tex4ht并且lwarp都latexml采用 latex 文件并生成 HTML。
他们使用不同的方法。
我想比较一些包含大量数学和图形的示例文档的 HTML 输出，看看这 3 个系统有何不同。tex4ht到目前为止（以及latexhtml之前），我只使用过它们。
我知道我可以学习lwarp和安装latexml（这是一个有点复杂的过程）并弄清楚如何使用它们并尝试自己做到这一点。
但我想先问问是否有人已经做过类似的事情并可以分享他们的结果。我在谷歌上搜索了一下，没有找到这样的网站可以在同一文档上比较这三个系统。
我知道的唯一一个由以下网站生成的 HTMLlwarp是这但其中并没有太多的数学内容。有人知道其他网站吗？
因为latexml我知道https://dlmf.nist.gov/它使用 Latexml。有谁知道其他网站？
更新和 MWE
这是我刚刚编写的 MWE。我尝试在其中添加一些有用的东西（我还应该添加表格，但我忘了）。
它包含一些图像、代码列表、一些数学知识，使用包含一些章节和部分的书籍类别。
我将展示 MWE 和tex4ht用于编译它的命令。我将源代码和小型 tex4ht .cfg 文件以及 HTML 输出放在一个 zip 文件中。所以它们都在一个文件夹中。编译命令是
make4ht -u -c ./nma.cfg foo.tex "html,pic-align"
哪里nma.cfg
\Preamble{xhtml,p-width}
%need this below for MATH.
\Configure{Picture}{.svg}
\begin{document}
\Configure{$}{\PicMath}{\EndPicMath}{}
\Configure{PicMath}{}{}{}{class="math";align="absmiddle"}
\EndPreamble
使用的图像是 texlive 发行版中的标准图像example-image。但有时 tex4ht 看不到此图像文件，因此.png如果需要，我还将此图像包含在 zip 文件中。
该 zip 文件位于我自己的网页上的这个文件夹中 这里
这是 MWE 的清单。这也可以通过 lualatex 编译。生成的 HTML 和 PDF 文件也位于代码清单下方：
\documentclass[11pt]{book}
\raggedbottom
\usepackage{amsmath,mathtools,amssymb}
\usepackage[activate={true,nocompatibility},final,tracking=true,factor=1100,stretch=10,shrink=10]{microtype}
\usepackage{mwe}
\usepackage{graphicx}
\DeclareGraphicsExtensions{.pdf,.PDF,.png,.PNG,.jpg,.jpeg,.JPG,.JPEG}
\usepackage{color}
\usepackage{listings}
\lstset{language=Matlab,%
breaklines=true,%
morekeywords={matlab2tikz},
keywordstyle=\color{blue},%
morekeywords=[2]{1}, keywordstyle=[2]{\color{black}},
identifierstyle=\color{black},%
showstringspaces=false,%without this there will be a symbol in the places where there is a space
numbers=left,%
numberstyle={\tiny \color{black}},% size of the numbers
numbersep=9pt, % this defines how far the numbers are from the text
emph=[1]{for,end,break},emphstyle=[1]\color{red}, %some words to emphasise
}
\usepackage{fancyvrb}
\usepackage{etex} %adds more registers
\usepackage{upquote} %to handle correct tex4ht to html conversion of `
\newcommand{\authorMe}[0]
{\author{\footnotesize \href{mailto:user@comain}{John Doe}}}%
\usepackage[us,12hr]{datetime}
\usepackage{ntheorem}
\newtheorem{theorem}{Theorem}
\usepackage[english]{babel}
\usepackage{blindtext}
\usepackage{caption}
\usepackage{float}
\usepackage{bera}
\usepackage[T1]{fontenc}
\usepackage{hyperref}
\usepackage[letterpaper,bindingoffset=0.2in,%
left=1.2in,right=1.2in,top=.8in,bottom=.8in,%
footskip=.25in]{geometry}
\DeclareMathOperator{\Res}{Res}
\begin{document}
\begin{description}
\item[] \href{/index.htm}{home}
\item[] \href{../index.htm}{up}
\end{description}
\title{My big title}
\authorMe
\date{Spring 2018 \hspace{.2in} \tiny{Compiled on \today\ at \currenttime}}
\maketitle
\tableofcontents
\chapter{Introduction}
This is main chapter. Here is a nice image
\includegraphics[width=0.7\textwidth]{example-image}
\section{some math}
\blindtext
\pagestyle{empty}
\begin{theorem}[Residue Theorem]
Let $f$ be analytic in the region $G$ except for the isolated
singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed
rectifiable curve in $G$ which does not pass through any of the
points $a_k$ and if $\gamma\approx 0$ in $G$, then
\[
\frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m
n(\gamma;a_k)\Res(f;a_k)\,.
