我看到了这个问题，答案是这里。我尝试使用\\\nopagebreak，但无法获得良好的长表。我的代码
\documentclass[12pt]{article}
\usepackage{tabularray}
\UseTblrLibrary{diagbox}
\UseTblrLibrary{varwidth}
\UseTblrLibrary{booktabs}
\UseTblrLibrary{counter}
\usepackage{enumitem}
\usepackage{ninecolors}
\UseTblrLibrary{amsmath}
%\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{siunitx}
\sisetup{output-decimal-marker={,}}
\UseTblrLibrary{siunitx}
\usepackage{tikz}
\usepackage[paperwidth=19cm, paperheight=26.5cm, left=1.7cm,right=1.7cm,top=1.8cm,bottom=1.7cm]{geometry}
\DefTblrTemplate{contfoot-text}{normal}{Continued on next page}
\SetTblrTemplate{contfoot-text}{normal}
\DefTblrTemplate{conthead-text}{normal}{(Continued)}
\SetTblrTemplate{conthead-text}{normal}
\SetTblrTemplate{conthead-text}{normal}
\newcounter{mycnta}
\newcommand{\mycnta}{\stepcounter{mycnta}\arabic{mycnta}}
\newcommand{\startproblem}[1]{
\SetCell[r=#1]{m}\SetRow{bg=teal9}\SetCell{bg=gray9}\mycnta
}
\begin{document}
\begin{longtblr}[
expand=\startproblem,
caption={Some text}]{
colspec = {Q[c,gray9]X[l,valign=m]Q[c]},
rowhead = 1,
vlines,
hlines,
row{1}={yellow9,font=\bfseries},
cell{1}{2-3}={halign=c},
column{1}={font=\bfseries},
}
Problem & Content & Point \\
\startproblem{5} & Solve the equation $ x^2 - 5x + 6 = 0 $ & \num{1.00} \\
& $ \Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & \num{0.25} \\
& $ x = \dfrac{-(-5) -1}{2} = 2$. & \num{0.25} \\
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & \num{0.25} \\
& The given equation has two solutions $x=2$ and $x = 3$. & \num{0.25} \\
\startproblem{4} & Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ $ y = \dfrac{2x+1}{x-2} $ $ y = \dfrac{2x+1}{x-2} $ $ y = \dfrac{2x+1}{x-2} $ $ y = \dfrac{2x+1}{x-2} $ $ y = \dfrac{2x+1}{x-2} $& \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
\startproblem{3} & Is this $ x^2 + 3x +4 >0$ true or fail? why? & \num{0.75} \\
& $ x^2 + 3x +4 >0$ true. & \num{0.25} \\
& $ x^2 + 3x +4 $ have $ \Delta = -7<0 $ and coefficient of $ x^2 $ is $ 1 >0$. & \num{0.50}
\\
\startproblem{5} & Solve the equation $ x^2 - 5x + 6 = 0 $ & \num{1.00} \\
& $ \Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & \num{0.25} \\
& $ x = \dfrac{-(-5) -1}{2} = 2$. & \num{0.25} \\
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & \num{0.25} \\
& The given equation has two solutions $x=2$ and $x = 3$. & \num{0.25} \\
\startproblem{4} & Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
\startproblem{3} & Is this $ x^2 + 3x +4 >0$ true or fail? why? & \num{0.75} \\\nopagebreak
& $ x^2 + 3x +4 >0$ true. & \num{0.25} \\
& $ x^2 + 3x +4 $ have $ \Delta = -7<0 $ and coefficient of $ x^2 $ is $ 1 >0$. & \num{0.50}
\\
\startproblem{5} & Solve the equation $ x^2 - 5x + 6 = 0 $ & \num{1.00} \\
& $ \Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & \num{0.25} \\
& $ x = \dfrac{-(-5) -1}{2} = 2$. & \num{0.25} \\
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & \num{0.25} \\
& The given equation has two solutions $x=2$ and $x = 3$. & \num{0.25} \\
\startproblem{20} & Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\\nopagebreak
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $ & \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & \num{0.25} \\
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & \num{0.25} \\
& $y'=-\dfrac{5}{(x-2)^2}$. & \num{0,25} \\
\end{longtblr}
\end{document}
答案1
- 问题出
\startproblem在表格中的最后一条命令：它跨越的行数太多，所以它突出在页面下方。 - 解决方案是手动将其分为两部分，
8-a例如8-b - 我擅自将代码稍微缩短了一些……
