条件部分标题 [ifthenelse] 不起作用

Question

我有以下代码：

\documentclass{article}

\usepackage{ifthen}
\usepackage{amsmath,amsfonts,amsthm,amssymb}

\newcommand{\probName}{}%
\newcounter{probCounter}%
\setcounter{probCounter}{1}
\newcounter{partCounter}%

\makeatletter
\newcommand{\howmany}[2][subsection]{%
  \begingroup
  \@namedef{the#1}{\arabic{#1}}%
  \addtocounter{#1}{\m@ne}%
  \refstepcounter{#1}%
  \label{#2}%
  \endgroup}
\makeatother

\newenvironment{prob}[1][]%
  {\setcounter{partCounter}{0}
   \renewcommand{\probName}{#1}%
   \newcommand{\abc}{\ref{noPart:\arabic{probCounter}}}
   \section{{Problem \arabic{probCounter}}{: \probName} (contains
   \abc\
   %\ifthenelse{\abc=0}{1}{\abc}\  %%%%%%%%%%%%%%%%%%%%%% does not work 
    parts) }% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% wanna conditional parts/part

  }{\howmany{noPart:\arabic{probCounter}}
   \stepcounter{probCounter}%
  }

\newcommand{\partName}{}%
\renewenvironment{part}[1][]%
  {\stepcounter{partCounter}%
   \renewcommand{\partName}{#1}%
   \subsection{{Part \Alph{partCounter}}{: \partName}}
  }{}

\newenvironment{ques}[1][]{
    \ifthenelse{
        \equal{#1}{}
    }{}{
        \ifthenelse{
            \equal{#1}{1}
        }{
            \textbf{(#1 point)}
        }{
            \textbf{(#1 points)}
        }}
    }{}

\begin{document}

\begin{prob}[Prob 1]
\begin{part}
\end{part}

\begin{part}
\end{part}

\begin{part}
\end{part}
\end{prob}


\begin{prob}[Prob 2] %%%%%%%%%%% force to show => (contains 1 part)
\end{prob}

\begin{prob}[Prob 3] %%%%%%%%%%% correct the grammar => (contains 1 part)
\begin{part}
\end{part}
\end{prob}

\begin{prob}[Prob 4]
\begin{part}
\end{part}
\begin{part}
\end{part}
\begin{part}
\end{part}
\begin{part}
\end{part}
\end{prob}

\end{document}

现在我想要的是：

...

问题 2（包含1部分）%% 强制为 1 部分

问题 3（包含1部分）%% 从措辞中删除“s”（部分“s”）

...

我首先想到的是条件 ifthenelse 函数，它也包含在我的代码中，但它没有按预期工作。

Answer 1

问题是这\ref{...}不是一个数字，而是一组指令印刷一个号码。

如果引用尚未建立，则可以使用refcount提供扩展为 0 的包，否则扩展为引用本身。\getrefnumber

我做了一些其他小改动（特别是保护了几个行尾）。对于部分/部分，语法\ifthenelse很麻烦，我更喜欢直接使用\ifnum。

\documentclass{article}

\usepackage{ifthen,refcount}
\usepackage{amsmath,amsfonts,amsthm,amssymb}

\newcommand{\probName}{}%
\newcounter{probCounter}%
\setcounter{probCounter}{1}
\newcounter{partCounter}%

\makeatletter
\newcommand{\howmany}[2][subsection]{%
  \begingroup
  \@namedef{the#1}{\arabic{#1}}%
  \addtocounter{#1}{\m@ne}%
  \refstepcounter{#1}%
  \label{#2}%
  \endgroup}
\makeatother

\newenvironment{prob}[1][]%
  {\setcounter{partCounter}{0}
   \renewcommand{\probName}{#1}%
   \edef\numberofparts{\getrefnumber{noPart:\arabic{probCounter}}}%
   \newcommand{\abc}{\ref{noPart:\arabic{probCounter}}}%
   \section{Problem \arabic{probCounter}: \probName} (contains
   \ifnum\numberofparts=0
    1 part%
   \else
     \ifnum\numberofparts=1
       1 part%
     \else
       \numberofparts\ parts%
     \fi
   \fi)%
  }{\howmany{noPart:\arabic{probCounter}}%
   \stepcounter{probCounter}%
  }

\newcommand{\partName}{}%
\renewenvironment{part}[1][]%
  {\stepcounter{partCounter}%
   \renewcommand{\partName}{#1}%
   \subsection{{Part \Alph{partCounter}}{: \partName}}%
  }{}

\newenvironment{ques}[1][]{%
    \ifthenelse{%
        \equal{#1}{}
    }{}{%
        \ifthenelse{%
            \equal{#1}{1}%
        }{%
            \textbf{(#1 point)}%
        }{%
            \textbf{(#1 points)}%
        }}%
    }{}

\begin{document}

\begin{prob}[Prob 1]
\begin{part}
\end{part}

\begin{part}
\end{part}

\begin{part}
\end{part}
\end{prob}


\begin{prob}[Prob 2] %%%%%%%%%%% force to show => (contains 1 part)
\end{prob}

\begin{prob}[Prob 3] %%%%%%%%%%% correct the grammar => (contains 1 part)
\begin{part}
\end{part}
\end{prob}

\begin{prob}[Prob 4]
\begin{part}
\end{part}
\begin{part}
\end{part}
\begin{part}
\end{part}
\begin{part}
\end{part}
\end{prob}

\end{document}