\]
\end{theorem}
\subsection{this is a subsection}
Here is some code in minipage
\begin{minipage}{0.9\textwidth}
\begin{lstlisting}[basicstyle=\footnotesize]
%check if we converged or not
if k>opt.MAX_ITER || gradientNormTol(k)<=opt.gradientNormTol ...
|| (k>1 && levelSets(k)>levelSets(k-1))% check for getting worst
keepRunning = false;
else
....
end
\end{lstlisting}
\end{minipage}
Here is example using listings
\begin{lstlisting}
%Evaluate J(u) at u
function f = objectiveFunc(u)
u=u(:);
N = size(u,1);
f = 0;
for i = 1:N-1
f = f + 100*(u(i+1)-u(i)^2)^2 + (1-u(i))^2;
end
end
\end{lstlisting}
\subsubsection{This is subsubsection with images}
These two images should be side by side
\begin{figure}[H]
\centering
\begin{minipage}{0.48\textwidth}
\centering
\captionsetup{width=.8\textwidth}
\includegraphics[width=1\textwidth]{example-image}
\caption{Contour $J(u)$}
\end{minipage}
\hfill\begin{minipage}{0.48\textwidth}
\centering
\captionsetup{width=.8\textwidth}
\includegraphics[width=1\textwidth]{example-image}
\caption{Zooming on $J(u)$}
\end{minipage}
\end{figure}
\chapter{This is a new chapter}
Here is some verbatim
\begin{Verbatim}
K>> gradientNormTol(end-6:end)
....
16.1440020280613
17.487837406306
16.092991548592
17.4442963174089
\end{Verbatim}
\section{This is a section for more math}
\subsection{problem 1}
\underline{problem} Transform the following problem or system to set of first
order ODE $t^{2}x^{\prime\prime}+tx^{\prime}+\left( t^{2}-1\right) x=0$
\underline{solution} Since this is second order ODE, we need two state
variables, say $x_{1},x$
Let $x_{1}=x,x_{2}=x^{\prime}$, hence%
\[
\left.
\begin{array}
[c]{c}%
x_{1}=x\\
x_{2}=x^{\prime}%
\end{array}
\right\} \overset{\text{take derivative}}{\longrightarrow}\left.
\begin{array}
[c]{c}%
x_{1}^{\prime}=x^{\prime}\\
x_{2}^{\prime}=x^{\prime\prime}%
\end{array}
\right\} \overset{\text{replace RHS}}{\longrightarrow}%
\begin{array}
[c]{c}%
x_{1}^{\prime}=x_{2}\\
x_{2}^{\prime}=-\frac{x^{\prime}}{t}-\frac{\left( t^{2}-1\right) x}%
{t}=-\frac{x_{2}}{t}-\frac{\left( t^{2}-1\right) x_{1}}{t}%
\end{array}
\]
Hence the two first order ODE's are (now coupled)%
\begin{align*}
x_{1}^{\prime} & =x_{2}\\
x_{2}^{\prime} & =-\frac{x_{2}}{t}-\frac{\left( t^{2}-1\right) x_{1}}{t}%
\end{align*}
The matrix form of the above is%
\begin{align*}
\mathbf{x}^{\prime} & =A\mathbf{x}\\%
\begin{pmatrix}
x_{1}^{\prime}\\
x_{2}^{\prime}%
\end{pmatrix}
& =%
\begin{pmatrix}
0 & 1\\
-\frac{t^{2}-1}{t} & -\frac{1}{t}%
\end{pmatrix}%
\begin{pmatrix}
x_{1}\\
x_{2}%
\end{pmatrix}
\end{align*}