梅威瑟：
\documentclass[12pt]{article}
\usepackage[paperwidth=19cm, paperheight=26.5cm,
hmargin=1.7cm,
vmargin={1.8cm,1.7cm}]{geometry}
\usepackage{ninecolors}
\usepackage{tabularray}
\UseTblrLibrary{amsmath,
booktabs,
counter,
diagbox,
siunitx,
varwidth}
\sisetup{output-decimal-marker={,}}
\usepackage{amssymb}
\usepackage{enumitem}
\newcounter{mycnta}
\newcommand{\mycnta}{\stepcounter{mycnta}\arabic{mycnta}}
\newcommand{\startproblem}[1]%
{
\SetCell[r=#1]{m,bg=gray9, font=\bfseries}\mycnta
}
\begin{document}
\begin{longtblr}[
expand=\startproblem,
caption={Some text} ]{vlines,hlines,
colspec = {Q[c,gray9]
X[l,valign=m]
Q[c, si={table-format=1.2}]},
rowhead = 1,
row{1}={yellow9,font=\bfseries},
cell{1}{2-3}={halign=c},
}
Problem & Content & Point \\
\startproblem{5}
& Solve the equation $ x^2 - 5x + 6 = 0 $ & 1.00 \\*
& $\Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & 0.25 \\*
& $ x = \dfrac{-(-5) -1}{2} = 2$. & 0.25 \\*
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & 0.25 \\*
& The given equation has two solutions $x=2$ and $x = 3$.
& 0.25 \\
\startproblem{4}
& Find the derivate of the functions
$y = \dfrac{2x+1}{x-2},\ y = \dfrac{2x+1}{x-2},$
$y = \dfrac{2x+1}{x-2},\ y = \dfrac{2x+1}{x-2},\
y = \dfrac{2x+1}{x-2},\ y = \dfrac{2x+1}{x-2}$
& 0.75 \\*
& $y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\
\startproblem{3}
& Is this $ x^2 + 3x +4 >0$ true or fail? Why?
& 0.75 \\*
& $ x^2 + 3x +4 >0$ true. & 0.25 \\*
& $ x^2 + 3x +4 $ have $ \Delta = -7<0 $ and coefficient of $ x^2 $ is $ 1 >0$.
& 0.50 \\
\startproblem{5}
& Solve the equation $ x^2 - 5x + 6 = 0 $ & 1.00 \\*
& $\Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & 0.25 \\*
& $ x = \dfrac{-(-5) -1}{2} = 2$. & 0.25 \\*
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & 0.25 \\*
& The given equation has two solutions $x=2$ and $x = 3$.
& 0.25 \\
\startproblem{4}
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& 0.75 \\*
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\
\startproblem{3} & Is this $ x^2 + 3x +4 >0$ true or fail? why?
& 0.75 \\*
& $ x^2 + 3x +4 >0$ true. & 0.25 \\*
& $ x^2 + 3x +4 $ have $ \Delta = -7<0 $ and coefficient of $ x^2 $ is $ 1 >0$.
& 0.50 \\
\startproblem{5}
& Solve the equation $ x^2 - 5x + 6 = 0 $ & 1.00 \\*
& $\Delta =(-5)^2 - 4 \cdot 1 \cdot 6 = 1$, & 0.25 \\*
& $ x = \dfrac{-(-5) -1}{2} = 2$. & 0.25 \\*
& $ x = \dfrac{-(-5) + 1}{2} = 3$ & 0.25 \\*
& The given equation has two solutions $x=2$ and $x = 3$.
& 0.25 \\
\startproblem{13}-a % <--------------- changed
%\SetCell[r=13]{m,bg=gray9, font=\bfseries}\mycnta-a
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& 0.75 \\*
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\*
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& 0.75 \\*
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\*
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& 0.75 \\*
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\*
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& 0.75 \\
\SetCell[r=7]{m,bg=gray9, font=\bfseries}\themycnta-b % <--- inserted, changed
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$.
& 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\
& Find the derivaty of the function $ y = \dfrac{2x+1}{x-2} $
& \num{0.75} \\
& $ y' = \dfrac{(2x+1)'(x-2)-(2x+1)(x-2)'}{(x-2)^2}$. & 0.25 \\*
& $ y'=\dfrac{2(x-2) -(2x+1)}{(x-2)^2}$. & 0.25 \\*
& $y'=-\dfrac{5}{(x-2)^2}$. & 0,25 \\
\end{longtblr}
\end{document}
