독일의 법률 보고서는 종종 다소 이상한 구조 방식을 사용합니다. 일반적인 숫자 대신
ㅏ.
(B., CD ...) 첫 번째 레벨의 경우,
나.
-- 두 번째 수준의 로마 숫자
1.
세 번째 수준의 아라비아 숫자
ㅏ)
b), c), d) ... 네 번째 수준의 경우
아아)
bb), cc), dd)는 다섯 번째 수준입니다.
다음과 같은 거의 최소한의 예제에서 이를 구현했습니다. 간략한 참고자료를 갖고 싶습니다. 섹션 »B.<를 읽고 있고 하위 섹션 »B.II.<에 대한 참조가 있는 경우에는 »II.<를 인쇄하는 것으로 충분합니다. 그러나 섹션 A.를 읽으려면 »B.II.«가 필요합니다. B.II.3.a)cc)에서 B.II.3.a)bb)의 항목을 참조하는 경우 출력은 \ref분명히 B.II.3.a)bb)가 되어서는 안 되며 단순히 bb ). B.II.4.에 있는 경우 참조에는 3.a)bb)가 나와야 합니다.
그러나 CI1.의 어딘가에서 참조하는 경우 B.II.3.a)bb)에 대한 전체 참조가 필요합니다.
특별 요청: 솔루션은 Varioref 패키지에도 적용되어야 합니다.
jura 문서 클래스는 이러한 종류의 참조를 제공합니다. 하지만 나는 여러 가지 이유로 KOMA 스크립트에 의존합니다. 그래서 jura.dtx에서 코드를 가져와서 적용하려고 했지만 이해할 수 없는 오류 메시지가 나타납니다. 아래를 참조하세요.
\ref그래서: 텍스트의 위치에 따라 참조할 수 있습니까 ? 모든 솔루션을 환영합니다.
Ulrike의 답변을 편집하십시오.
이 코드는 항상 "II"를 생성합니다. 하위 섹션보다 낮은 수준:
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{babel, blindtext}
\setcounter{secnumdepth}{5} % fünf Ebenen
\renewcommand*{\thesection}{\Alph{section}.}
\renewcommand*{\thesubsection}{\Roman{subsection}.}
\renewcommand*{\thesubsubsection}{\arabic{subsubsection}.}
\renewcommand*{\theparagraph}{\alph{paragraph})}%)
\renewcommand*{\thesubparagraph}{\alph{subparagraph}\alph{subparagraph})}
\usepackage{zref-user,zref-counter}
\usepackage{expl3}
\makeatletter
\zref@newprop{section}[1]{\the\c@section}
\zref@newprop{subsection}[1]{\the\c@subsection}
\zref@addprop{main}{section}
\zref@addprop{main}{subsection}
\ExplSyntaxOn
\newcommand\myref[1]{%
\hyperref[#1]{
\int_compare:nTF
{
\zref@extractdefault {#1} {section} {0} = \the\c@section
}
{
%ref is from current section
\int_compare:nTF
{\zref@extractdefault {#1} {subsection} {0} = 0 }
{Was~nun?}
{\int_to_Roman:n{\zref@extractdefault {#1} {subsection} {0}}.}
}
{
%ref is from other section
\int_to_Alph:n {\zref@extractdefault {#1} {section} {0}}.
\int_to_Roman:n {\zref@extractdefault {#1} {subsection} {0}}
}}}
\ExplSyntaxOff
\makeatother
%\usepackage{xcolor}
\usepackage[]{hyperref}
\newcommand\Label[1]{\zlabel{#1}\label{#1}}
\begin{document}
\section{First}
\Label{sec:first}
I'm refering to the second section and I'd like to have a full
reference like this: \myref{sec:secondofsecond}.
\section{Second}
\Label{sec:second}
\myref{sec:secondofsecond}, \myref{sec:first}, \myref{sec:second} And lets test the
new references, see \myref{CLA:first-paragraph} or \myref{CLA:first-second.second}
\subsection{Firstofsecond}
\Label{sec:firstofsecond}
Text.
\subsection{Secondofsecond}
\Label{sec:secondofsecond}
Now down the tree:
\subsubsection{first of second.second}
\Label{CLA:first-second.second}
Just some letters.
