
我試圖僅在循環的某些迭代中使用 Circuitikz 中的連接點,但它失敗了。
\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 正確地沒有點,但我希望 R2 像 R0 一樣有點。我怎樣才能讓它發揮作用? (我的真實例子更複雜,做這種事情會節省我很多重複的程式碼)
答案1
這裡的問題是你的\tikzset
(儘管使用舊的、最好避免的語法)正在設定鍵/tikz/-*
(箭頭),正如你可能在錯誤中註意到的那樣:
prova.tex|20 error| Package pgf Error: Unknown arrow tip kind '*'.
prova.tex|20 error| Package pgf Error: Unknown arrow tip kind '*'.
極點的正確鍵是/tikz/circuitikz/-*
,所以這有效:
\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}
在循環中混合\if
事物\foreach
是相當危險的(儘管問題是這裡的另一個問題)。我會\ifthenelse
在這裡使用幾個巨集而不是樣式:
\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}
如果不想加載ifthen
,則測試
\ifnum\i=1\edef\maybedot{}\else\edef\maybedot{-*}\fi
也有效。