计算节点边界和路径之间的交点（或者，如何概括 pgfpointshapeborder）

Question 1

这是一个使用新形状的解决方案rounded rectangle X，当“外部点”在左侧或右侧时，它会返回圆角上的点。我希望我已经抓住了这里大多数奇怪的情况，但我不确定。

我当然没有处理过concave弧线，但也可以为此添加检查。

rounded rectangle X现在，如果我们只将连接到一个坐标或一个非节点，这就是我们要做的事情，rounded rectangle X因为当 TikZ 要求形状的边界点时，它会给出另一个节点中心的方向。但这不是我们想要的！

这就是为什么rr path样式检查给定的两个坐标是否都是形状的节点rounded rectangle X，然后才构造to path

-- (<point between the opposite centers of both nodes) -- (\tikztotarget)

此时，我假设节点要么直接垂直对齐，要么（因为您想在矩阵/CD 中使用它）在不同的列中。

否则，它会做出一些有趣的事情：

该拐点正好位于center west上节点和center east下节点的中间。我们需要检查圆弧的中心是否与节点本身处于相同的左右方向。

您可以使用不同的rounded rectangle arc lengths，因为形状会检查计算点是否实际位于边框的圆弧部分（通过测试对角线罗盘锚点）。

再次，对于仅将这些节点（不旋转）放置在网格中的 CD，这应该有效。

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{cd,shapes.misc}
\makeatletter
\def\pgfaddtoshape#1#2{%
  \begingroup
    \def\pgf@sm@shape@name{#1}%
    #2%
  \endgroup}
% Let's add two new anchors:
%  1. the center of the arc in the west
%  2. and the center of the arc in the east
\pgfaddtoshape{rounded rectangle}{%
  \pgf@sh@anchor{center west}{%
    \roundedrectanglepoints
    \ifx\westarc\pgf@lib@sh@misc@rr@text@convex
      \pgf@process{\csname pgf@anchor@rounded rectangle@west\endcsname}
      \advance\pgf@x by \radius
      \advance\pgf@x by \outerxsep
    \else
       \pgfpointadd{\centerpoint}{\pgfpoint{-\halfwidth-\radius}{+0pt}}%
    \fi
  }%
  \pgf@sh@anchor{center east}{%
    \roundedrectanglepoints
    \ifx\eastarc\pgf@lib@sh@misc@rr@text@convex
      \pgf@process{\csname pgf@anchor@rounded rectangle@east\endcsname}
      \advance\pgf@x by -\radius
      \advance\pgf@x by -\outerxsep
    \else
       \pgfpointadd{\centerpoint}{\pgfpoint{\halfwidth+\radius}{+0pt}}%
    \fi
  }%
}
% Let's declare a new shape and inherit everything from the original
% except for the anchorborder
\pgfdeclareshape{rounded rectangle X}{%
  \inheritsavedanchors[from=rounded rectangle]
  \inheritbehindbackgroundpath[from=rounded rectangle]
  \inheritbackgroundpath[from=rounded rectangle]
  \inheritbeforebackgroundpath[from=rounded rectangle]
  \inheritbehindforegroundpath[from=rounded rectangle]
  \inheritforegroundpath[from=rounded rectangle]
  \inheritbeforeforegroundpath[from=rounded rectangle]
  \inheritanchor[from=rounded rectangle]{center}
  \inheritanchor[from=rounded rectangle]{text}
  \inheritanchor[from=rounded rectangle]{mid}
  \inheritanchor[from=rounded rectangle]{mid west}
  \inheritanchor[from=rounded rectangle]{mid east}
  \inheritanchor[from=rounded rectangle]{base}
  \inheritanchor[from=rounded rectangle]{base west}
  \inheritanchor[from=rounded rectangle]{base east}
  \inheritanchor[from=rounded rectangle]{north}
  \inheritanchor[from=rounded rectangle]{south}
  \inheritanchor[from=rounded rectangle]{east}
  \inheritanchor[from=rounded rectangle]{west}
  \inheritanchor[from=rounded rectangle]{north west}
  \inheritanchor[from=rounded rectangle]{north east}
  \inheritanchor[from=rounded rectangle]{south west}
  \inheritanchor[from=rounded rectangle]{south east}
  \inheritanchor[from=rounded rectangle]{rect west}
  \inheritanchor[from=rounded rectangle]{rect east}
  \inheritanchor[from=rounded rectangle]{center west}
  \inheritanchor[from=rounded rectangle]{center east}
  \inheritanchorborder[from=rounded rectangle]
  % Approach:
  % we check whether the external point is to the left or to the right
  % if it's to the left we calculate the borderpoint of the circle
  % (or ellipse when outer seps are different)
  % with center .center west and radius \radius + \outer[xy]sep
  \anchorborder{%
    % save both the given external point (just a direction)
    \pgfextract@process\externalPoint{}%
    % as well as the one in the cs of the node
    \pgfextract@process\externalpoint{% this is weird but okay
      \pgfpointadd{\centerpoint}{\externalPoint}%
    }%
    %
    \roundedrectanglepoints
    \pgf@xa=\pgf@x \pgf@ya=\pgf@y % externalpoint is in xa, ya
    \pgf@process{\csname pgf@anchor@rounded rectangle X@north west\endcsname}%
    \pgf@xb=\pgf@x % west point is in xb
    \pgf@process{\csname pgf@anchor@rounded rectangle X@north east\endcsname}%
    \pgf@xc=\pgf@x % east point is in xc, yc
    \ifdim\pgf@xa<\pgf@xb % external point is to the left
      \pgfpointborderellipse{%
        \pgfpointdiff
          {\csname pgf@anchor@rounded rectangle X@center west\endcsname}
          {\externalpoint}%
        }{%
          \pgfpoint{\radius+\outerxsep}{\radius+\outerysep}%
        }%
        \pgfpointadd{}{\csname pgf@anchor@rounded rectangle X@center west\endcsname}%
        % is the calculated point even on the drawn arc?
        \ifdim\pgf@x<\pgf@xb
          % ok, proceed
        \else
          % let's use the normal border instead
          \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
        \fi
    \else
      \ifdim\pgf@xa>\pgf@xc % to the right
        \pgfpointborderellipse{%
          \pgfpointdiff
            {\csname pgf@anchor@rounded rectangle X@center east\endcsname}
            {\externalpoint}%
          }{%
            \pgfpoint{\radius+\outerxsep}{\radius+\outerysep}%
          }%
          \pgfpointadd{}{\csname pgf@anchor@rounded rectangle X@center east\endcsname}%
          % is the calculated point even on the drawn arc?
          \ifdim\pgf@x>\pgf@xc
            % ok, proceed
          \else
            % let's use the normal border instead
            \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
          \fi
      \else
        % let's use the normal border
        \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
      \fi
    \fi
  }
}

