在 tikz 中的矩形节点自定义形状的锚点处添加额外的节点？

Question 1

最终解决方案：我将圆圈移到边界内，创建了新的锚点并绘制了开关。所有这些 \pgfscope 都是必需的。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{calc}
\usetikzlibrary{arrows.meta}

\newlength{\circsize}
\newlength{\spdtright}
\newlength{\spdtup}
\newlength{\spdtleft}
\newlength{\spdtdown}
\newlength{\spdtlen}

\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}

\pgfkeys{/tikz/spdtcircsize/.initial = 0.08cm}
\pgfdeclareshape{spdt}{
  \inheritsavedanchors[from={rectangle}]
  %\inheritbackgroundpath[from={rectangle}]
  \inheritanchorborder[from={rectangle}]
  \foreach \x in {center,text,north east,north west,north,south,south east,south west,east,west}{
    \inheritanchor[from={rectangle}]{\x}
  }
\anchor{out}{
  \pgfextractx{\spdtright}{\northeast}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextracty{\spdtdown}{\southwest}%
  \pgfpoint{\spdtright}{0.5\spdtup+0.5\spdtdown}
 }
\anchor{off}{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextractx{\spdtleft}{\southwest}%
    \pgfextracty{\spdtdown}{\southwest}%
    \pgfpoint{\spdtleft}{\spdtdown+\circsize}%
}
\anchor{on}{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextractx{\spdtleft}{\southwest}%
   \pgfpoint{\spdtleft}{\spdtup-\circsize}%
}
\foregroundpath{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextractx{\spdtright}{\northeast}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextractx{\spdtleft}{\southwest}%
  \pgfextracty{\spdtdown}{\southwest}%
  % compute sin and cos for sloped line
  \pgfscope
    \spdtlen=0.5\spdtup-0.5\spdtdown-\circsize% length
    \pgfmathmultiply{\spdtlen}{\spdtlen}%
    \let\spdtcos=\pgfmathresult% macro
    \spdtlen=\spdtright-\spdtleft-2\circsize%
    \pgfmathparse{sqrt(\spdtlen*\spdtlen+\spdtcos)}
    \let\spdtsin=\pgfmathresult%
    \pgfmathdivide{\spdtlen}{\spdtsin}%
    \global\let\spdtcos=\pgfmathresult%
    \spdtlen=0.5\spdtup-0.5\spdtdown-\circsize%
    \pgfmathdivide{\spdtlen}{\spdtsin}%
    \global\let\spdtsin=\pgfmathresult%
  \endpgfscope
  % draw circles
  \pgfscope
    \pgfpathcircle{\pgfpoint{\spdtleft+\circsize}{\spdtdown+\circsize}}{\circsize}%
    \pgfpathcircle{\pgfpoint{\spdtleft+\circsize}{\spdtup-\circsize}}{\circsize}%
    \pgfpathcircle{\pgfpoint{\spdtright-\circsize}{0.5\spdtup+0.5\spdtdown}}{\circsize}%
    \pgfusepath{stroke}
  \endpgfscope
  % draw arrow
  \pgfscope
    \pgfsetarrowsend{Triangle[open]}%
    \pgfpathmoveto{\pgfpoint{\spdtright-\circsize-\spdtcos\circsize}{0.5\spdtup+0.5\spdtdown+\spdtsin\circsize}}%
    \pgfpathlineto{\pgfpoint{\spdtleft+\circsize+\spdtcos\circsize}{\spdtup-\circsize-\spdtsin\circsize}}%
    \pgfusepath{stroke}%
    \endpgfscope
  }
}
\begin{document}
\begin{tikzpicture}
\node[draw] (n1) at (1,0) {Testing};
\node[spdt,draw,minimum width=20pt,minimum height=20pt] (n2) [above=10pt of n1] {};
\draw (n2.on) -- +(-0.2,0)
      (n2.off) -- +(-0.2,0)
      (n2.out) -- +(0.2,0);
\node[spdt,draw,minimum width=20pt,minimum height=20pt] (n3) [below=10pt of n1] {X};
\end{tikzpicture}
\end{document}

Answer

最终解决方案：我将圆圈移到边界内，创建了新的锚点并绘制了开关。所有这些 \pgfscope 都是必需的。

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{calc}
\usetikzlibrary{arrows.meta}

