如何连接具有 45 度和 0 度线的节点（可能通过样式？）

Question

我想按以下方式连接节点：

\documentclass{standalone}

\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\node (foo) at (0,0) {(0,0)};
\node (bar) at (5,3) {(5,3)};

\node (baz) at (0,-1) {(0,-1)};
\node (boo) at (5,1) {(5,1)};

% What I want to happen
\draw (foo) -- ++(3,3) -- (bar);
\draw (baz) -- ++(2,2) -- (boo);

% The kind of syntax I would like, but if there is a more
% 'correct' syntax I'm happy to use it
%\draw[**magic**] (foo) -- (bar);
%\draw[**magic**] (baz) -- (boo);

\end{tikzpicture}
\end{document}

我知道一定有一个具有样式的解决方案，但我的 TikZ 技能目前不足以知道如何或在哪里查找。

我怎样才能按照上述风格连接节点？

Answer 1

我改编了代码保罗·加博里茨 回答使其更加通用。

我使用abs并sign允许其他用例。

更通用的版本：

选择to[diagonal line]连接后，首先对角线（45°）然后水平或垂直线绘制一条直线。如果起点和终点之间的角度是 45° 的倍数，则只绘制一条直线。

代码：

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}

\tikzset{
    diagonal line/.style={
        to path={
            let \p{start}=(\tikztostart),
                \p{target}=(\tikztotarget),
                \p{diff}=({\x{target}-\x{start}}, {\y{target}-\y{start}}),
                \p{absdiff}=({abs(\x{diff})}, {abs(\y{diff})}),
                \n{mindiff}={min(\x{absdiff}, \y{absdiff})},
                \p{inter}=(
                    {sign(\x{diff}) * \n{mindiff}},
                    {sign(\y{diff}) * \n{mindiff}}
                )
            in 
                \ifnum \ifdim\x{target}=\x{start} 1
                \else \ifdim\y{target}=\y{start} 1
                \else \ifdim\x{absdiff}=\y{absdiff} 1
                \else 0\fi\fi\fi=1 %primitive tex or condition
                    -- (\tikztotarget)
                \else
                    --++ (\p{inter}) -- (\tikztotarget)
                \fi
        },
    },
}

\begin{document}
    \begin{tikzpicture} 
        \foreach \d in {1,...,16} {
            \node (a\d) at ($(0,0)+(90-\d*22.5:12mm)$) {\footnotesize $a_{\d}$};
            \node (b\d) at ($(0,0)+(90-\d*22.5:35mm)$) {\footnotesize $b_{\d}$};
            \draw (a\d) to[diagonal line] (b\d);
        }
    \end{tikzpicture}
\end{document}

结果：

较旧的非通用版本

代码：

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\tikzset{
    special line/.style={
        to path={
            let \p{start}=(\tikztostart), \p{target}=(\tikztotarget),
            \p{inter}=({\x{start} + sign(\x{target}-\x{start}) * abs(\y{target}-\y{start})}, \y{target})
            in -- (\p{inter}) -- (\tikztotarget)
        },
    },
}

\begin{document}
    \begin{tikzpicture}
        \node (x) at (0,0) {(0,0)};
        \node (y1) at (5,2) {(5,2)};
        \node (y2) at (5,-2) {(5,-2)};
        \node (y3) at (-5,-2) {(-5,-2)};
        \node (y4) at (-5,2) {(-5,2)};
        
        \draw (x) to[special line] (y1);
        \draw (x) to[special line] (y2);
        \draw (x) to[special line] (y3);
        \draw (x) to[special line] (y4);
    \end{tikzpicture}
\end{document}

结果：

使用 Pauls 或 Ignasis 代码的结果是：

Answer 2

calc以下是使用TikZ 库、let操作和样式的解决方案to path：

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\tikzset{
  special line/.style={
    to path={
      let \p{start}=(\tikztostart), \p{target}=(\tikztotarget),
      \p{inter}=(\x{start}+\y{target}-\y{start},\y{target})
      in -- (\p{inter}) -- (\tikztotarget)
    },
  },
}

\begin{document}
\begin{tikzpicture}

  \node (foo) at (0,0) {(0,0)};
  \node (bar) at (5,3) {(5,3)};

  \node (baz) at (0,-1) {(0,-1)};
  \node (boo) at (5,1) {(5,1)};

  \draw (foo) to[special line] (bar);
  \draw (baz) to[special line] (boo);
\end{tikzpicture}
\end{document}

Answer 3

style与 Paul 的解决方案不同，而是command：

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

\newcommand{\connect}[3][]{%
\draw[#1] (#2) let \p1 = ($(#3)-(#2)$) in --++(\y1,\y1)--(#3); 
}

\begin{document}
\begin{tikzpicture}

\node (a) at (0,0) {A};
\node (b) at (5,3) {B};

\connect{a}{b}

\node (c) at (0,-1) {C};
\node (d) at (5,1) {D};

\connect{c}{d}
\connect[red]{a}{d}
\connect[dashed, blue]{c}{b}
\end{tikzpicture}
\end{document}