\paragraph{first-paragraph}
\Label{CLA:first-paragraph}
And more text.
\end{document}
이것은 나의 초기 MWE이며 이제는 더 이상 사용되지 않습니다.
\documentclass[english, parskip=half, numbers=noendperiod]{scrartcl}
\usepackage[T1]{fontenc}\usepackage[utf8]{inputenc}
\usepackage{babel, blindtext}
\setcounter{secnumdepth}{5} % five levels
\renewcommand*{\thesection}{\Alph{section}.}
\renewcommand*{\thesubsection}{\Roman{subsection}.}
\renewcommand*{\thesubsubsection}{\arabic{subsubsection}.}
\renewcommand*{\theparagraph}{\alph{paragraph})}%)
\renewcommand*{\thesubparagraph}{\alph{subparagraph}\alph{subparagraph})}
\renewcommand*{\paragraphformat}{\theparagraph{}\enskip}
\renewcommand*{\subparagraphformat}{%
\thesubparagraph{}\enskip}
\makeatletter
\renewcommand{\p@subsection}{\thesection}
\renewcommand{\p@subsubsection}{\thesection\thesubsection}
\renewcommand{\p@paragraph}{\thesection\thesubsection\thesubsubsection}
\renewcommand{\p@subparagraph}{\thesection\thesubsection\thesubsubsection\theparagraph\,}
\makeatother{}
\begin{document}
\section{First}
\label{sec:first}
I'm refering to the second section and I'd like to have a full
reference like this: \ref{sec:secondofsecond}.
\section{Second}
\label{sec:second}
\subsection{Firstofsecond}
\label{sec:firstofsecond}
Text.
\subsection{Secondofsecond}
\label{sec:secondofsecond}
But now I'm refering to »Firstofsecond« and my readers do not need the
full reference (\ref{sec:firstofsecond}), simply »I.« would be sufficient.
From here a reference to »first of second / second« should look like
this: »1.« and not \ref{sec:first-second-second}
\subsubsection{First of second / second}
\label{sec:first-second-second}
text.
\end{document}
좋아요, 이제 더 이상 사용되지 않는 jura.dtx의 코드를 재활용하는 아이디어로 넘어갑니다.
jknappen의 조언에 따라 나는 내가 원하는 동작을 정확히 구현하기 위해 jura.dtx의 코드를 사용하려고 했습니다. 코드를 다음과 같이 변경했습니다.
%\documentclass[english, parskip=half, numbers=noendperiod]{scrartcl}
\documentclass[english]{article}
\usepackage[T1]{fontenc}\usepackage[utf8]{inputenc}
\usepackage{babel, blindtext}
\setcounter{secnumdepth}{5} % fünf Ebenen
\renewcommand*{\thesection}{\Alph{section}.}
\renewcommand*{\thesubsection}{\Roman{subsection}.}
\renewcommand*{\thesubsubsection}{\arabic{subsubsection}.}
\renewcommand*{\theparagraph}{\alph{paragraph})}%)
\renewcommand*{\thesubparagraph}{\alph{subparagraph}\alph{subparagraph})}
%% Code stolen from jura.dtx
\makeatletter{}
\newcommand*{\@greek}[1]{\ifcase#1\relax\or$\alpha$\or$\beta$\or
$\gamma$\or$\delta$\or$\varepsilon$\or$\zeta$\or$\eta$\or$\vartheta$\or
$\iota$\or$\kappa$\or$\lambda$\or$\mu$\or$\nu$\or$\xi$\or$o$\or$\pi$\or
$\rho$\or$\sigma$\or$\tau$\or$\upsilon$\or$\varphi$\or$\chi$\or$\psi$\or
$\omega$\else\@ctrerr\fi}
\newcounter{lvla}
\newcounter{lvlb}
\newcounter{lvlc}
\newcounter{lvld}
\newcounter{lvle}
\newcounter{lvlf}
\newcounter{lvlg}
\newcounter{lvlh}
\newcounter{lvli}
\newcounter{lvlj}
\newcounter{lvlk}
\newcounter{lvll}