\newlength{\circsize}
\newlength{\spdtright}
\newlength{\spdtup}
\newlength{\spdtleft}
\newlength{\spdtdown}
\newlength{\spdtlen}

\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}

\pgfkeys{/tikz/spdtcircsize/.initial = 0.08cm}
\pgfdeclareshape{spdt}{
  \inheritsavedanchors[from={rectangle}]
  %\inheritbackgroundpath[from={rectangle}]
  \inheritanchorborder[from={rectangle}]
  \foreach \x in {center,text,north east,north west,north,south,south east,south west,east,west}{
    \inheritanchor[from={rectangle}]{\x}
  }
\anchor{out}{
  \pgfextractx{\spdtright}{\northeast}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextracty{\spdtdown}{\southwest}%
  \pgfpoint{\spdtright}{0.5\spdtup+0.5\spdtdown}
 }
\anchor{off}{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextractx{\spdtleft}{\southwest}%
    \pgfextracty{\spdtdown}{\southwest}%
    \pgfpoint{\spdtleft}{\spdtdown+\circsize}%
}
\anchor{on}{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextractx{\spdtleft}{\southwest}%
   \pgfpoint{\spdtleft}{\spdtup-\circsize}%
}
\foregroundpath{
  \circsize=\pgfkeysvalueof{/tikz/spdtcircsize}%
  \pgfextractx{\spdtright}{\northeast}%
  \pgfextracty{\spdtup}{\northeast}%
  \pgfextractx{\spdtleft}{\southwest}%
  \pgfextracty{\spdtdown}{\southwest}%
  % compute sin and cos for sloped line
  \pgfscope
    \spdtlen=0.5\spdtup-0.5\spdtdown-\circsize% length
    \pgfmathmultiply{\spdtlen}{\spdtlen}%
    \let\spdtcos=\pgfmathresult% macro
    \spdtlen=\spdtright-\spdtleft-2\circsize%
    \pgfmathparse{sqrt(\spdtlen*\spdtlen+\spdtcos)}
    \let\spdtsin=\pgfmathresult%
    \pgfmathdivide{\spdtlen}{\spdtsin}%
    \global\let\spdtcos=\pgfmathresult%
    \spdtlen=0.5\spdtup-0.5\spdtdown-\circsize%
    \pgfmathdivide{\spdtlen}{\spdtsin}%
    \global\let\spdtsin=\pgfmathresult%
  \endpgfscope
  % draw circles
  \pgfscope
    \pgfpathcircle{\pgfpoint{\spdtleft+\circsize}{\spdtdown+\circsize}}{\circsize}%
    \pgfpathcircle{\pgfpoint{\spdtleft+\circsize}{\spdtup-\circsize}}{\circsize}%
    \pgfpathcircle{\pgfpoint{\spdtright-\circsize}{0.5\spdtup+0.5\spdtdown}}{\circsize}%
    \pgfusepath{stroke}
  \endpgfscope
  % draw arrow
  \pgfscope
    \pgfsetarrowsend{Triangle[open]}%
    \pgfpathmoveto{\pgfpoint{\spdtright-\circsize-\spdtcos\circsize}{0.5\spdtup+0.5\spdtdown+\spdtsin\circsize}}%
    \pgfpathlineto{\pgfpoint{\spdtleft+\circsize+\spdtcos\circsize}{\spdtup-\circsize-\spdtsin\circsize}}%
    \pgfusepath{stroke}%
    \endpgfscope
  }
}
\begin{document}
\begin{tikzpicture}
\node[draw] (n1) at (1,0) {Testing};
\node[spdt,draw,minimum width=20pt,minimum height=20pt] (n2) [above=10pt of n1] {};
\draw (n2.on) -- +(-0.2,0)
      (n2.off) -- +(-0.2,0)
      (n2.out) -- +(0.2,0);
\node[spdt,draw,minimum width=20pt,minimum height=20pt] (n3) [below=10pt of n1] {X};
\end{tikzpicture}
\end{document}

Question 2

好的，我这里遇到了几个问题。

首先，我同时绘制了圆路径和圆节点，以确保我理解正确，这样我就可以根据需要使用其中任何一种。事实证明，必须小心\pgfdeclareshape 中圆形路径与圆形节点的大小- in\pgfpathcircle有一个半径参数，但\pgfnodeviaminumum size有一个直径参数。当然，在这种情况下，我只需要节点，因为我想在外部使用它们。

