为什么我无法让条件命令在 tikzpicture 中调用的宏中起作用？

Question

我想创建一个可以在 tikz 路径中间调用的宏，并且根据其中一个参数表现出不同的行为。

以下是一个示例文档：

\documentclass[a4paper]{amsart}

\usepackage{ifthen, tikz}

\newcommand{\nameandlabel}[2]{%
    node [midway, \ifthenelse{\equal{#1}{a}}{above}{below}] {#2}%
}

\begin{document}

\begin{tikzpicture}[xscale = 25mm]
    \node (a) at (0, 0) {$A$};
    \node (b) at (1, 0) {$B$};
    \node (c) at (2, 0) {$C$};
    \draw (a) -- (b) node [midway, above] {$f$};
    \draw (b) -- (c) \nameandlabel{a}{$g$};
\end{tikzpicture}

\end{document}

本文档无法编译。它导致出现此错误消息，这令人惊讶，因为\equal这是 ifthen 提供的宏。

! Undefined control sequence.
<argument> \equal 
                  {a}{a}
l.16     \draw (b) -- (c) \nameandlabel{a}{$g$}
                                               ;

我在 TeX-LaTeX Stack Exchange 上看到有人讨论说 ifthen 已经过时了，\IfStrEq可以使用 xstring 包中的宏来代替。所以我将示例更改为以下内容：

\documentclass[a4paper]{amsart}

\usepackage{xstring, tikz}

\newcommand{\nameandlabel}[2]{%
    node [midway, \IfStrEq{#1}{a}{above}{below}] {#2}%
}

\begin{document}

\begin{tikzpicture}[xscale = 25mm]
    \node (a) at (0, 0) {$A$};
    \node (b) at (1, 0) {$B$};
    \node (c) at (2, 0) {$C$};
    \draw (a) -- (b) node [midway, above] {$f$};
    \draw (b) -- (c) \nameandlabel{a}{$g$};
\end{tikzpicture}

\end{document}

但这会导致更难以理解的错误信息：

! Argument of \XC@definec@lor has an extra }.
<inserted text> 
                \par 
l.16     \draw (b) -- (c) \nameandlabel{a}{$g$}
                                               ;

为什么这仍然不起作用？

Answer 1

测试ifthen都不起作用xstring扩张它们都做出了内部定义。

如果您使用纯可扩展测试，它就会按您预期的方式工作。

\documentclass[a4paper]{amsart}

\usepackage{ifthen, tikz}

\newcommand{\nameandlabel}[2]{%
    node [midway, \ifx a#1above\else below\fi] {#2}%
}

\begin{document}

\begin{tikzpicture}[x = 25mm]
    \node (a) at (0, 0) {$A$};
    \node (b) at (1, 0) {$B$};
    \node (c) at (2, 0) {$C$};
    \draw (a) -- (b) node [midway, above] {$f$};
    \draw (b) -- (c) \nameandlabel{a}{$g$};
\end{tikzpicture}

\begin{tikzpicture}[x = 25mm]
    \node (a) at (0, 0) {$A$};
    \node (b) at (1, 0) {$B$};
    \node (c) at (2, 0) {$C$};
    \draw (a) -- (b) node [midway, above] {$f$};
    \draw (b) -- (c) \nameandlabel{b}{$g$};
\end{tikzpicture}

\end{document}

Answer 2

David Carlisle 的回答直接回答了你的问题。但我认为，在你的情况下，使用条件句更符合 Ti 的精神钾Z 使用.is choice密钥处理程序。下面是操作方法的示例。

\documentclass[varwidth]{standalone}
\usepackage{tikz}

\tikzset{
  ab/.is choice,
  ab/a/.style={above},
  ab/b/.style={below},
}
\newcommand{\nameandlabel}[2]{%
    node [midway,ab=#1] {#2}%
}

\begin{document}

  \begin{tikzpicture}[x = 25mm]
      \node (a) at (0, 0) {$A$};
      \node (b) at (1, 0) {$B$};
      \node (c) at (2, 0) {$C$};
      \draw (a) -- (b) node [midway, above] {$f$};
      \draw (b) -- (c) \nameandlabel{a}{$g$};
  \end{tikzpicture}

  \begin{tikzpicture}[x = 25mm]
      \node (a) at (0, 0) {$A$};
      \node (b) at (1, 0) {$B$};
      \node (c) at (2, 0) {$C$};
      \draw (a) -- (b) node [midway, above] {$f$};
      \draw (b) -- (c) \nameandlabel{b}{$g$};
  \end{tikzpicture}

\end{document}