\subsection{Example on page 500, textbook (Edwards\&Penny, 3rd edition)}
\underline{problem} This problem was solved in textbook using matrix
exponential. Here is solved using the fundamental matrix only. Use the method
of variation of parameters to solve $\mathbf{x}^{\prime}=A\mathbf{x}%
+\mathbf{f}\left( t\right) $.%
\begin{align*}
A & =%
\begin{pmatrix}
4 & 2\\
3 & -1
\end{pmatrix}
\\
\bar{f}\left( t\right) & =%
\begin{pmatrix}
-15\\
4
\end{pmatrix}
te^{-2t}\\
\bar{x}\left( 0\right) & =%
\begin{pmatrix}
7\\
3
\end{pmatrix}
\end{align*}
\underline{Solution}
The homogeneous solution was found in the book as%
\[
\bar{x}_{h}=c_{1}%
\begin{pmatrix}
1\\
-2
\end{pmatrix}
e^{-2t}+c_{2}%
\begin{pmatrix}
2\\
1
\end{pmatrix}
e^{5t}%
\]
Following scalar case, the guess would be $\bar{x}_{p}=\left( \bar{b}+\bar
{a}t\right) e^{-2t}$ but since $e^{-2t}$ is in the homogeneous, we have to
adjust to be $\bar{x}_{p}=\left( \bar{b}t+\bar{a}t^{2}\right) e^{-2t}%
+\bar{c}e^{5t}$. Notice we had to add $\bar{c}e^{5t}$, else it will not work
if we just guessed $\bar{x}_{p}=\left( \bar{b}t+\bar{a}t^{2}\right)
e^{-2t}\,$\ based on what we would do in scalar case, we will find we get
$\bar{a}=\bar{b}=0$. This seems to be a trial and error stage and one just
have to try to find out. This is why undermined coefficients for systems is
not as easy to use as with scalar case. Hence
\[
\bar{x}_{p}=\left( \bar{b}t+\bar{a}t^{2}\right) e^{-2t}+\bar{c}e^{5t}%
\]
Now we plug-in this back into the ODE\ and solve for $\bar{a},\bar{b},\bar{c}%
$. But an easier method is to use Variation of parameters. The fundamental
matrix is%
\begin{align*}
\Phi & =%
\begin{pmatrix}
\bar{x}_{1} & \bar{x}_{2}%
\end{pmatrix}
\\
& =%
\begin{pmatrix}
e^{-2t} & 2e^{5t}\\
-2e^{-2t} & e^{5t}%
\end{pmatrix}
\end{align*}
And
\[
\Phi^{-1}=\frac{%
\begin{pmatrix}
e^{5t} & 2e^{-2t}\\
-2e^{5t} & e^{-2t}%
\end{pmatrix}
^{T}}{\left\vert \Phi\right\vert }=\frac{%
\begin{pmatrix}
e^{5t} & -2e^{5t}\\
2e^{-2t} & e^{-2t}%
\end{pmatrix}
}{e^{3t}+4e^{3t}}=\frac{1}{5}%
\begin{pmatrix}
e^{2t} & -2e^{2t}\\
2e^{-5t} & e^{-5t}%
\end{pmatrix}
\]
Hence using
\begin{align*}
\bar{x}_{p} & =\Phi\int\Phi^{-1}\bar{f}\left( t\right) dt\\
& =\frac{1}{5}\Phi\int%
\begin{pmatrix}
e^{2t} & -2e^{2t}\\
2e^{-5t} & e^{-5t}%
\end{pmatrix}%
\begin{pmatrix}
-15te^{-2t}\\
4te^{-2t}%
\end{pmatrix}
dt\\
& =\frac{1}{5}\Phi\int%
\begin{pmatrix}
-23t\\
-26te^{-7t}%
\end{pmatrix}
dt
\end{align*}
The integral of $\allowbreak\int-23tdt=\frac{-23}{2}t^{2}$ and $\int%
-26te^{-7t}dt=\allowbreak\frac{26}{49}e^{-7t}\left( 7t+1\right) $ (using
