如何从列表中选择使用 \ref 中的结果？

Question 1

包xstring知道\StrBetween最后一个可选参数，该参数需要一个命令标记来存储结果。因此\variation可以给出一个可选参数（在开头以避免在结尾处丢失空格）。使用可选参数，结果将存储在给出的宏中。

\documentclass{article}
\usepackage{xstring}

\newcommand*{\variation}[3][]{%
  \edef\variationNext{%
    \noexpand\StrBetween[%
      \the\numexpr(#2)\relax,%
      \the\numexpr(#2)+1\relax
    ]%
  }%
  \variationNext{;#3;}{;}{;}[#1]\relax
}
\newcommand*{\variationref}[2]{%
  \variation[\tmp]{#1}{#2}\ref{\tmp}%
}

\begin{document}

\begin{equation} 
2+2=4 \label{aaa}
\end{equation}

\begin{equation}  
 3+3=6 \label{bbb}
\end{equation}

I can refer to these equations as equation~\ref{aaa} and
equation~\ref{bbb}.

Now I have the string ``aaa;bbb''. If I refer to these equations as
equation~\variation[\tmp]{1}{aaa;bbb}\ref{\tmp} and
equation~\variationref{2}{aaa;bbb}.                

\end{document}

Answer

包xstring知道\StrBetween最后一个可选参数，该参数需要一个命令标记来存储结果。因此\variation可以给出一个可选参数（在开头以避免在结尾处丢失空格）。使用可选参数，结果将存储在给出的宏中。

\documentclass{article}
\usepackage{xstring}

\newcommand*{\variation}[3][]{%
  \edef\variationNext{%
    \noexpand\StrBetween[%
      \the\numexpr(#2)\relax,%
      \the\numexpr(#2)+1\relax
    ]%
  }%
  \variationNext{;#3;}{;}{;}[#1]\relax
}
\newcommand*{\variationref}[2]{%
  \variation[\tmp]{#1}{#2}\ref{\tmp}%
}

\begin{document}

\begin{equation} 
2+2=4 \label{aaa}
\end{equation}

\begin{equation}  
 3+3=6 \label{bbb}
\end{equation}

I can refer to these equations as equation~\ref{aaa} and
equation~\ref{bbb}.

Now I have the string ``aaa;bbb''. If I refer to these equations as
equation~\variation[\tmp]{1}{aaa;bbb}\ref{\tmp} and
equation~\variationref{2}{aaa;bbb}.                

\end{document}

Question 2

问题在于\ref它需要它的参数是一个字符串，而不是生成该字符串的指令集。

最好对此类引用使用不同的命令，例如

\varref{1}{aaa;bbb}

对我的答案进行几乎直接的修改就可以实现您想要的效果：

\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn

\NewDocumentCommand{\variation}{ s O{;} m m }
 {
  \IfBooleanTF{#1}
   {
    \geoff_variation_check:nnn { #2 } { #3 } { #4 }
   }
   {
    \geoff_variation_nocheck:nnn { #2 } { #3 } { #4 }
   }
   \tl_use:N \l__geoff_item_tl
 }
\NewDocumentCommand{\varref}{s O{;} m m }
 {
  \IfBooleanTF{#1}
   {
    \geoff_variation_check:nnn { #2 } { #3 } { #4 }
   }
   {
    \geoff_variation_nocheck:nnn { #2 } { #3 } { #4 }
   }
   \exp_args:NV \ref \l__geoff_item_tl
 }

\msg_new:nnn { variation }
 { index~out~of~bounds }
 { You~have~requested~an~item~that~doesn't~exist }
\seq_new:N \l__geoff_item_list_seq
\tl_new:N \l__geoff_item_tl

\cs_new_protected:Npn \geoff_variation_nocheck:nnn #1 #2 #3
 {
  \seq_set_split:Nnn \l__geoff_item_list_seq { #1 } { #3 }
  \tl_set:Nx \l__geoff_item_tl { \seq_item:Nn \l__geoff_item_list_seq { #2 } }
 }

\cs_new_protected:Npn \geoff_variation_check:nnn #1 #2 #3
 {
  \seq_set_split:Nnn \l__geoff_item_list_seq { #1 } { #3 }
  \int_compare:nTF { #2 > \seq_count:N \l__geoff_item_list_seq }
   {
    \msg_warning:nn { variation }{ index~out~of~bounds }
    \tl_set:Nn \l__geoff_item_tl { --inexistentitem-- }
   }
   {
    \tl_set:Nx \l__geoff_item_tl { \seq_item:Nn \l__geoff_item_list_seq { #2 } }
   }
 }

\ExplSyntaxOff

\begin{document}

\variation{3}{aaa;bbb;ccc;ddd;eee}

\variation[,]{3}{aaa,bbb,ccc,ddd,eee}

\variation{1}{aaa}

\variation{2}{aaa}

\variation*{2}{aaa}

\begin{equation}
2+2=4 \label{aaa}
\end{equation}

\begin{equation}
 3+3=6 \label{bbb}
\end{equation}

I can refer to these equations as equation~\ref{aaa} and
equation~\ref{bbb}.

If I refer to these equations as equation~\varref{1}{aaa;bbb} and
equation~\varref{2}{aaa;bbb}, I get correct numbers.

\end{document}

我保留了原始\variation命令，尽管它看起来并不是真正需要的。

不同之处在于，现在\geoff_variation_nocheck:nnn将结果存储在标记列表变量中并\varref调用它（或check变体），然后执行

\exp_args:NV \ref \l__geoff_item_tl

如果没有溢出，\varref和的行为是相同的；如果，您会收到警告并且会产生；如果，您只会得到。\varref*\varref*{3}{aaa;bbb}\ref{--inexistentitem--}\varref{3}{aaa;bbb}\ref{}

Answer