\renewcommand*{\thelvla}{\J@Number}
\renewcommand*{\thelvlb}{\J@Number}
\renewcommand*{\thelvlc}{\J@Number}
\renewcommand*{\thelvld}{\J@Number}
\renewcommand*{\thelvle}{\J@Number}
\renewcommand*{\thelvlf}{\J@Number}
\renewcommand*{\thelvlg}{\J@Number}
\renewcommand*{\thelvlh}{\J@Number}
\renewcommand*{\thelvli}{\J@Number}
\renewcommand*{\thelvlj}{\J@Number}
\renewcommand*{\thelvlk}{\J@Number}
\renewcommand*{\thelvll}{\J@Number}
\newcounter{tiefe}\setcounter{tiefe}{1}
%\setcounter{tocdepth}{12}
%\setcounter{secnumdepth}{12}
\newcommand*{\J@Number}{\J@NumberRoot{\value{tiefe}}%
\J@NumberEnd{\value{tiefe}}}
\newcommand*{\J@NumberRoot}[1]{\J@INumberRoot{#1}%
{\csname c@lvl\@alph{#1}\endcsname}}
\newcommand*{\J@INumberRoot}[2]{%
\ifcase#1\or
\@Alph{#2}\or
\@Roman{#2}\or
\@arabic{#2}\or
\@alph{#2}\or
\@alph{#2}\@alph{#2}\or
\@greek{#2}\or
\@greek{#2}\@greek{#2}\or
(\@arabic{#2})\or
(\@alph{#2})\or
(\@alph{#2}\@alph{#2})\or
(\@greek{#2})\or
(\@greek{#2}\@greek{#2})\fi}
\newcommand*{\J@NumberEnd}[1]{\ifnum#1<8\ifnum#1<4.\else)\fi\fi}
\newcommand*{\J@SetCurrent}{\xdef\@currentlabel{{\arabic{lvla}}%
{\arabic{lvlb}}{\arabic{lvlc}}{\arabic{lvld}}{\arabic{lvle}}%
{\arabic{lvlf}}{\arabic{lvlg}}{\arabic{lvlh}}{\arabic{lvli}}%
{\arabic{lvlj}}{\arabic{lvlk}}{\arabic{lvll}}}}
\ifnum\value{lvl\alph{tiefe}}=1%
\PackageWarning{alphanum}{only one subitem in level "\J@Number" ending}%
\fi
\setcounter{lvl\alph{tiefe}}{0}
\addtocounter{tiefe}{-1}
\renewcommand*{\ref}{%
\@ifstar{\@tempswatrue\J@ref}{\@tempswafalse\J@ref}}
\newcommand*{\J@ref}[1]{%
\expandafter\ifx\csname r@#1\endcsname\relax
\protect\G@refundefinedtrue
\nfss@text{\reset@font\bfseries ??}
\@latex@warning{Reference '#1' on page \thepage \space undefined}%
\else
\expandafter\J@refX\csname r@#1\endcsname
\fi
}
\newcommand*{\J@refX}[1]{\expandafter\J@refXX#1}
\newcommand*{\J@refXX}[2]{%
\ifcat_\ifnum9<1#1_\else X\fi
#1%
\else
\expandafter\J@refA#1%
\fi}
\newcommand*{\J@refA}[6]{\@tempcnta\@ne%
\J@refP1{#1}0\J@refP2{#2}{#1}\J@refP3{#3}{#2}\J@refP4{#4}{#3}%
\J@refP5{#5}{#4}\J@refP6{#6}{#5}\J@refB#6}
\newcommand*{\J@refB}[7]{\J@refP7{#2}{#1}\J@refP8{#3}{#2}%
\J@refP9{#4}{#3}\J@refP{10}{#5}{#4}\J@refP{11}{#6}{#5}%
\J@refP{12}{#7}{#6}\J@NumberEnd\@tempcnta}
\newcommand*{\J@refP}[3]{%
\if@tempswa%
\ifnum#2>0\@tempcnta#1\,\J@INumberRoot{#1}{#2}\fi
\else%
\ifnum#2=\value{lvl\@alph{#1}}%
\ifnum#1=12\PackageError{alphanum}{Self-reference detected!}{You used
a \protect\ref\space in the same section as the referenced
\protect\label\MessageBreak That makes no sense!}\fi
\else
\ifnum#2>0\@tempcnta#1\@tempswatrue\J@INumberRoot{#1}{#2}\else
\ifnum#3>0%
\@tempcnta#1\advance\@tempcnta-1\relax\J@INumberRoot{\@tempcnta}{#3}%
\fi
\fi
\fi
\fi}
\makeatother{}
\begin{document}
\section{First}
\label{sec:first}
I'm refering to the second section and I'd like to have a full
reference like this: \ref{sec:secondofsecond}.
\section{Second}
\label{sec:second}
\subsection{Firstofsecond}
\label{sec:firstofsecond}
Text.
\subsection{Secondofsecond}
\label{sec:secondofsecond}
But now I'm refering to »Firstofsecond« and my readers do not need the
full reference (\ref{sec:firstofsecond}), simply »I.« would be sufficient.
From here a reference to »first of second / second« should look like
this: »1.« and not \ref{sec:first-second-second}
\subsubsection{First of second / second}
\label{sec:first-second-second}
text.
\end{document}
안타깝게도 오류가 있습니다(파일 이름: kurze-verweise-mwe.tex).
./kurze-verweise-mwe.tex:134: Missing number, treated as zero.
<to be read again>
I
l.134 ...ence like this: \ref{sec:secondofsecond}.
./kurze-verweise-mwe.tex:134: Missing = inserted for \ifnum.
<to be read again>
여기 두 가지 예 중 하나에 대한 해결책을 찾을 시간과 재미를 갖고 있는 사람이 있습니까?
답변1
이론적으로 이는 그리 어렵지 않습니다. zref를 사용하여 레이블의 섹션, 하위 섹션 등의 다양한 값을 모두 저장할 수 있으며, 그런 다음 해당 값을 테스트하고 현재 값과 비교하여 어떤 일이 발생해야 하는지 결정하기만 하면 됩니다.
실제로는 꽤 많은 테스트가 있고 일반적으로 모든 분기를 올바르게 수행하려면 큰 종이가 필요하기 때문에 모든 레벨에 대해 실제로 수행하려는 경우 약간 지루할 수 있습니다. (저는 첫 번째 시도에서는 절대 수행하지 않습니다.) .
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{babel, blindtext}
\setcounter{secnumdepth}{5} % fünf Ebenen
\renewcommand*{\thesection}{\Alph{section}.}
\renewcommand*{\thesubsection}{\Roman{subsection}.}
\renewcommand*{\thesubsubsection}{\arabic{subsubsection}.}
\renewcommand*{\theparagraph}{\alph{paragraph})}%)
\renewcommand*{\thesubparagraph}{\alph{subparagraph}\alph{subparagraph})}
\usepackage{zref-user,zref-counter}
\usepackage{expl3}
\makeatletter
\zref@newprop{section}[1]{\the\c@section}
\zref@newprop{subsection}[1]{\the\c@subsection}
\zref@addprop{main}{section}
\zref@addprop{main}{subsection}
\ExplSyntaxOn
\newcommand\myref[1]{%
\hyperref[#1]{
\int_compare:nTF
{
\zref@extractdefault {#1} {section} {0} = \the\c@section
}
{
%ref is from current section
\int_compare:nTF
{\zref@extractdefault {#1} {subsection} {0} = 0 }
{Was~nun?}
{\int_to_Roman:n{\zref@extractdefault {#1} {subsection} {0}}.}
}
{
%ref is from other section
\int_to_Alph:n {\zref@extractdefault {#1} {section} {0}}.
\int_to_Roman:n {\zref@extractdefault {#1} {subsection} {0}}
}}}
\ExplSyntaxOff
\makeatother
%\usepackage{xcolor}
\usepackage[]{hyperref}
\newcommand\Label[1]{\zlabel{#1}\label{#1}}
\begin{document}
\section{First}
\Label{sec:first}
I'm refering to the second section and I'd like to have a full
reference like this: \myref{sec:secondofsecond}.
\section{Second}
\Label{sec:second}
\myref{sec:secondofsecond}, \myref{sec:first}, \myref{sec:second}
\subsection{Firstofsecond}
\Label{sec:firstofsecond}
Text.
\subsection{Secondofsecond}
\Label{sec:secondofsecond}
\end{document}
답변2
당신은 알고 있습니까?쥐라 패키지? 그러나 꽤 오래되고 지원되지 않으며 스케치한 미세한 등급 수준 구조를 제공합니다(그리고 α) β) 및 αα) ββ)로 표시된 수준까지 더 내려갑니다.
jura편집: (또는 레벨 구조를 포함하는) 도입된 문서 마크업은 \usepackage{alphanum}일반적인 LaTeX 섹션 명령(예: \section등)과 호환되지 않습니다. , , 를 jura사용하여 단면화하는 스타일을 사용해야 합니다 .\toc\sub\levelup
답변3
그만큼답변~의울리케 피셔나에게 길을 보여줬고 하루 종일 고민한 끝에 다음과 같은 해결책을 찾았습니다.
아마도 그것은 당신이 읽게 될 코딩 수준 중 가장 낮은 수준일 것이므로 주의하십시오. 누군가가 더 나은 해결책을 생각해낸다면, 좋습니다!
하지만 울리케 님의 아낌없는 도움이 없었다면 불가능했을 것입니다. 정말 감사합니다!
일부 댓글은 독일어로 되어 있습니다. 죄송합니다. 지금은 늦었고 오늘 제가 쓴 내용을 내일 이해할 수 있을지 의문입니다! 아, 그리고 주석에 있는 다양하고 쓸모없는 "("는 Emacs를 행복하게 만들기 위한 것입니다. 그렇지 않으면 "{"와 "}"가 일치하지 않습니다.
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{babel, blindtext}
\setcounter{secnumdepth}{5} % fünf Ebenen
\renewcommand*{\thesection}{\Alph{section}.}
\renewcommand*{\thesubsection}{\Roman{subsection}.}
\renewcommand*{\thesubsubsection}{\arabic{subsubsection}.}
\renewcommand*{\theparagraph}{\alph{paragraph})}%)
\renewcommand*{\thesubparagraph}{\alph{subparagraph}\alph{subparagraph})}
\usepackage{zref-user, zref-counter}
\usepackage{expl3}
\makeatletter
\zref@newprop{section}[1]{\the\c@section}
\zref@newprop{subsection}[1]{\the\c@subsection}
\zref@newprop{subsubsection}[1]{\the\c@subsubsection}
\zref@newprop{paragraph}[1]{\the\c@paragraph}
\zref@newprop{subparagraph}[1]{\the\c@subparagraph}
\zref@addprop{main}{section}
\zref@addprop{main}{subsection}
\zref@addprop{main}{subsubsection}
\zref@addprop{main}{paragraph}
\zref@addprop{main}{subparagraph}
\ExplSyntaxOn
\newcommand\myref[1]{%
\hyperref[#1]{%
\int_compare:nTF
{% A.-Ebene prüfen: sind wir auf dem gleichen Zweig?
\zref@extractdefault {#1} {section} {0} = \the\c@section
}
{ % Ja, wir sind auf dem gleichen Zweig
% Existiert eine weitere Ebene?
\int_compare:nTF
{\zref@extractdefault {#1} {subsection} {0} = 0}
% Wenn subsection 0 ist, also nicht existiert, dann muss section ausgegeben
% werden.
{
% \emph{Fehlerhafter Verweis}
\int_to_Alph:n {\zref@extractdefault {#1} {section} {0}}.
}
% Wenn dagegen subsection existiert, dann geht die Prüfung weiter:
{
% OK, weitere ebene existiert.
\int_compare:nTF % I.-Ebene
{
\zref@extractdefault {#1} {subsection} {0} = \the\c@subsection
}% Sind wir auf der gleichen I.-Ebene?
{% I.-Ebene: Gleiche Ebene
% Prüfung: gibt es eine weitere Ebene?
\int_compare:nTF
{\zref@extractdefault {#1} {subsubsection} {0} = 0}
{% Wenn 0, dann keine weitere Ebene und subsection ausgeben:
\int_to_Roman:n {\zref@extractdefault {#1} {subsection} {0}}.
}
{% OK, weitere Ebene existiert
\int_compare:nTF % subsubsection-Ebene
{% 1.-Ebene prüfen
\zref@extractdefault {#1} {subsubsection} {0} = \the\c@subsubsection
}
{% 1.-Ebene ist gleich
%Prüfung: gibt es eine weitere Ebene?
\int_compare:nTF % Gibt es eine a-Ebene?
{\zref@extractdefault {#1} {paragraph} {0} = 0}
{% Wenn 0, dann keine weitere Ebene und subsubsection ausgeben:
\int_to_arabic:n { \zref@extractdefault {#1} {subsubsection} {0}}.
}
{% Paragraph existiert:
\int_compare:nTF % a-Ebene prüfen
{
\zref@extractdefault {#1} {paragraph} {0} = \the\c@paragraph
}
{% a-Ebene ist gleich
% Prüfung: aa-Ebene vorhanden?
\int_compare:nTF %
{ \zref@extractdefault {#1} {subparagraph} {0} = 0}
%
{%( damit emacs happy ist
\int_to_alph:n {\zref@extractdefault {#1} {paragraph} {0} })
}
{% aa-Ebene vorhanden
\int_compare:nTF
{\zref@extractdefault {#1} {subparagraph} {0} = \the\c@subparagraph}
{\emph{Fehlerhafter VERWEIS}}
%( make emacs happy
{\int_to_alph:n { \zref@extractdefault {#1} {subparagraph} {0} }
\int_to_alph:n { \zref@extractdefault {#1} {subparagraph} {0} })
}
}
}
{% a-Ebene ist unterschiedlich (
\int_to_alph:n{\zref@extractdefault {#1} {paragraph} {0}})
}
}
}
{% 1.-Ebene ist unterschiedlich
\int_to_arabic:n{\zref@extractdefault {#1} {subsubsection} {0}}.
}
}% 3. Klammer Prüfung, ob subsection existiert (also wenn ja)
}
{% I.-Ebene: ungleich
% Also römische Nummer ausgeben
\int_to_Roman:n{\zref@extractdefault {#1} {subsection} {0}}.%
\int_compare:nTF%existiert die nächste Ebene?
{ \zref@extractdefault {#1} {subsubsection} {0} = 0 }
{}% = 0, subsubsection nicht vorhanden
{% ungleich 0
% also arabische Zahl ausgeben
\int_to_arabic:n { \zref@extractdefault {#1} {subsubsection} {0}}.
\int_compare:nTF%existiert die nächste Ebene paragraph?
{ \zref@extractdefault {#1} {paragraph} {0} = 0 }
{}% = 0, subsubsection nicht vorhanden
{% ungleich 0 (
\int_to_alph:n { \zref@extractdefault {#1} {paragraph} {0}})
}
% subpar Ebene vorhanden?
\int_compare:nTF
{ \zref@extractdefault {#1} {subparagraph} {0} = 0}
{}% ja, also existiert subpar nicht
{ %(
\int_to_alph:n {\zref@extractdefault {#1} {subparagraph} {0}}
\int_to_alph:n {\zref@extractdefault {#1} {subparagraph} {0}})
}
}
}% 3. Klammer I.-Ebene
}% 3. Klammer Prüfung, ob I. Ebene besteht.
}% 2. Klammer A.-Ebene
{% A.-Ebene: Ungleich
%ref is from other section
\int_to_Alph:n {\zref@extractdefault {#1} {section} {0}}.
\int_compare:nTF%existiert die nächste Ebene?
{ \zref@extractdefault {#1} {subsection} {0} = 0 }
{}% ja, erfüllt, keine subsection
{% subsection existiert:
\int_to_Roman:n {\zref@extractdefault {#1} {subsection} {0}}.
\int_compare:nTF%existiert die nächste Ebene?
{ \zref@extractdefault {#1} {subsubsection} {0} = 0 }
{}% = 0, subsubsection nicht vorhanden
{% ungleich 0
\int_to_arabic:n { \zref@extractdefault {#1} {subsubsection} {0} }.
