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如何使用边界框的高度创建自定义节点？

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如何使用边界框的高度创建自定义节点？

我一直在尝试自定义圆弧，但似乎不知道如何自动设置圆弧半径。目标是创建一个节点，该节点限制两个半圆并根据需要变宽

末尾有所需图片的链接

https://i.stack.imgur.com/E4qag.jpg

\usepackage{xparse}
\NewDocumentCommand{\mynode}{%
O{}
m
m
m
O{}
}{
{
\node [#1] (#2)  at #3 {#4};

\def\ra{.3cm}

\draw  [#5]  (#2.north west) arc
    [
        start angle=90,
        end angle=270,
        x radius=\ra,
        y radius=\ra
    ] ;
    

\draw [#5] (#2.north east)to [in=0,out=180] (#2.north west);
\draw [#5] (#2.south east)to [in=0,out=180] (#2.south west);

\draw  [#5]  (#2.south east) arc
    [
        start angle=270,
        end angle=450,
        x radius=\ra,
        y radius=\ra
    ] ;

 
}
}


[![This is what I want to make][2]][2]


  [1]: https://i.stack.imgur.com/E4qag.jpg
  [2]: https://i.stack.imgur.com/iuZgn.jpg

答案1

这使用\pgfextracty等来获取节点的高度。我还将所有绘图合并到一条路径中。

\documentclass{standalone}
\usepackage{tikz}
\usepackage{xparse}
\newlength{\mydiameter}
\NewDocumentCommand{\mynode}{% #1 = node options (optional), #2 = node name, #3 = coordinates, #4 = text, #5 = draw options (optional)
O{}
m
m
m
O{}
}{%
{%
\node [#1] (#2)  at #3 {#4};

\pgfextracty{\mydiameter}{\pgfpointdiff{\pgfpointanchor{#2}{south}}{\pgfpointanchor{#2}{north}}}%

\draw  [#5]  (#2.north west) arc
    [
        start angle=90,
        end angle=270,
        radius={0.5\mydiameter}
    ] -- (#2.south east) arc
    [
        start angle=-90,
        end angle=90,
        radius={0.5\mydiameter}
    ] -- cycle;
}%
}

\begin{document}
\begin{tikzpicture}
\mynode{A}{(0,0)}{Test}
\end{tikzpicture}
\end{document}

演示

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