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最后一列的宽度不符合要求

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最后一列的宽度不符合要求

我正在尝试使用包创建一个表格tabularx。我希望第一列具有一定的宽度（38 毫米），而所有其他列具有另一个宽度（15 毫米）。但最后一列似乎有一个固定的宽度值，如图 1 所示，无论我为其设置的值是什么。

该图说明了正在发生的事情

代码如下：

\documentclass[12pt]{report}  
\usepackage{multirow,multicol}  
\usepackage{tabularx}

\begin{document}    

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

    \renewcommand{\arraystretch}{1.5} 

    % Definning column specifications:
    \renewcommand{\tabularxcolumn}[1]{>{\normalsize}m{#1}}
    \newcolumntype{A}{>{\normalsize\centering\arraybackslash}m{38mm}}
    \newcolumntype{B}{>{\normalsize\centering\arraybackslash}m{15mm}}

    \begin{table}[!hbt]
        \centering
        \caption{Shape functions for quadrilateral elements.} \label{tab.:2DQuadShapeFunc}~\\[-3mm]
        %\begin{tabular}{lc*{4}c}    
        \begin{tabularx}{\textwidth}{ABBBBB}
            \multicolumn{1}{c}{}                     & \multicolumn{5}{c}{Included only if node $i$ is defined} \\
            \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{$i=5$}             & \multicolumn{1}{c|}{$i=6$}   
            & \multicolumn{1}{c|}{$i=7$}             & \multicolumn{1}{c|}{$i=8$} 
            & \multicolumn{1}{c|}{$i=9$}             \\ \hline
            \multicolumn{1}{l|}{$h_1=\frac{1}{4}(1-r)(1-s)$} & \multicolumn{1}{c|}{$-\frac{1}{2}h_5$}   & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{}                  &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_8$}   &
            \multicolumn{1}{c|}{$-\frac{1}{4}h_9$}   \\
            \multicolumn{1}{l|}{$h_2=\frac{1}{4}(1+r)(1-s)$} & \multicolumn{1}{c|}{$-\frac{1}{2}h_5$}   &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_6$}   & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{}                  &
            \multicolumn{1}{c|}{$-\frac{1}{4}h_9$}   \\
            \multicolumn{1}{l|}{$h_3=\frac{1}{4}(1+r)(1+s)$} & \multicolumn{1}{c|}{}                    &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_6$}   &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_7$}   & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{$-\frac{1}{4}h_9$} \\
            \multicolumn{1}{l|}{$h_4=\frac{1}{4}(1-r)(1+s)$} & \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{$-\frac{1}{2}h_7$} &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_8$}   &
            \multicolumn{1}{c|}{$-\frac{1}{4}h_9$}   \\ \cline{1-1}
            $h_5=\frac{1}{2}(1-r^2)(1-s)$                    & \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{}                  & \multicolumn{1}{c|}{}      
            & \multicolumn{1}{c|}{$-\frac{1}{2}h_9$} \\ \cline{1-2}
            $h_6=\frac{1}{2}(1-s^2)(1+r)$                    &                                        & \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}      
            & \multicolumn{1}{c|}{}                  &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_9$}   \\ \cline{1-3}
            $h_7=\frac{1}{2}(1-r^2)(1+s)$                    &                                        &                   & \multicolumn{1}{c|}{}      
            & \multicolumn{1}{c|}{}                  &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_9$}   \\ \cline{1-4}
            $h_8=\frac{1}{2}(1-s^2)(1-r)$                    &                                        &                   &                            
            & \multicolumn{1}{c|}{}                  &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_9$}   \\ \cline{1-5}
            $h_9=(1-r^2)(1-s^2)$                             &                                        &                   &                            
            &                                        & \multicolumn{1}{c|}{}      
            \\ \hline
        \end{tabularx}
    \end{table}

    \renewcommand{\arraystretch}{1}

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

\end{document}

谁能帮我吗？

答案1

不需要tabularx！您将用 覆盖每一行的最后一列规范c，它不使用 的原因是m{1.5cm}：

\documentclass[12pt]{report}  
\usepackage{multirow,multicol}  
\usepackage{ragged2e,array}
\begin{document}    

