))
,,,,,,,{
,,,,,,,,,intersectionList.push (pl);
,,,,,,,,,subList.push (pl);
,,,,,,,}
,,,,,,,
,,,,,,,{
,,,,,,,,,unionList.push (pl);
,,,,,,,}
,,,,,})
,,,}
,,,
,,,{
,,,,,if (isContainer)
,,,,,{
,,,,,,,if (sourceContainerTarget)
,,,,,,,{
,,,,,,,,,intersectionList.push(目标);
,,,,,,,,,subList.push(源,目标);
,,,,,,,,,unionList.push(源);
,,,,,,,}
,,,,,,,
,,,,,,,{
,,,,,,,,,unionList.push(目标);
,,,,,,,,,intersectionList.push(源);
,,,,,,,}
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这篇文章将为大家详细讲解有关JavaScript如何实现封闭区域布尔运算,小编觉得挺实用的,因此分享给大家做个参考,希望大家阅读完这篇文章后可以有所收获。
代码如下:
