垂直对齐两个矩形分割水平节点

Question

我希望能够垂直对齐两个或多个rectangle split horizontal节点，使中心垂直线对齐。据我所知，直到后节点已创建。这是我目前得到的结果，north west底部节点的锚点与顶部节点中心垂直线的底部对齐。

\documentclass{article}

\usepackage[]{geometry}
\usepackage{tikz}
\usepackage{xparse}
\usepackage{etoolbox}
\usepackage{varwidth}

\newsavebox{\leftbox}
\newsavebox{\rightbox}
\newcounter{blocknum}

\usetikzlibrary{calc,positioning,shapes.multipart}

%% Thanks to egreg, make lrbox global (requires etoolbox.sty):
\cslet{glrbox}\lrbox
\expandafter\patchcmd\csname glrbox\endcsname{\setbox}{\global\setbox}{}{}
\cslet{endglrbox}\endlrbox

\NewDocumentEnvironment{Recipe}{}{%
    \begin{glrbox}{\leftbox}
        \begin{varwidth}{0.45\textwidth}
}{%
        \end{varwidth}%
    \end{glrbox}%
    %%% More...
    \ifnum\theblocknum=0
        \tikz[remember picture,overlay] {\node[name=block-\theblocknum,
            draw,
            rectangle split,
            rectangle split parts=2,
            rectangle split horizontal]
            {\nodepart{one}\usebox{\leftbox}\nodepart{two}\usebox{\rightbox}};
            \coordinate (putit-\theblocknum) at (block-\theblocknum.text split |- block-\theblocknum.two split south);
        }
    \else
        \tikz[remember picture,overlay] {\node[name=block-\theblocknum,
            draw,
            anchor=north west,
            rectangle split,
            rectangle split parts=2,
            rectangle split horizontal] at  (putit-\previous)
            {\nodepart{one}\usebox{\leftbox}\nodepart{two}\usebox{\rightbox}};
            \coordinate (putit-\theblocknum) at (block-\theblocknum.text split |- block-\theblocknum.two split south);
        }
    \fi
    \xdef\previous{\theblocknum}
    \stepcounter{blocknum}%
}

\NewDocumentCommand{\righttext}{}{%
        \end{varwidth}%
    \end{glrbox}%
    \begin{glrbox}{\rightbox}
        \begin{varwidth}{0.45\textwidth}
}

%\setmainfont{}

\begin{document}

\begin{Recipe}
Left\strut
\righttext
Right\\1\\2\\3\\4\strut
\end{Recipe}

\begin{Recipe}
Left which is longer\strut
\righttext
Right which is certainly quite long\\1\\2\\3\\4\strut
\end{Recipe}

\end{document}

Answer 1

text split或者one split锚点对我有用：

\documentclass[tikz,border=2mm]{standalone}

\usetikzlibrary{calc,positioning,shapes.multipart}

\begin{document}

\begin{tikzpicture}
\node[ draw, align=left,
        rectangle split,
        rectangle split parts=2,
        rectangle split horizontal] (A) 
        {Left\strut \nodepart{two} Right\\1\\2\\3\\4\strut};

\node[ draw, align=left,
        rectangle split,
        rectangle split parts=2,
        rectangle split horizontal,
        below= -\pgflinewidth of A.one split south,
        anchor=one split north
        ] (B) 
        {Left which is longer\strut \nodepart{two} Right which is certainly quite long\\1\\2\\3\\4\strut};
\end{tikzpicture}

\end{document}

Answer 2

这并不是尝试以您尝试的方式解决问题，而只是告诉您可以使用获得所需的（？）输出fit。

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc,positioning,fit}
\begin{document}
\begin{tikzpicture}
\node[align=left] (L1) {Left};
\node[align=left,right=0pt of L1] (R1) {Right\\1\\2\\3\\4\strut};
\node[fit=(L1)(R1),draw,inner sep=0pt](fit1){};
\draw (fit1.north -| R1.west) -- (fit1.south -| R1.west) coordinate (aux1);
%
\node[align=left,below=0pt of aux1,anchor=north west] (R2) {Right which is certainly quite long\\1\\2\\3\\4\strut};
\node[align=left,left=0pt of R2] (L2) {Left which is longer};
\node[fit=(L2)(R2),draw,inner sep=0pt](fit2){};
\draw (fit2.north -| R2.west) -- (fit2.south -| R2.west) coordinate (aux2);
\end{tikzpicture}
\end{document}

您似乎非常擅长定义新环境，所以我甚至没有尝试在这里这样做。