\def\tikz@roundedrectanglex@text{rounded rectangle X}%
\tikzset{
  rr path/.style={
    to path={%
      \pgfextra
      \def\tikz@to@path{-- (\tikztotarget)}% default
      \pgf@process{\tikz@scan@one@point\pgfutil@firstofone(\tikztostart)\relax}%
      \pgf@xa=\pgf@x \pgf@ya=\pgf@y % start is xa, ya
      \iftikz@shapeborder % is start a node?
        \expandafter\ifx\csname pgf@sh@ns@\tikz@pp@name{\tikztostart}\endcsname\tikz@roundedrectanglex@text
          % this is a rrX!
          \pgf@process{\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget)\relax}%
          \pgf@xb=\pgf@x \pgf@yb=\pgf@y % target is xb, yb
          \iftikz@shapeborder % is target a node?
            \expandafter\ifx\csname pgf@sh@ns@\tikz@pp@name{\tikztotarget}\endcsname\tikz@roundedrectanglex@text
              % they're both a rrX
              % check if they're vertical
              \pgf@xc=\pgf@xb\advance\pgf@xc-\pgf@xa
              \ifdim\pgf@xc<0pt \pgf@xc=-\pgf@xc\fi
              \ifdim\pgf@xc<1pt
                % they're vertical!
              \else
                \ifdim\pgf@xa<\pgf@xb % start left of target
                  \pgf@process{\pgfpointlineattime{+.5}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztostart.center east)\relax}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget.center west)\relax}}%
                  \edef\middlepoint{\the\pgf@x,\the\pgf@y}%
                \else
                  \pgf@process{\pgfpointlineattime{+.5}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztostart.center west)\relax}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget.center east)\relax}}%
                  \edef\middlepoint{\the\pgf@x,\the\pgf@y}%
                \fi
                \edef\tikz@to@path{--(\middlepoint)--(\tikztotarget)}%
              \fi
            \fi
          \fi
        \fi
      \fi
      \expandafter\endpgfextra\tikz@to@path\tikztonodes
    }
  }
}
\makeatother
\tikzset{n/.style={shape=rounded rectangle X, draw, ultra thick, rounded rectangle arc length=180}}
\begin{document}
\begin{tikzcd}
  |[n]| ABCD \ar[rd,rr path] \ar[r,rr path] \ar[d, rr path] & |[n]| EFG\\
  |[n]| ABCDEFGH & |[n]| EFG
\end{tikzcd}
\end{document}