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

    \renewcommand{\arraystretch}{1.5} 

    \newcolumntype{A}{>{\Centering}m{38mm}}
    \newcolumntype{B}{>{\Centering}m{15mm}}

\begin{table}[!hbt]
\centering
\caption{Shape functions for quadrilateral elements.}\label{tab.:2DQuadShapeFunc}~\\[-3mm]
        %\begin{tabular}{lc*{4}c}    
\begin{tabular}{A*5B | }
    & \multicolumn{5}{c}{Included only if node $i$ is defined} \\
 \multicolumn{1}{c|}{}   & 
\multicolumn{1}{c|}{$i=5$}             & \multicolumn{1}{c|}{$i=6$}   
    & \multicolumn{1}{c|}{$i=7$}             & \multicolumn{1}{c|}{$i=8$} 
    & $i=9$             \\ \hline
    \multicolumn{1}{l|}{$h_1=\frac{1}{4}(1-r)(1-s)$} & 
    \multicolumn{1}{c|}{$-\frac{1}{2}h_5$}   & \multicolumn{1}{c|}{}        
    & \multicolumn{1}{c|}{}                  &
    \multicolumn{1}{c|}{$-\frac{1}{2}h_8$}   &
    $-\frac{1}{4}h_9$   \\
\multicolumn{1}{l|}{$h_2=\frac{1}{4}(1+r)(1-s)$} & 
        \multicolumn{1}{c|}{$-\frac{1}{2}h_5$}   &
    \multicolumn{1}{c|}{$-\frac{1}{2}h_6$}   & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{}                  &
            $-\frac{1}{4}h_9$   \\
            \multicolumn{1}{l|}{$h_3=\frac{1}{4}(1+r)(1+s)$} & 
            \multicolumn{1}{c|}{}                    &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_6$}   &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_7$}   & \multicolumn{1}{c|}{}        
            &  $-\frac{1}{4}h_9$ \\
            \multicolumn{1}{l|}{$h_4=\frac{1}{4}(1-r)(1+s)$} & 
            \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{$-\frac{1}{2}h_7$} &
            \multicolumn{1}{c|}{$-\frac{1}{2}h_8$}   &
            $-\frac{1}{4}h_9$   \\ \cline{1-1}
            $h_5=\frac{1}{2}(1-r^2)(1-s)$                    & 
            \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}        
            & \multicolumn{1}{c|}{}                  & \multicolumn{1}{c|}{}      
            & $-\frac{1}{2}h_9$ \\ \cline{1-2}
            $h_6=\frac{1}{2}(1-s^2)(1+r)$                    
            &                                        & 
            \multicolumn{1}{c|}{}                    & \multicolumn{1}{c|}{}      
            & \multicolumn{1}{c|}{}                  &
            $-\frac{1}{2}h_9$   \\ \cline{1-3}
            $h_7=\frac{1}{2}(1-r^2)(1+s)$                    
            &                                        &                   & 
            \multicolumn{1}{c|}{}      
            & \multicolumn{1}{c|}{}                  &
            $-\frac{1}{2}h_9$   \\ \cline{1-4}
$h_8=\frac{1}{2}(1-s^2)(1-r)$ & & & &\multicolumn{1}{c|}{} &    $-\frac{1}{2}h_9$   \\ 
\cline{1-5}
            $h_9=(1-r^2)(1-s^2)$                             
            &                                        &                   
            &                            
            &                                        & \\ \hline
        \end{tabular}
    \end{table}

    \renewcommand{\arraystretch}{1}

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

\end{document}

在此处输入图片描述

也许定义垂直线然后在\multicolumn不应该出现的地方使用它更容易！这里我展示了前七行：

\documentclass[12pt]{report}  
\usepackage{multirow,multicol}  
\usepackage{ragged2e,array}
\begin{document}    

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

    \renewcommand{\arraystretch}{1.5} 

\newcolumntype{A}{m{38mm}}
\newcolumntype{B}{>{\Centering}m{15mm}}
\let\MC\multicolumn