其次，我尝试计算主节点的高度和宽度（\pgfpointdiff{\northeast}{\southwest}），因为我不知道如何访问它的锚点；结果发现有一个命令，\pgfpointanchor-并且\tikz@fig@name可以与它一起用来引用主节点。

第三，我不太了解 PGF 引擎——我的理解是，如果你调用类似的命令\pgfpointanchor，那么你就会在内部寄存器中获得结果\pgf@x，\pgf@y然后你可以将其分配给临时寄存器等\pgf@xa。\pgf@ya因此，我认为这可以获取锚点坐标并绘制圆圈：

{
\pgftransformreset
\pgfpointanchor{\tikz@fig@name}{south west}
\xdef\mycoordinate{\noexpand\pgfpoint{\the\pgf@x}{\the\pgf@y}}
}
\pgfpathcircle{\mycoordinate}{\circsize}

... 但出于某种原因，它没有 -\mycoordinate结果错了。但是 - 这是奇怪的部分 - 如果我把整个放在\pgfpointanchor{\tikz@fig@name}{south west}而\pgfpathcircle不是中\mycoordinate，那么定位就没问题了？！

因此，考虑到所有这些因素，我得出了下面发布的 MWE，其结果如下：

测试11.png

红色节点(n2)仅用于测试样式和位置；否则(n3)其右侧就是如何使用它。

但还有一个问题 - 我从某处注意到自定义形状（通过\pgfdeclareshape）应该只使用 PGF 基元，而不是 TikZ 绘图命令 - 但我必须使用\draw，这样我才能轻松指定箭头尖端，并且该线连接圆形节点的边界 - 而不是它们的中心。如果有人知道可以使用 PGF 命令实现相同解决方案的解决方案，请发布答案。

除此之外，看来这个 MWE 为我做了这件事：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{calc}
\usetikzlibrary{arrows}

\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}

\makeatletter
\pgfkeys{/tikz/.cd,
  mycircsize/.store in=\circsize,
  mycircsize=0.2cm
}
\pgfdeclareshape{spdt}{
  \inheritsavedanchors[from={rectangle}]
  \inheritbackgroundpath[from={rectangle}]
  \inheritanchorborder[from={rectangle}]
  \foreach \x in {center,north,north east,north west,south,south east,south west,east,west}{
    \inheritanchor[from={rectangle}]{\x}
  }
%   \backgroundpath{
  \foregroundpath{
    \pgfkeys{/pgf/minimum size = \circsize}
    \pgfset{inner sep=0pt,outer sep=0pt}
    % draw connector nodes
    \southwest
    \pgfpathcircle{\southwest}{0.5*\circsize}% just a test
    {
    \pgftransformshift{\southwest}
    \pgfnode{circle}{center}{}{\tikz@fig@name-c1}{\pgfusepath{draw}}
    %\pgfusepath{draw}
    }
    {
    \pgftransformshift{\pgfpointanchor{\tikz@fig@name}{north west}}
    \pgfnode{circle}{center}{}{\tikz@fig@name-c2}{\pgfusepath{draw}}
    }
    {
    \pgftransformshift{\pgfpointanchor{\tikz@fig@name}{east}}
    \pgfnode{circle}{center}{}{\tikz@fig@name-c3}{\pgfusepath{draw}}
    }
    % draw switch arrow line
    % with PGF primitives, line goes from center of nodes, not borders
    %\pgfpathmoveto{\pgfpointanchor{\tikz@fig@name}{east}}
    %\pgfpathlineto{\pgfpointanchor{\tikz@fig@name}{north west}}
    % so using a TikZ draw command:
    \draw[-open triangle 60] (\tikz@fig@name-c3) -- (\tikz@fig@name-c2);

  }
}
\makeatother

\begin{document}
\begin{tikzpicture}
\node[draw] (n1) at (1,0) {Testing};
\node[spdt,draw=red,line width=2pt,minimum width=20pt,minimum height=20pt] (n2) [above=10pt of n1] {X};
\node[spdt,minimum width=20pt,minimum height=20pt] (n3) [right=40pt of n2] {};

\draw (n1.west) -- ++(-15pt,0) |- (n2-c1);
\draw (n1.east) -- ++(15pt,0) |- (n2-c3);

\draw (n1.east) -- (n3-c3|-n1.east) -- ++(15pt,0) |- (n3-c3);

\end{tikzpicture}
\end{document}

Answer