输出

Answer

这是一个使用新形状的解决方案rounded rectangle X，当“外部点”在左侧或右侧时，它会返回圆角上的点。我希望我已经抓住了这里大多数奇怪的情况，但我不确定。

我当然没有处理过concave弧线，但也可以为此添加检查。

rounded rectangle X现在，如果我们只将连接到一个坐标或一个非节点，这就是我们要做的事情，rounded rectangle X因为当 TikZ 要求形状的边界点时，它会给出另一个节点中心的方向。但这不是我们想要的！

这就是为什么rr path样式检查给定的两个坐标是否都是形状的节点rounded rectangle X，然后才构造to path

-- (<point between the opposite centers of both nodes) -- (\tikztotarget)

此时，我假设节点要么直接垂直对齐，要么（因为您想在矩阵/CD 中使用它）在不同的列中。

否则，它会做出一些有趣的事情：

该拐点正好位于center west上节点和center east下节点的中间。我们需要检查圆弧的中心是否与节点本身处于相同的左右方向。

您可以使用不同的rounded rectangle arc lengths，因为形状会检查计算点是否实际位于边框的圆弧部分（通过测试对角线罗盘锚点）。

再次，对于仅将这些节点（不旋转）放置在网格中的 CD，这应该有效。

代码

\documentclass[tikz]{standalone}
\usetikzlibrary{cd,shapes.misc}
\makeatletter
\def\pgfaddtoshape#1#2{%
  \begingroup
    \def\pgf@sm@shape@name{#1}%
    #2%
  \endgroup}
% Let's add two new anchors:
%  1. the center of the arc in the west
%  2. and the center of the arc in the east
\pgfaddtoshape{rounded rectangle}{%
  \pgf@sh@anchor{center west}{%
    \roundedrectanglepoints
    \ifx\westarc\pgf@lib@sh@misc@rr@text@convex
      \pgf@process{\csname pgf@anchor@rounded rectangle@west\endcsname}
      \advance\pgf@x by \radius
      \advance\pgf@x by \outerxsep
    \else
       \pgfpointadd{\centerpoint}{\pgfpoint{-\halfwidth-\radius}{+0pt}}%
    \fi
  }%
  \pgf@sh@anchor{center east}{%
    \roundedrectanglepoints
    \ifx\eastarc\pgf@lib@sh@misc@rr@text@convex
      \pgf@process{\csname pgf@anchor@rounded rectangle@east\endcsname}
      \advance\pgf@x by -\radius
      \advance\pgf@x by -\outerxsep
    \else
       \pgfpointadd{\centerpoint}{\pgfpoint{\halfwidth+\radius}{+0pt}}%
    \fi
  }%
}
% Let's declare a new shape and inherit everything from the original
% except for the anchorborder
\pgfdeclareshape{rounded rectangle X}{%
  \inheritsavedanchors[from=rounded rectangle]
  \inheritbehindbackgroundpath[from=rounded rectangle]
  \inheritbackgroundpath[from=rounded rectangle]
  \inheritbeforebackgroundpath[from=rounded rectangle]
  \inheritbehindforegroundpath[from=rounded rectangle]
  \inheritforegroundpath[from=rounded rectangle]
  \inheritbeforeforegroundpath[from=rounded rectangle]
  \inheritanchor[from=rounded rectangle]{center}
  \inheritanchor[from=rounded rectangle]{text}
  \inheritanchor[from=rounded rectangle]{mid}
  \inheritanchor[from=rounded rectangle]{mid west}
  \inheritanchor[from=rounded rectangle]{mid east}
  \inheritanchor[from=rounded rectangle]{base}
  \inheritanchor[from=rounded rectangle]{base west}
  \inheritanchor[from=rounded rectangle]{base east}
  \inheritanchor[from=rounded rectangle]{north}
  \inheritanchor[from=rounded rectangle]{south}
  \inheritanchor[from=rounded rectangle]{east}
  \inheritanchor[from=rounded rectangle]{west}
  \inheritanchor[from=rounded rectangle]{north west}
