我使用 mdframed 环境来处理我的命题/定理等。但现在我已经这样做了,标记的定理不能在文本中引用,就像
代码
\documentclass[a4paper,11pt]{scrreprt}
\usepackage{tikz}
\usepackage[framemethod=tikz]{mdframed}
\usetikzlibrary{calc}
\usepackage{ntheorem}
\newcounter{mydef}[chapter]
\renewcommand\themydef{\arabic{chapter}.\arabic{mydef}}
\makeatletter
\def\mdf@mytitle{}
\define@key{mdf}{mytitle}{%
\def\mdf@mytitle{#1}}
\mdf@do@stringoption{digressiontitle=={Digression}}
\tikzset{
excursus line/.style={%
line width=2pt,
draw=gray!40,
rounded corners=2ex,
},
excursus head/.style={%
fill=white,
font=\bfseries\sffamily,
text=blue!80,
anchor=base west,
},
}
\mdfdefinestyle{digressionarrows}{%
extra={\stepcounter{mydef}},%
singleextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path let \p1=(Q), \p2=(O) in (\x1,{(\y1-\y2)/2}) coordinate (M);
\path [excursus line]
($(O)+(5em,0ex)$) -| (M) |- %
($(Q)+(30em,0ex)$);
%\node at ($(M)$) {M};
%\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {Digression};},
\ifx\empty\mdf@mytitle\empty
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef};
\else
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef:\space\mdf@mytitle};
\fi},
firstextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow,-to]
(O) |- %
($(Q)+(12em,0ex)$) .. controls +(0:16em) and +(185:6em) .. %
++(23em,2ex);
%\node [excursus head] at ($(Q)+(2.5em,-2pt)$) {Digression};},
\ifx\empty\mdf@mytitle\empty
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef};
\else
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef:\space\mdf@mytitle};
\fi},
secondextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow,round cap-]
($(O)+(5em,0ex)$) -| (Q);},
middleextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow]
(O) -- (Q);},
middlelinewidth=2.5em,middlelinecolor=white,
hidealllines=true,topline=true,
innertopmargin=0.5ex,
innerbottommargin=2.5ex,
innerrightmargin=2pt,
innerleftmargin=2ex,
skipabove=0.87\baselineskip,
skipbelow=0.62\baselineskip,
}
\makeatother
\newmdenv[style=digressionarrows,digressiontitle=Definition]{df}
\newmdenv[style=digressionarrows,digressiontitle=Proposition]{prop}
\begin{document}
\chapter{Chapter}
\section{test}
\begin{df}[mytitle = Pythagoraen Theorem]
test text
\label{pythag}
\end{df}
In the pythagoraen theorem there are $a,b$ and $c$ such that $a^2+b^2 = c^2$ (see \ref{pythag}).
\begin{df}
test text.
\end{df}
\begin{prop}[Proposition bla]
blablab.
\label{propbla}
\end{prop}
In proposition \ref{propbla} there are things.
\end{document}
我在这里做错了什么?提前谢谢您。
技术
答案1
明白了!我删除了 mfdefinestyle 中的“额外”条目,并添加了
settings = \refstepcounter{mydef}
在 \newmdenv 选项中。然后文本中的引用就可以正常工作了。
\documentclass[a4paper,11pt]{scrreprt}
\usepackage{tikz}
\usepackage[framemethod=tikz]{mdframed}
\usetikzlibrary{calc}
\usepackage{ntheorem}
\newcounter{mydef}[chapter]
\renewcommand\themydef{\thechapter.\arabic{mydef}}
\makeatletter
\def\mdf@mytitle{}
\define@key{mdf}{mytitle}{%
\def\mdf@mytitle{#1}}
\mdf@do@stringoption{digressiontitle=={Digression}}
\tikzset{
excursus line/.style={%
line width=2pt,
draw=gray!40,
rounded corners=2ex,
},
excursus head/.style={%
fill=white,
font=\bfseries\sffamily,
text=blue!80,
anchor=base west,
},
}
\mdfdefinestyle{digressionarrows}{%
%extra={\refstepcounter{mydef}},%
singleextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path let \p1=(Q), \p2=(O) in (\x1,{(\y1-\y2)/2}) coordinate (M);
\path [excursus line]
($(O)+(5em,0ex)$) -| (M) |- %
($(Q)+(30em,0ex)$);
%\node at ($(M)$) {M};
%\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {Digression};},
\ifx\empty\mdf@mytitle\empty
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef};
\else
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef:\space\mdf@mytitle};
\fi},
firstextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow,-to]
(O) |- %
($(Q)+(12em,0ex)$) .. controls +(0:16em) and +(185:6em) .. %
++(23em,2ex);
%\node [excursus head] at ($(Q)+(2.5em,-2pt)$) {Digression};},
\ifx\empty\mdf@mytitle\empty
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef};
\else
\node [excursus head] at ($(Q)+(2.5em,-0.75pt)$) {\mdf@digressiontitle~\themydef:\space\mdf@mytitle};
\fi},
secondextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow,round cap-]
($(O)+(5em,0ex)$) -| (Q);},
middleextra={%
\path let \p1=(P), \p2=(O) in (\x2,\y1) coordinate (Q);
\path [excursus arrow]
(O) -- (Q);},
middlelinewidth=2.5em,middlelinecolor=white,
hidealllines=true,topline=true,
innertopmargin=0.5ex,
innerbottommargin=2.5ex,
innerrightmargin=2pt,
innerleftmargin=2ex,
skipabove=0.87\baselineskip,
skipbelow=0.62\baselineskip,
}
\makeatother
\newmdenv[style=digressionarrows, settings = \refstepcounter{mydef}, digressiontitle=Definition]{df}
\newmdenv[style=digressionarrows, settings = \refstepcounter{mydef}, digressiontitle=Proposition]{prop}
\begin{document}
\chapter{Chapter}
\section{test}
\begin{df}[mytitle = Pythagoraen Theorem]
test text
\label{pythag}
\end{df}
In the pythagoraen theorem there are $a,b$ and $c$ such that $a^2+b^2 = c^2$ (see \ref{pythag}).
\begin{df}
test text.
\end{df}
\begin{prop}
blablab.
\label{propbla}
\end{prop}
In proposition \ref{propbla} there are things.
\end{document}