integration by parts) hence the above simplifies to%
\begin{align*}
\bar{x}_{p} & =\Phi%
\begin{pmatrix}
\frac{-23}{10}t^{2}\\
\frac{26}{245}e^{-7t}+\frac{26}{35}te^{-7t}%
\end{pmatrix}
\\
& =%
\begin{pmatrix}
e^{-2t} & 2e^{5t}\\
-2e^{-2t} & e^{5t}%
\end{pmatrix}%
\begin{pmatrix}
\frac{-23}{10}t^{2}\\
\frac{26}{245}e^{-7t}+\frac{26}{35}te^{-7t}%
\end{pmatrix}
\\
& =%
\begin{pmatrix}
\frac{52}{245}e^{-2t}+\frac{52}{35}te^{-2t}-\frac{23}{10}t^{2}e^{-2t}\\
\frac{26}{245}e^{-2t}+\frac{26}{35}te^{-2t}+\frac{23}{5}t^{2}e^{-2t}%
\end{pmatrix}
\\
& =%
\begin{pmatrix}
\frac{1}{490}e^{-2t}\left( -1127t^{2}+728t+104\right) \\
\frac{1}{245}e^{-2t}\left( 1127t^{2}+182t+26\right)
\end{pmatrix}
\end{align*}
Hence the complete solution is%
\begin{align*}
\bar{x} & =\bar{x}_{h}+\bar{x}_{p}\\
& =c_{1}%
\begin{pmatrix}
1\\
-2
\end{pmatrix}
e^{-2t}+c_{2}%
\begin{pmatrix}
2\\
1
\end{pmatrix}
e^{5t}+%
\begin{pmatrix}
\frac{1}{490}e^{-2t}\left( -1127t^{2}+728t+104\right) \\
\frac{1}{245}e^{-2t}\left( 1127t^{2}+182t+26\right)
\end{pmatrix}
\end{align*}
To find the constants, we apply initial conditions. At $t=0$%
\begin{align*}%
\begin{pmatrix}
7\\
3
\end{pmatrix}
& =c_{1}%
\begin{pmatrix}
1\\
-2
\end{pmatrix}
+c_{2}%
\begin{pmatrix}
2\\
1
\end{pmatrix}
+%
\begin{pmatrix}
\frac{52}{245}\\
\frac{26}{245}%
\end{pmatrix}
\\
c_{1}%
\begin{pmatrix}
1\\
-2
\end{pmatrix}
+c_{2}%
\begin{pmatrix}
2\\
1
\end{pmatrix}
& =%
\begin{pmatrix}
7\\
3
\end{pmatrix}
-%
\begin{pmatrix}
\frac{52}{245}\\
\frac{26}{245}%
\end{pmatrix}
\\%
\begin{pmatrix}
1 & 2\\
-2 & 1
\end{pmatrix}%
\begin{pmatrix}
c_{1}\\
c_{2}%
\end{pmatrix}
& =%
\begin{pmatrix}
\frac{1663}{245}\\
\frac{709}{245}%
\end{pmatrix}
\\%
\begin{pmatrix}
1 & 2\\
0 & 5
\end{pmatrix}%
\begin{pmatrix}
c_{1}\\
c_{2}%
\end{pmatrix}
& =%
\begin{pmatrix}
\frac{1663}{245}\\
\frac{807}{49}%
\end{pmatrix}
\end{align*}
Hence $5c_{2}=\frac{807}{49}$ or $c_{2}=\frac{807}{245}$ and $c_{1}%
+2c_{2}=\frac{1663}{245}$, hence $c_{1}=\frac{1663}{245}-2\left( \frac
{807}{245}\right) =\frac{1}{5}$. Therefore the solution becomes%
\[
\bar{x}=\frac{1}{5}%
\begin{pmatrix}
1\\
-2
\end{pmatrix}
e^{-2t}+\frac{807}{245}%
\begin{pmatrix}
2\\
1
\end{pmatrix}
e^{5t}+%
\begin{pmatrix}
\frac{1}{490}e^{-2t}\left( -1127t^{2}+728t+104\right) \\
\frac{1}{245}e^{-2t}\left( 1127t^{2}+182t+26\right)
\end{pmatrix}
\]
\end{document}
这就是HTML 看起来像由 tex4ht 使用上述命令生成
这个文件在 latexml HTML 中看起来怎么样？lwarp HTML？
这是PDF 文件如果需要，由 lualatex 生成。
答案1
一个实质的lwarp例子是这里：这是我教的一门课程的 65 页讲义。它包括tikz图形、cleveref参考资料和大量数学知识。我必须提供一些 js 和 css 来调整最终输出，但lwarp所有繁重的工作都由我完成。
我曾尝试latexml处理同一文档但无法cleveref工作。