\begin{table}[!hbt]
\centering
\caption{Shape functions for quadrilateral elements.}\label{tab.:2DQuadShapeFunc}~\\[-3mm]
        %\begin{tabular}{lc*{4}c}    
\begin{tabular}{A | *5{B|} }
\MC{1}{c}{} & \multicolumn{5}{c}{Included only if node $i$ is defined} \\
    & $i=5$ & $i=6$ & $i=7$ & $i=8$ & $i=9$ \\ \hline
$h_1=\frac{1}{4}(1-r)(1-s)$ & $-\frac{1}{2}h_5$ & & & $-\frac{1}{2}h_8$ & 
    $-\frac{1}{4}h_9$ \\
$h_2=\frac{1}{4}(1+r)(1-s)$ & $-\frac{1}{2}h_5$ & $-\frac{1}{2}h_6$ & & & 
    $-\frac{1}{4}h_9$   \\
$h_3=\frac{1}{4}(1+r)(1+s)$ & & $-\frac{1}{2}h_6$ & $-\frac{1}{2}h_7$ & &   
    $-\frac{1}{4}h_9$ \\
$h_4=\frac{1}{4}(1-r)(1+s)$ & & & $-\frac{1}{2}h_7$ & $-\frac{1}{2}h_8$ &
    $-\frac{1}{4}h_9$   \\ \cline{1-1}
\MC{1}{l}{$h_5=\frac{1}{2}(1-r^2)(1-s)$} & & & & & $-\frac{1}{2}h_9$ \\ \cline{1-2}
\MC{1}{l}{$h_6=\frac{1}{2}(1-s^2)(1+r)$} & \MC{1}{c}{} & & & & $-\frac{1}{2}h_9$   \\ 
\cline{1-3}
\MC{1}{l}{$h_7=\frac{1}{2}(1-r^2)(1+s)$} & \MC{1}{c}{}& \MC{1}{c}{}& & & 
$-\frac{1}{2}h_9$   \\ \cline{1-4}
\end{tabular}
    \end{table}

    \renewcommand{\arraystretch}{1}

    Some text! Some text! Some text! Some text! Some text! Some text! Some 
    text! Some text!

\end{document}

在此处输入图片描述

答案2

:tabularx第一列宽度设置为所需的 38 毫米（不过，我的意见是最好使用列类型l，其余列的宽度留给tabularx计算：

\documentclass[12pt]{report}
\usepackage{makecell, multirow, tabularx}
\setcellgapes{3pt}
\let\MC\multicolumn
\usepackage{lipsum}

\begin{document}
\lipsum[11]


    \begin{table}[!hbt]
    \centering
\caption{Shape functions for quadrilateral elements.}
\label{tab.:2DQuadShapeFunc}
\makegapedcells
\begin{tabularx}{\linewidth}{>{\centering $}p{38 mm}<{$} |
                        *{5}{>{\centering\arraybackslash $}X<{$}|}}
\MC{1}{c}{} & \MC{5}{c}{Included only if node i is defined} \\
            & i=5               & i=6 & i=7 & i=8 & i=9                 \\
    \hline
h_1=\frac{1}{4}(1-r)(1-s)
            & -\frac{1}{2}h_5   & & & -\frac{1}{2}h_8 & -\frac{1}{4}h_9 \\
h_2=\frac{1}{4}(1+r)(1-s)
            & -\frac{1}{2}h_5   & -\frac{1}{2}h_6 & & & -\frac{1}{4}h_9 \\
h_3=\frac{1}{4}(1+r)(1+s)
            & & -\frac{1}{2}h_6 & -\frac{1}{2}h_7 &   & -\frac{1}{4}h_9 \\
h_4=\frac{1}{4}(1-r)(1+s)
            & & & -\frac{1}{2}h_7 & -\frac{1}{2}h_8   & -\frac{1}{4}h_9 \\
    \cline{1-1}
\MC{1}{l}{$h_5=\frac{1}{2}(1-r^2)(1-s)$}
            & & & & & -\frac{1}{2}h_9                                   \\
    \cline{1-2}
\MC{1}{l}{$h_6=\frac{1}{2}(1-s^2)(1+r)$}
            & \MC{1}{c}{} & & & & -\frac{1}{2}h_9                       \\
    \cline{1-3}
\MC{1}{l}{$h_7=\frac{1}{2}(1-r^2)(1+s)$}
            & \MC{2}{c}{}      & & & -\frac{1}{2}h_9           \\
    \cline{1-4}
    \end{tabularx}
\end{table}
\lipsum[12]

\end{document}

在此处输入图片描述

\makegapedcells为了在单元格中使用更多的垂直空间，从包中使用宏makecell，删除了列定义（因为它们没有使用）

笔记：

  • 如果你使用tabularx至少一个列必须是X类型
  • 由于宏\makegapedcells与列类型不兼容，因此删除了列m的重新定义。X\renewcommand{\tabularxcolumn}[1]{>{\normalsize}m{#1}}
  • 如果新命令旨在全局使用（在整个文档中），则应在序言中定义它们（这样更容易找到它们），如果它们仅在环境中有效，则在本地定义它们是合理的，在哪里定义

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