Path svg将形状放置在路径上
我正在用SVG创建一个游戏,并要求沿一条名为道路的路径放置方形瓷砖。Path svg将形状放置在路径上,path,svg,shape,css-position,Path,Svg,Shape,Css Position,我正在用SVG创建一个游戏,并要求沿一条名为道路的路径放置方形瓷砖。 路径可以弯曲或以一定角度放置,我想相对于路径旋转方形瓷砖 我当前的结构是这样的:听起来你应该考虑使用标记来标记瓷砖,然后分割路径,这样你就有了顶点,你希望矩形显示为标记中点。你可以通过标记免费获得方向。拉斐尔提供了路径通过路径的任意点的角度。getPointAtLenght(x)。SVG有一个本机实现,但它不提供角度。你可以看看拉斐尔的消息来源,看看发生了什么。以下是相关位: <defs> <rec
路径可以弯曲或以一定角度放置,我想相对于路径旋转方形瓷砖
我当前的结构是这样的:听起来你应该考虑使用标记来标记瓷砖,然后分割路径,这样你就有了顶点,你希望矩形显示为标记中点。你可以通过标记免费获得方向。拉斐尔提供了路径通过路径的任意点的角度。getPointAtLenght(x)。SVG有一个本机实现,但它不提供角度。你可以看看拉斐尔的消息来源,看看发生了什么。以下是相关位:
<defs>
<rect id="tile" width="200" height="200" stroke-width="5" stroke="red"/>
</defs>
getLengthFactory = function (istotal, subpath) {
return function (path, length, onlystart) {
path = path2curve(path);
var x, y, p, l, sp = "", subpaths = {}, point,
len = 0;
for (var i = 0, ii = path.length; i < ii; i++) {
p = path[i];
if (p[0] == "M") {
x = +p[1];
y = +p[2];
} else {
l = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6]);
if (len + l > length) {
if (subpath && !subpaths.start) {
point = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6], length - len);
sp += ["C" + point.start.x, point.start.y, point.m.x, point.m.y, point.x, point.y];
if (onlystart) {return sp;}
subpaths.start = sp;
sp = ["M" + point.x, point.y + "C" + point.n.x, point.n.y, point.end.x, point.end.y, p[5], p[6]].join();
len += l;
x = +p[5];
y = +p[6];
continue;
}
if (!istotal && !subpath) {
point = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6], length - len);
return {x: point.x, y: point.y, alpha: point.alpha};
}
}
len += l;
x = +p[5];
y = +p[6];
}
sp += p.shift() + p;
}
subpaths.end = sp;
point = istotal ? len : subpath ? subpaths : R.findDotsAtSegment(x, y, p[0], p[1], p[2], p[3], p[4], p[5], 1);
point.alpha && (point = {x: point.x, y: point.y, alpha: point.alpha});
return point;
};
getLengthFactory=函数(istotal,子路径){
返回函数(路径、长度、仅开始){
路径=路径2曲线(路径);
变量x,y,p,l,sp=“”,子路径={},点,
len=0;
对于(变量i=0,ii=path.length;i长度){
if(子路径&!子路径开始){
点=getPointAtSegmentLength(x,y,p[1],p[2],p[3],p[4],p[5],p[6],长度-len);
sp+=[“C”+点.start.x,点.start.y,点.m.x,点.m.y,点.x,点.y];
if(onlystart){return sp;}
子路径开始=sp;
sp=[“M”+点.x,点.y+“C”+点.n.x,点.n.y,点.end.x,点.end.y,p[5],p[6]].join();
len+=l;
x=+p[5];
y=+p[6];
继续;
}
如果(!istotal&&!子路径){
点=getPointAtSegmentLength(x,y,p[1],p[2],p[3],p[4],p[5],p[6],长度-len);
返回{x:point.x,y:point.y,alpha:point.alpha};
}
}
len+=l;
x=+p[5];
y=+p[6];
}
sp+=p.shift()+p;
}
子路径结束=sp;
point=istotal?len:子路径?子路径:R.findDotsAtSegment(x,y,p[0],p[1],p[2],p[3],p[4],p[5],1);
point.alpha&(point={x:point.x,y:point.y,alpha:point.alpha});
返回点;
};
其中point.alpha是旋转角度。但据我所知,您无法将事件或侦听器链接到标记,因为它们实际上不是DOM/结构的一部分,而是更多的东西的表示端。这是我上次尝试SVGy时记得的。您的问题没有提到这一要求。
id = #
vector2D = {
x1 = #
y1 = #
x2 = #
y2 = #
}
tiles = # // number of tiles in road
getLengthFactory = function (istotal, subpath) {
return function (path, length, onlystart) {
path = path2curve(path);
var x, y, p, l, sp = "", subpaths = {}, point,
len = 0;
for (var i = 0, ii = path.length; i < ii; i++) {
p = path[i];
if (p[0] == "M") {
x = +p[1];
y = +p[2];
} else {
l = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6]);
if (len + l > length) {
if (subpath && !subpaths.start) {
point = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6], length - len);
sp += ["C" + point.start.x, point.start.y, point.m.x, point.m.y, point.x, point.y];
if (onlystart) {return sp;}
subpaths.start = sp;
sp = ["M" + point.x, point.y + "C" + point.n.x, point.n.y, point.end.x, point.end.y, p[5], p[6]].join();
len += l;
x = +p[5];
y = +p[6];
continue;
}
if (!istotal && !subpath) {
point = getPointAtSegmentLength(x, y, p[1], p[2], p[3], p[4], p[5], p[6], length - len);
return {x: point.x, y: point.y, alpha: point.alpha};
}
}
len += l;
x = +p[5];
y = +p[6];
}
sp += p.shift() + p;
}
subpaths.end = sp;
point = istotal ? len : subpath ? subpaths : R.findDotsAtSegment(x, y, p[0], p[1], p[2], p[3], p[4], p[5], 1);
point.alpha && (point = {x: point.x, y: point.y, alpha: point.alpha});
return point;
};