Tikz，我可以在表格或矩陣中使用多重網格嗎？

Question

我負責宿舍裡的小商店，我們需要一些會計系統。人們用信用支付。

通常它是在紙上完成的，人們拿起啤酒並在他們的名字行和啤酒欄中畫一條線。我厭倦了數線，因為人們真的不擅長畫直線（尤其是當他們喝醉的時候），所以我想製作一個新的佈局，讓人們劃過一個小方塊而不是畫線。

這就是我現在所擁有的：

\documentclass{article}
\usepackage{tikz,tabularx}
\usepackage{mathpazo}
\usetikzlibrary{matrix}
\newcounter{row}
\newcounter{col}
\usepackage{geometry}
 \geometry{
 a4paper,
 total={210mm,297mm},
 left=10mm,
 right=10mm,
 top=10mm,
 bottom=10mm,
 }


\begin{document}
\begin{tikzpicture}[scale=.27]
\begin{scope}[yshift=+50.8cm,xshift=+30.8cm]
\matrix(dict)[matrix of nodes, nodes={align=center,text width=3cm},
        row 1/.style={anchor=south},
        column 1/.style={nodes={text width=2.5cm,align=right,text height=0.57cm}}
    ]{
        ~ &  $b33r - 5kr$  & $Candy - 2kr $ & $b_1b_2$ & $b_1b_3$ & $b_2b_3$ & $b_1b_2b_3$\\ \hline
        Pilliam \\ \cline{1-2}
        Rilliam \\ \cline{1-2}
        Tilliam \\ \cline{1-2}
        Nilliam \\ \cline{1-2}
        Milliam \\ \cline{1-2}
        Qilliam \\ \cline{1-2}
        Pilliam \\ \cline{1-2}
        Rilliam \\ \cline{1-2}
        Tilliam \\ \cline{1-2}
        Nilliam \\ \cline{1-2}
        Milliam \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        Qilliam \\ \cline{1-2}
        Qilliam \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        Pilliam \\ \cline{1-2}
        Rilliam \\ \cline{1-2}
        Tilliam \\ \cline{1-2}
        Nilliam \\ \cline{1-2}
        Milliam \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\ \cline{1-2}
        William \\
    }; 
    \end{scope}
    %\draw(dict-1-1.south west)--(dict-1-8.south east);
    \draw(dict-1-1.north east)--(dict-33-1.south east);
    \draw(dict-1-2.north east)--(dict-1-2.south east);
    \draw(dict-1-3.north east)--(dict-1-3.south east);

    %[xshift=-15.9cm,yshift=-28.8cm]

    \begin{scope}
  \draw (0, 0) grid (60, 99);
    \draw[very thick, scale=3] (0, 0) grid (20, 33);
    \end{scope}
  \end{tikzpicture}
\end{document}

一個大正方形裡面有九個小正方形，所以我知道當一個大正方形穿過時，它就算九個。

問題是行線與網格線不符。

請幫忙隨意製作比我更棒的東西。

Answer 1

我不太接受你的做法。但我認為這會給你想要的結果。

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

%% create a box to save the grid in (saves some time since 
%% the grid won't have to be recreated for each cell).     
\def\aeunit{0.25cm}
\newsavebox\tictactoebox
\begin{lrbox}\tictactoebox
  \begin{tikzpicture}[x=\aeunit-0.4pt/3,
                      y=\aeunit-0.4pt/3,
                      every node/.style={outer sep=0pt,inner sep=0pt,minimum size=\aeunit-0.4pt},
                      baseline={($(current bounding box.center)+(0,-\aeunit/2)$)},
                     ]
    \draw    (0,0) rectangle (3,3);
    \foreach \myn in {0,1,...,3}
      {    \draw (0,\myn) -- (3,\myn); }
    \foreach \myn in {0,1,...,3}
      {    \draw (\myn,0) -- (\myn,3); }
  \end{tikzpicture}
\end{lrbox}

%% draw the extended vertical lines between marked columns
%% #1 column where to draw vertical line                  
\newcommand\aeDrawBorderAt[1]{%%
  \coordinate (B/C#1/top) at (R1/C#1);
  \coordinate (B/C#1/bot) at (R\aeMaxRowCount/C#1|-R\aeMaxRowCount/bot);
  \draw[blue,line width=0.6pt]
    (B/C#1/top)
    ++
    (0,4pt+\baselineskip)
    --
    (B/C#1/bot);
  }

%% label the columns              
%% #1 left column number          
%% #2 right column number         
%% #3 content for labeling columns
\newcommand\aeLabelColumns[3]{%%
  \path (R1/C#1) -- (R1/C#2) node[midway,above,yshift=0pt] {#3};
  \aeDrawBorderAt{#1}
  \aeDrawBorderAt{#2}
}

%% #1 = number of columns to create
%% #2 = list of names              
\newcommand\aebuildtable[2]{%%
  \foreach \myn [count=\mycnt] in {#2}
  {
    %% previous value 
    \pgfmathsetmacro\myp{int(\mycnt-1)}
    %% set nodes that fix each row
    \ifnum\mycnt=1\relax
      \coordinate (R\mycnt) at (0,\mycnt-1);
    \else                   
      \coordinate (R\mycnt) at (R\myp/bot);
    \fi
    %% a coordinate for the bottom of each row
    \coordinate (R\mycnt/bot)  at ($(R\mycnt)+(0,-3*\aeunit)$);
    %% node for name:
    \node[anchor=north east,outer sep=0pt] (N\mycnt) at ($(R\mycnt)+(-4pt,-\aeunit/2)$) {\myn};
    \foreach \mycol in {0,...,#1}
    {
      \coordinate (R\mycnt/C\mycol) at ($(R\mycnt)+({\mycol*(\aeunit*3)},0)$);
      \node[outer sep=0pt,
            inner sep=0pt,
            anchor=north west,]  
           at (R\mycnt/C\mycol.north west) {\usebox\tictactoebox};
    }
    \xdef\aeMaxRowCount{\mycnt}
  }
}

\begin{document}

\begin{tikzpicture}

  \aebuildtable{10}{Mary,Jane,Scott,William,Todd}

  \aeLabelColumns{0}{4}{$b33r-5kr$}
  \aeLabelColumns{4}{8}{$Candy-2kr$}

\end{tikzpicture}

\end{document}