\int_compare:nTF%existiert die nächste Ebene paragraph?
{ \zref@extractdefault {#1} {paragraph} {0} = 0 }
{}% = 0, paragraph nicht vorhanden
{% ungleich 0 (
\int_to_alph:n { \zref@extractdefault {#1} {paragraph} {0} })
\int_compare:nTF
{ \zref@extractdefault {#1} {subparagraph} {0} = 0}
{}% ja, also existiert subpar nicht
{% (
\int_to_alph:n {\zref@extractdefault {#1} {subparagraph} {0}}
\int_to_alph:n {\zref@extractdefault {#1} {subparagraph} {0}})
}
}
}
}
}% 3. Klammer A.-Ebene
}% hyperref Klammer zu
}%Ende Befehlsdefinition myref
\ExplSyntaxOff
\makeatother
\usepackage[]{hyperref}
\newcommand\Label[1]{\zlabel{#1}\label{#1}}
\begin{document}
\textbf{Präambel}
\begin{enumerate}
\item Reference to first section: \myref{sec:first}
\item Reference to second section: \myref{sec:second}
\item Reference to snd section / first subsection: \myref{sec:firstofsecond}
\item Ref. to snd sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to snd sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to snd sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
\section{First}
\Label{sec:first}
Sytematically:
\begin{enumerate}
\item Reference to this section should display a warning: \myref{sec:first}.
\item Reference to next section: \myref{sec:second}
\item Reference to next section / first subsection: \myref{sec:firstofsecond}
\item Ref. to next sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to next sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to next sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
\section{Second}
\Label{sec:second}
Sytematically:
\begin{enumerate}
\item Reference to previous section: \myref{sec:first}.
\item Reference to this section should display a warning: \myref{sec:second}
\item Reference to this section / first subsection: \myref{sec:firstofsecond}
\item Ref. to this sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to this sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to this sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
\subsection{Firstofsecond}
\Label{sec:firstofsecond}
Sytematically:
\begin{enumerate}
\item Reference to previous section: \myref{sec:first}.
\item Reference to this section: \myref{sec:second}
\item Reference to this section / first subsection: \myref{sec:firstofsecond}
\item Ref. to this sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to this sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to this sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
\subsection{Secondofsecond}
\Label{sec:secondofsecond}
Now down the tree:
\subsubsection{first of second.second}
\Label{CLA:first-second.second}
Just some letters.
\paragraph{first-paragraph}
\Label{CLA:first-paragraph}
And more text. Ref to next paragraph, should be a b): \myref{CLA:second-paragr}
\begin{enumerate}
\item Reference to previous section: \myref{sec:first}.
\item Reference to this section: \myref{sec:second}
\item Reference to this section / first subsection: \myref{sec:firstofsecond}
\item Ref. to this sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to this sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to this sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
\paragraph{second paragr}
\Label{CLA:second-paragr}
\begin{enumerate}
\item Reference to previous section: \myref{sec:first}.
\item Reference to this section: \myref{sec:second}
\item Reference to this section / first subsection: \myref{sec:firstofsecond}
\item Ref. to this sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to this sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to this sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\end{enumerate}
\subparagraph{first sub}
\Label{CLA:first-sub}
\begin{enumerate}
\item Reference to previous section: \myref{sec:first}.
\item Reference to this section: \myref{sec:second}
\item Reference to this section / first subsection: \myref{sec:firstofsecond}
\item Ref. to this sect. / second subsect: \myref{sec:secondofsecond}
\item Ref to this sect. / second subsect / first arabic:
\myref{CLA:first-second.second}
\item Ref. to this sect. / second subsect / first arabic / second par:
\myref{CLA:second-paragr}
\item Subparagraph: \myref{CLA:first-sub}
\end{enumerate}
dsfsdf
\subsubsection{Last one}
\Label{CLA:last-one}
text
\paragraph{paragraph somewhere}
\Label{CLA:paragraph-somewhere}
OK, the par above should display a). And now we refer to the next subpar:
\myref{CLA:second-subpar}.
And now a subpar:
\subparagraph{second subpar}
\Label{CLA:second-subpar}
text
\subparagraph{third subpar}
\end{document}