更新

Question

我正在尝试使用 xsim 为学生设计一本练习册，并为导师提供第二版。其中将包含练习/问题/家庭作业（无论你想如何称呼它）及其各自的解决方案。

从我目前的评论来看，模拟，似乎是实现此目的的最合适的软件包。但是，有一条信息我在文档中找不到：如何在空的答案空间/空白/框和打印的解决方案之间切换？

理想情况下，我希望答案空间能够精确占据排版解决方案的幻像空间。

附件中您可以找到我希望它如何工作的 MWE，但我希望该过程能够自动化。

以下 MWE 产生（注意solution/print = true）

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage{xsim}
\usepackage{tcolorbox}



\DeclareExerciseEnvironmentTemplate{tcolorbox}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    \GetExercisePropertyT{subtitle}{ \textit{\PropertyValue}}%
    ]%
}
{\endtcolorbox}

\DeclareExerciseType{problem}{
exercise-env = problem ,
solution-env = answer ,
exercise-name = Problem ,
solution-name = Answer ,
exercise-template = tcolorbox ,
solution-template = tcolorbox
}

\xsimsetup{
    answer/print=true
}


\begin{document}
    \begin{problem}
        Show that $(\mathbb{R}, +, \cdot)$ is a field.
    \end{problem}

    \begin{answer}
        Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
        \begin{description}
            \item[Commutativity of addition]  $a + b = b+ a$.
            \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
            \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
            \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
            \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
            \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
            \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
            \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.      
            \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$. 
        \end{description}
        These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
    \end{answer}
\end{document}

但我想要当答案被禁用时，尺寸相同的空框。

更新

最终，我放弃了xsim，选择了一条不同的道路，并提出了自己的替代解决方案，具体如下在另一个问题和答案中。但是，这仍然不是这个问题的答案。

更新

再过一步，我又重新开始使用xsim。请参阅下面的解决方案。

Answer 1

与此同时，我已经想到了一个答案，但是这个答案取决于xsim正在使用的版本。

xsim < v0.19b

例如，请参阅更新后的 MWE：

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage[]{xsim}
\usepackage{tcolorbox}

\xsimsetup{
    solution/print = true,
}

\SetExerciseParameters{exercise}{
    exercise-name = \XSIMtranslate{exercise},
    solution-name = \XSIMtranslate{solution},
    exercise-template = exercise,
    solution-template = solution,
    counter=section,
}



\DeclareExerciseEnvironmentTemplate{exercise}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{%
\endtcolorbox
% at this point, add invisible solution
\IfSolutionPrintTF{}{%
\tcolorbox[%
    upperbox=invisible,%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}%
}%

\DeclareExerciseEnvironmentTemplate{solution}%
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{SOL: \GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{\endtcolorbox}%


\begin{document}

\section{A section title}

\begin{exercise}
    Show that $(\mathbb{R}, +, \cdot)$ is a field.
\end{exercise}

\begin{solution}
    Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
    \begin{description}
        \item[Commutativity of addition]  $a + b = b+ a$.
        \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
        \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
        \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
        \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
        \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
        \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
        \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.
        \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$.
    \end{description}
    These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
\end{solution}
\end{document}

产生输出（取决于属性printsolutions）：

xsim v0.20

理想情况下，相同的代码应该可以工作。然而，直到这个错误已解决。问题评论中的解决方法是使用

\xsimsetup{collect=true}

然后在文档主体中使用

\printcollection[print=both]{all exercises}

solutions/print=false不幸的是，如果设置了该选项，则会附加解决方案的填充副本。

因此，目前针对此答案的解决方法是：将解决方案主体移到练习中进行打印：

\documentclass{article}
\usepackage{mwe}
\usepackage{amssymb, amsmath}
\usepackage[]{xsim}
\usepackage{tcolorbox}

\xsimsetup{
    solution/print = true,  % toggle this line
    collect=true,
}

\SetExerciseParameters{exercise}{
    exercise-name = \XSIMtranslate{exercise},
    solution-name = \XSIMtranslate{solution},
    exercise-template = exercise,
    solution-template = solution,
    counter=section,
}



\DeclareExerciseEnvironmentTemplate{exercise}
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{%
\endtcolorbox
% at this point, add invisible solution
\IfSolutionPrintTF{%
\tcolorbox[%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}{%
\tcolorbox[%
    upperbox=invisible,%
    beforeafter skip = .5\baselineskip ,%
    title =
    \textbf{\GetExerciseParameter{solution-name}~\GetExerciseProperty{counter}}%
    ]%
    \GetExerciseBody{solution}%
    \endtcolorbox%
}%
}%

\DeclareExerciseEnvironmentTemplate{solution}%
{%
\tcolorbox[
    beforeafter skip = .5\baselineskip ,
    title =
    \textbf{\GetExerciseName~\GetExerciseProperty{counter}}%
    ]%
}%
{\endtcolorbox}%


\begin{document}
\section{A section title}


\begin{exercise}
    Show that $(\mathbb{R}, +, \cdot)$ is a field.
\end{exercise}

\begin{solution}
    Let $a, b, c \in \mathbb{R}$ arbitrary. Then, we show the following properties:
    \begin{description}
        \item[Commutativity of addition]  $a + b = b+ a$.
        \item[Associativity of addition] $a +(b + c) = (a+b) +c$.
        \item[Additive identity] $0 \in \mathbb{R}$, and $a +0 = a $.
        \item[Additive inverse] for each $a \in \mathbb{R}$, there exists $-a \in \mathbb{R}$, sch that $a + (-a) = 0$.
        \item[Commutativity of multiplication] $a \cdot b = b \cdot a$.
        \item[Associativity of multiplications] $a \cdot (b \dot c) = (a \cdot b) \cdot c$.
        \item[Multiplicative identity] $1 \in \mathbb{R}$, such that $1 \cdot a = a$.
        \item[Multiplicative inverse] for each $a \in \mathbb{R}$, there exists $a^{-1}\in \mathbb{R}$, such that $a \cdot a^{-1} = 1$.
        \item[Distributivity] $a\cdot (b + c) = (a \cdot b) + (a \cdot c)$.
    \end{description}
    These are the necessary and sufficient properties of a field, hence $(\mathbb{R}, +, \cdot)$ is a field.
\end{solution}



\printcollection[print=exercises]{all exercises}
\end{document}

这会产生以下所需的输出（与以前的版本相同，但代码不太简单并且可能存在额外的未观察到的缺点）：