如何使用 tikz 获取节点之间的矩形线？

Question

抱歉，如果重复了（我确定是重复了），但我真的找不到手动操作的方法。我已经创建了下一个区块链。

这里你可以看到最后一个块和第三个块之间的曲线。但我想要的是这样的：

问题是：我怎样才能以这种方式连接它们？我相信这很简单。MWE：

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{shapes,arrows,positioning,automata}

\begin{document}
\begin{tikzpicture}
\node[draw, thick, rectangle] (0) {Data arrival};
\node[draw, thick, rectangle, below of=0] (1) {Coks backoff procedure};
\node[draw, thick, rectangle, below of= 1] (2) {Get trigger-frame?};
\node[draw, thick, rectangle, below of= 2] (3) {Performs backoff procedure};
\node[draw, thick, shape aspect=2.7, diamond, below =0.5cm of 3] (4) {$b<0$?};

 \path[>=latex, auto = right,every loop]

(0) edge[] node {} (1)
(1) edge node {} (2)
(2) edge node {} (3)
(3) edge node {} (4)
(4.east) edge[in=0, out=0, looseness=3] node[right] {N} (2.east)
;     
\end{tikzpicture}
\end{document}

Answer 1

下面的代码使用以下路径构造。还有更多，请参阅 tikz 手册或任何互联网上有很多例子）。

• (a) -- (b)a描述从到 的直线路径b。

• (a) -| (b)a描述从到 的路径b，首先水平移动直到低于或高于b然后垂直移动。

• (a) |- (b)是一样的，但是从垂直方向开始，然后继续水平方向。

• ++(1,-2)表示相对于前一个位置的位置，向右移动 1 个单位，向下移动 2 个单位。因此，(a) -| ++(1,-2)路径从 开始a，向右移动 1 个单位，向下移动 2 个单位。作为副作用，新的起始位置已移动到路径的末尾。因此

  \draw (a) -| ++(1,-2) -| ++(-1,2);
  

  绘制一个矩形，其最终位置再次为a。

• +(1,-2)和 基本相同++(1,-2)，只是位置不动。

  \draw (a) -| +(1,-2) -| +(-1,2);
  

  绘制两条各有一定角度的线，一条在 的右边a，一条在 的左边a。

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{shapes,arrows,positioning,automata}

\begin{document}
\begin{tikzpicture}
  [>=latex,
   action/.style={draw,thick},
   test/.style={draw, thick, shape aspect=2.7, diamond}
  ]
\node[action] (0) {Data arrival};
\node[action, below=of 0] (1) {Coks backoff procedure};
\node[action, below=of 1] (2) {Get trigger-frame?};
\node[action, below=of 2] (3) {Performs backoff procedure};
\node[test, below= 0.5cm of 3] (4) {$b<0$?};
\node[action, left=of 3] (5) {Do something};
\path[->]
  (0) edge node {} (1)
  (1) edge node {} (2)
  (2) edge node {} (3)
  (3) edge node {} (4);
\draw[->] (4) -- node[below right,pos=0.2]{N} ++(3,0) |- (2);     
\draw[->] (4) -| node[below left,pos=0.1]{Y} (5);
\draw[->] (5) |- (2); 
\draw[->] (4) --node[right] {maybe} +(0,-1.5);     
\end{tikzpicture}
\end{document}

Answer 2

略有修改格诺特答案（为了快乐和锻炼）：

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{arrows, chains, positioning, shapes}% added chains
\makeatletter
\tikzset{supress chain/.code={\def\tikz@after@path{}}}% added for suppress joining of nodes
\makeatother

\begin{document}
\begin{tikzpicture}[
            > = latex,
node distance = 5mm and 7mm,% added (not used default value)
  start chain = going below,% activation of chains
action/.style = {draw, thick, on chain, join= by ->},% nodes are in chain and connected by ->
  test/.style = {diamond, draw, thick, shape aspect=2.4, on chain, join= by ->}% node is in the chain and connected by -> with previous node
  ]
\node[action]   (n0) {Data arrival};
\node[action]   (n1) {Coks backoff procedure};
\node[action]   (n2) {Get trigger-frame?};
\node[action]   (n3) {Performs backoff procedure};
\node[test]     (n4) {$b<0$?};
\node[action, 
      supress chain, % this node is not connected with join
      left=of n3]   (n5) {Do something};
\draw[->] (n4) -| node[below,pos=0.25] {Y}  (n5); % left feedback loop
\draw[->] (n5) |- (n0); % left feedback loop
\draw[->] (n4) -|  ([xshift=5mm] n3.east) node[below,pos=0.25] {N} |- (n2); % right feedback loop
\draw[->] (n4.south) -- node[right] {maybe} ++ (0,-1.1);
\end{tikzpicture}
\end{document}

结果几乎是一样的：