我尝试仅在循环的某些迭代中使用 circuitikz 中的连接点，但失败了。
\documentclass[border=6mm]{standalone}
\usepackage[siunitx, american]{circuitikz}
\begin{document}
\usetikzlibrary{calc}
\begin{tikzpicture}[
font=\sffamily,
every node/.style = {align=center}
]
\foreach[count=\i] \x in {0,3}
{
\ifnum\i=1
\tikzstyle{maybedot} = []
\else
\tikzstyle{maybedot} = [-*]
\fi
\draw (\x,0) to [R, l_={$R_\i$}, maybedot] (\x, 3) to [short, maybedot](\x,5);
}
\draw (-2,0) to [R, l_={$R_0$ \\ noloop}, -*] (-2, 3) to [short, -*](-2,5);
\end{tikzpicture}
\end{document}
R1 正确地没有点，但我希望 R2 像 R0 一样有点。我该如何实现它？（我的真实示例更复杂，这样做可以节省大量重复代码）
答案1
这里的问题是，您\tikzset（尽管使用了旧的、最好避免的语法）正在设置键/tikz/-*（箭头），正如您可能在错误中注意到的那样：
prova.tex|20 error| Package pgf Error: Unknown arrow tip kind '*'.
prova.tex|20 error| Package pgf Error: Unknown arrow tip kind '*'.
杆子的正确键是/tikz/circuitikz/-*，因此其有效：
\documentclass[border=6mm]{standalone}
\usepackage[siunitx, american, RPvoltages]{circuitikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[
font=\sffamily,
every node/.style = {align=center},
]
\foreach[count=\i] \x in {0,3}
{
\ifnum\i=1
\ctikzset{maybedot/.style={}}
\else
\ctikzset{maybedot/.style={-*}}
\fi
\draw (\x,0) to [R, l_={$R_\i$}, maybedot] (\x, 3)
to [short, maybedot](\x,5);
}
\draw (-2,0) to [R, l_={$R_0$ \\ noloop}, -*] (-2, 3) to [short, -*](-2,5);
\end{tikzpicture}
\end{document}
在循环中混合使用\if东西\foreach非常危险（尽管这里还有另一个问题）。我会\ifthenelse在这里使用几个宏来代替样式：
\documentclass[border=6mm]{standalone}
\usepackage{ifthen}
\usepackage[siunitx, american, RPvoltages]{circuitikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[
font=\sffamily,
every node/.style = {align=center},
]
\foreach[count=\i] \x in {0,3}
{
\ifthenelse{\i = 1}{\edef\maybedot{}}{\edef\maybedot{-*}}
\draw (\x,0) to [R, l_={$R_\i$}, \maybedot] (\x, 3)
to [short, \maybedot](\x,5);
}
\draw (-2,0) to [R, l_={$R_0$ \\ noloop}, -*] (-2, 3) to [short, -*](-2,5);
\end{tikzpicture}
\end{document}
如果你不想加载ifthen，测试
\ifnum\i=1\edef\maybedot{}\else\edef\maybedot{-*}\fi
也有效。