  \inheritanchor[from=rounded rectangle]{north east}
  \inheritanchor[from=rounded rectangle]{south west}
  \inheritanchor[from=rounded rectangle]{south east}
  \inheritanchor[from=rounded rectangle]{rect west}
  \inheritanchor[from=rounded rectangle]{rect east}
  \inheritanchor[from=rounded rectangle]{center west}
  \inheritanchor[from=rounded rectangle]{center east}
  \inheritanchorborder[from=rounded rectangle]
  % Approach:
  % we check whether the external point is to the left or to the right
  % if it's to the left we calculate the borderpoint of the circle
  % (or ellipse when outer seps are different)
  % with center .center west and radius \radius + \outer[xy]sep
  \anchorborder{%
    % save both the given external point (just a direction)
    \pgfextract@process\externalPoint{}%
    % as well as the one in the cs of the node
    \pgfextract@process\externalpoint{% this is weird but okay
      \pgfpointadd{\centerpoint}{\externalPoint}%
    }%
    %
    \roundedrectanglepoints
    \pgf@xa=\pgf@x \pgf@ya=\pgf@y % externalpoint is in xa, ya
    \pgf@process{\csname pgf@anchor@rounded rectangle X@north west\endcsname}%
    \pgf@xb=\pgf@x % west point is in xb
    \pgf@process{\csname pgf@anchor@rounded rectangle X@north east\endcsname}%
    \pgf@xc=\pgf@x % east point is in xc, yc
    \ifdim\pgf@xa<\pgf@xb % external point is to the left
      \pgfpointborderellipse{%
        \pgfpointdiff
          {\csname pgf@anchor@rounded rectangle X@center west\endcsname}
          {\externalpoint}%
        }{%
          \pgfpoint{\radius+\outerxsep}{\radius+\outerysep}%
        }%
        \pgfpointadd{}{\csname pgf@anchor@rounded rectangle X@center west\endcsname}%
        % is the calculated point even on the drawn arc?
        \ifdim\pgf@x<\pgf@xb
          % ok, proceed
        \else
          % let's use the normal border instead
          \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
        \fi
    \else
      \ifdim\pgf@xa>\pgf@xc % to the right
        \pgfpointborderellipse{%
          \pgfpointdiff
            {\csname pgf@anchor@rounded rectangle X@center east\endcsname}
            {\externalpoint}%
          }{%
            \pgfpoint{\radius+\outerxsep}{\radius+\outerysep}%
          }%
          \pgfpointadd{}{\csname pgf@anchor@rounded rectangle X@center east\endcsname}%
          % is the calculated point even on the drawn arc?
          \ifdim\pgf@x>\pgf@xc
            % ok, proceed
          \else
            % let's use the normal border instead
            \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
          \fi
      \else
        % let's use the normal border
        \csname pgf@anchor@rounded rectangle@border\endcsname{\externalPoint}%
      \fi
    \fi
  }
}

