![flalign 环境中的行号](https://linux22.com/image/256660/flalign%20%E7%8E%AF%E5%A2%83%E4%B8%AD%E7%9A%84%E8%A1%8C%E5%8F%B7.png)
为什么以下示例中多了一个编号行?我想删除它,以便行号仅与指定的方程式相对应。我意识到这是由于我的错误造成的,但我看不出我做错了什么:
\documentclass{article}
\usepackage{amsmath}
\usepackage{lineno}
\linenumbers
\date{}
\begin{document}
\begin{linenomath*}
\begin{flalign}
&Q_{h} = \rho_{a}C_{pa}C_{h}U_{a}\left(T_{s} - T_{a}\right)&\\
&Q_{e} = \rho_{a}LC_{e}U_{a}\left(q_{s}^{*}-q_{s}\right)&\\
&Q_{b} = 0.985\sigma{T_{s}^{4}}\left(\left(0.39-0.05\right)e_{a}^{0.5}\right )\left(1-{0.6 {nc}}\right)
\end{flalign}
\end{linenomath*}
\end{document}
答案1
lineno
最终对amsmath
用于定位对齐的一些内部框进行编号。未经广泛测试,但此修改lineno
可检查它是否对零尺寸框进行编号。
\documentclass{article}
\usepackage{amsmath}
\usepackage{lineno}
\makeatletter
\def\MakeLineNo{%
\@LN@maybe@normalLineNumber % v4.31
\boxmaxdepth\maxdimen\setbox\z@\vbox{\unvbox\@cclv}%
\@tempdima\dp\z@ \unvbox\z@
\setbox2\lastbox
\nointerlineskip\copy2
\ifdim\ht2>\z@
\sbox\@tempboxa{\hb@xt@\z@{\makeLineNumber}}%
\stepLineNumber
\else
\setbox\@tempboxa\hbox{}%
\fi
\ht\@tempboxa\z@ \@LN@depthbox
%%
\@LN@do@vadjusts
\count@\lastpenalty
%%
\ifnum\outputpenalty=-\linenopenaltypar
\ifnum\count@=\z@ \else
%%
\xdef\@LN@parpgbrk{%
\penalty\the\count@
\global\let\noexpand\@LN@parpgbrk
\noexpand\@LN@screenoff@pen}% v4.4
%%
\fi
\else
%%
\@tempcnta\outputpenalty
\advance\@tempcnta -\linenopenalty
%%
\penalty \ifnum\count@<\@tempcnta \@tempcnta \else \count@ \fi
%%
\fi
}
\makeatletter
\linenumbers
\date{}
\begin{document}
\begin{linenomath*}
\begin{flalign}
&Q_{h} = \rho_{a}C_{pa}C_{h}U_{a}\left(T_{s} - T_{a}\right)&\\
&Q_{e} = \rho_{a}LC_{e}U_{a}\left(q_{s}^{*}-q_{s}\right)&\\
&Q_{b} = 0.985\sigma{T_{s}^{4}}\left(\left(0.39-0.05\right)e_{a}^{0.5}\right )\left(1-{0.6 {nc}}\right)
\end{flalign}%
aaa
\end{linenomath*}%
\end{document}