\def\tikz@roundedrectanglex@text{rounded rectangle X}%
\tikzset{
  rr path/.style={
    to path={%
      \pgfextra
      \def\tikz@to@path{-- (\tikztotarget)}% default
      \pgf@process{\tikz@scan@one@point\pgfutil@firstofone(\tikztostart)\relax}%
      \pgf@xa=\pgf@x \pgf@ya=\pgf@y % start is xa, ya
      \iftikz@shapeborder % is start a node?
        \expandafter\ifx\csname pgf@sh@ns@\tikz@pp@name{\tikztostart}\endcsname\tikz@roundedrectanglex@text
          % this is a rrX!
          \pgf@process{\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget)\relax}%
          \pgf@xb=\pgf@x \pgf@yb=\pgf@y % target is xb, yb
          \iftikz@shapeborder % is target a node?
            \expandafter\ifx\csname pgf@sh@ns@\tikz@pp@name{\tikztotarget}\endcsname\tikz@roundedrectanglex@text
              % they're both a rrX
              % check if they're vertical
              \pgf@xc=\pgf@xb\advance\pgf@xc-\pgf@xa
              \ifdim\pgf@xc<0pt \pgf@xc=-\pgf@xc\fi
              \ifdim\pgf@xc<1pt
                % they're vertical!
              \else
                \ifdim\pgf@xa<\pgf@xb % start left of target
                  \pgf@process{\pgfpointlineattime{+.5}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztostart.center east)\relax}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget.center west)\relax}}%
                  \edef\middlepoint{\the\pgf@x,\the\pgf@y}%
                \else
                  \pgf@process{\pgfpointlineattime{+.5}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztostart.center west)\relax}
                    {\tikz@scan@one@point\pgfutil@firstofone(\tikztotarget.center east)\relax}}%
                  \edef\middlepoint{\the\pgf@x,\the\pgf@y}%
                \fi
                \edef\tikz@to@path{--(\middlepoint)--(\tikztotarget)}%
              \fi
            \fi
          \fi
        \fi
      \fi
      \expandafter\endpgfextra\tikz@to@path\tikztonodes
    }
  }
}
\makeatother
\tikzset{n/.style={shape=rounded rectangle X, draw, ultra thick, rounded rectangle arc length=180}}
\begin{document}
\begin{tikzcd}
  |[n]| ABCD \ar[rd,rr path] \ar[r,rr path] \ar[d, rr path] & |[n]| EFG\\
  |[n]| ABCDEFGH & |[n]| EFG
\end{tikzcd}
\end{document}

输出

Question 2

这主要提供概念验证。它可能需要根据您的具体用例进行调整，因为我根据您的示例做出的假设可能在更一般的情况下无效。尽管如此，它表明这可以通过交叉点来实现。

它还展示了如何使用\subnodes（来自tikzmark) 定义节点的各部分，这些部分可用作路径的“目标”。在ABCD节点中，D被封装在中\subnode，然后可以将其作为独立的节点来引用。

有了这些，放置连接线的技术就很简单了：定义（并保存）所需目标之间的线（在本ABCD例中，这是D子节点，但它也可以是新的锚点），然后在它与每个节点的边界路径相交的地方将其切断。一旦切断，就可以移除外部部分，只留下节点之间的部分。

在这种tikz-cd情况下，所有这些都被打包成一种样式。在我的版本中，由于我使用\subnodes 作为目标，因此它需要两个参数，即子节点。这些可以留空，在这种情况下，它会默认返回主节点。这使用了一些将name prefix节点名称包装在一些额外绒毛中的功能。在撰写本文时，要name prefix与子节点一起使用，需要 tikzmark 的开发版本，可从上面的 github 链接获得 - 这将很快进入 CTAN。

唯一额外的麻烦tikz-cd是定义节点的路径稍后会移动，因此通过spath3库保存的路径不是实际路径。幸运的是，这种转变很容易解决（但这可能是我的一个假设），路径也会相应地进行调整。

\documentclass{article}
%\url{https://tex.stackexchange.com/q/619274/86}
\usepackage{tikz}
\usetikzlibrary{
  cd,
  shapes.misc,
  positioning,
  calc,
  intersections, % <- to calculate intersections of paths
  tikzmark, % <- for subnodes
  spath3 % <- for path manipulations
}

\makeatletter
\tikzset{
  % This is a useful diagnostic tool that I frequently use
  show node name/.code={%
    \show\tikz@fig@name%
  },
  defaultNodeStyle/.style={
    shape=rounded rectangle,
    draw,
    anchor=center,
    % Save the node's path, needs to be global as nodes
    % have their own scope. Use the node's name as the path name.
    spath/save global/.expand once=\tikz@fig@name,
    % Append the node's name to the subnode name
    name prefix/.append/.expanded=\tikz@fig@name
  },
  % The two arguments allow for subnodes to be used as targets
  % This could be easily be adapted to using the additional anchors instead
  N/.style 2 args={
    to path={%
      \pgfextra{%
        % The target is split into \tikzcd@ar@target and \tikzcd@endanchor
        % the target anchor doesn't have the node prefix/suffix applied
        % and as we need to use the absolute name to access the node path
        % then we need a version with that applied.
        \edef\@arrowtarget{\tikz@pp@name{\tikzcd@ar@target}}%
        % I'm not an expert with tikz-cd so I don't know the circumstances
        % in which \tikzcd@endanchor is *not* empty, but just in case it isn't
        % then we'd better use it. If it is empty then we default to center.
        \ifx\tikzcd@endanchor\pgfutil@empty
        \def\@arrowtargetanchor{center}%
        \else
        \let\@arrowtargetanchor=\tikzcd@endanchor
        \fi
        % Now we construct our path that will become the arrow path.
        % The 'overlay' is probably not needed, but let's be cautious.
        % There's something a bit weird going on with node name
        % prefix/suffices here since \tikztostart does have them applied
        % while \tikzcd@ar@target doesn't, but nevertheless this seems
        % to work. Weird.
        \path[overlay, spath/save=arrow path] (\tikztostart#1.center) -- (\tikzcd@ar@target#2.\@arrowtargetanchor);
        % Now we get to work with our intersections.
        \tikzset{
          % Clone the node paths
          spath/clone={source path}{\tikztostart},
          spath/clone={target path}{\@arrowtarget},
          % The node paths are currently centred at the origin, we
          % need to shift them to their eventual locations.
          % I suspect this is because of them being part of a matrix
          % so if this were used in a non-matrix context these two
          % lines would not be needed.  Presumably there's some
          % "if matrix" test that could be used here
          spath/transform={source path}{shift={(\tikztostart.center)}},
          spath/transform={target path}{shift={(\@arrowtarget.center)}},
          % Now we intersect the arrow path with the node outlines
          % and split the arrow path where it meets the node outlines
          spath/split at intersections with/.expanded={arrow path}{source path},
          spath/split at intersections with/.expanded={arrow path}{target path},
          % I've naively assumed that the arrow path crosses each node path
          % once, but for a proof-of-concept that doesn't seem too bad
          % an assumption
          spath/remove components={arrow path}{1,3}
        }
        % Finally, render the arrow
      } [spath/use=arrow path]
    }
  }
}
\makeatother

\begin{document}
\begin{tikzpicture}[
  % Needed for subnodes to work
  remember picture,
  % avoid name clashes with the tikz-cd environment
  name prefix=pic-,
  tikzmark prefix=pic-
]
% The subnode here basically makes the 'D' into a viable target
  \node[rounded rectangle, draw, spath/save global=abcd] (A) at (0,0) {ABC\subnode{D}{D}};
  \node[circle, draw, spath/save global=e] (E) at (1,-1) {E};
  \draw[ultra thick,red] (E.center) -- (A.center);
\draw (E) -- (A);
% This is the path that we'll use to draw our path, normally we
% wouldn't draw this particular path
\draw[ultra thick,red,spath/save=ed] (E.center) -- (D.center);
% This is the "desired outcome" path
\draw[thick,blue] ($(A.north east)!.5!(A.south east)$) -- (E.center);
% These keys split the connection path where it meets each node boundary
% and then throws away the first and third components, leaving only the middle
\tikzset{
  spath/split at intersections with={ed}{abcd},
  spath/split at intersections with={ed}{e},
  spath/remove components={ed}{1,3}
}
% Lastly, we use the remaining part of the path.
\draw[spath/use={ed},green];
\end{tikzpicture}

\begin{tikzcd}[
  remember picture,
  name prefix=cd-,
  tikzmark prefix=cd-,
]
  |[defaultNodeStyle]| ABC\subnode{D}{D} \ar[rd,N=D{}] & |[defaultNodeStyle]| EFG\\
  |[defaultNodeStyle]| ABCD & |[defaultNodeStyle]| EFG
\end{tikzcd}

\end{document}

结果如下：

Answer