I would like to add a rectangular box to my table to highlight some rows. I'm trying to use tikz with the second lot of code from  to a similar question. However, unlike the earlier question, I'm using beamer and the box appears in the wrong place after compiling several times. I'm attaching a picture with the desired result and what I obtain instead.      This is my code:  \documentclass{beamer} \usepackage[english]{babel} \usepackage{tabularx,booktabs,dcolumn} \usepackage{tikz} \usetikzlibrary{matrix,shapes,arrows,fit} \pdfmapfile{+sansmathaccent.map} \newcolumntype{d}{D..{-1}}  \newcounter{nodecount} \newcommand\tabnode[1]{\addtocounter{nodecount}{1} \tikz \node                (\arabic{nodecount}) {#1};}  % Some options common to all the nodes and paths \tikzstyle{every picture}+=[remember picture,baseline] \tikzstyle{every node}+=[inner sep=0pt,anchor=base, minimum width=1.8cm,align=center,text depth=.25ex,outer sep=1.5pt] \tikzstyle{every path}+=[thick, rounded corners]  \begin{document}  \begin{frame} \frametitle{Title of my slide} \begin{table} \resizebox{\textwidth}{!}{    \begin{tabular}{c*{7}{d}} \midrule  \multicolumn{1}{c}{$CCCCC$}  &amp; \multicolumn{1}{c}{$CCCC$} &amp; \multicolumn{1}{c}{$CCCC$} &amp; \multicolumn{1}{c}{$CCC$} &amp; \multicolumn{1}{c}{$CCCCCCC$} &amp; \multicolumn{1}{c}{$CCCCC$} &amp; \multicolumn{1}{c}{$CCCCC$} &amp; \multicolumn{1}{c}{$CCCCCCC$} \\ \midrule \multicolumn{8}{c}{\textbf{Panel A: AAA}} \\ \midrule \tabnode{0}     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp;            \tabnode{0.00} \\ \tabnode{}    &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp;          \tabnode{0.00} \\   &amp;       &amp;       &amp;       &amp;       &amp;       &amp;       &amp;  \\   0     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp; 0.00 \\   &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) \\   &amp;       &amp;       &amp;       &amp;       &amp;       &amp;       &amp;  \\   0     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp; 0.00 \\   &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) \\   \multicolumn{8}{c}{\textbf{Panel B: BBB}} \\  \midrule  0     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp; 0.00 \\    &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) \\    &amp;       &amp;       &amp;       &amp;       &amp;       &amp;       &amp;  \\  0     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp; 0.00 \\    &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) \\    &amp;       &amp;       &amp;       &amp;       &amp;       &amp;       &amp;  \\  0     &amp; 0.00  &amp; 0.00  &amp; 0.00 &amp; 0.00 &amp; 0.00  &amp; 0.00 &amp; 0.00 \\    &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) &amp; (0.00) \\   \midrule  \end{tabular}  }  \begin{tikzpicture}[overlay]  \node[draw=red,rounded corners = 1ex,fit=(1)(4),inner sep = 1pt] {};  \end{tikzpicture}  \end{table} \end{frame} \end{document}
I asked a question  but, rather stupidly, added extra information after it had been answered.  My additional question is this: I have a table with varying lengths of words. I want to encircle different groups of cells, including diagonals. If possible, it would be great if the rectangles don't overlap each other where possible.  I looked at the answer , which, whilst tedious, worked except for the overlap.    This is where I've got to so far:  \documentclass[fleqn,12pt,a4paper,landscape]{article}  \usepackage{tikz} \usepackage{tabularx}  \begin{document}  % Introduce a new counter for counting the nodes needed for circling \newcounter{nodecount} % Command for making a new node and naming it according to the nodecount     counter \newcommand\tabnode[1]{\addtocounter{nodecount}{1} \tikz \node     (\arabic{nodecount}) {#1};}  % Some options common to all the nodes and paths \tikzstyle{every picture}+=[remember picture,baseline] \tikzstyle{every node}+=[inner sep=0pt,anchor=base, minimum width=1.8cm,align=center,text depth=.25ex,outer sep=1.5pt] \tikzstyle{every path}+=[thick, rounded corners]  \begin{table}[ht]   \begin{minipage}[b]{1 \linewidth}\centering    \begin{tabular}{cccccc} % Turn the cells needed for defining the circling paths into nodes with our     custom command \tabnode{Long ago}&amp; \tabnode{A few days ago}&amp; \tabnode{Day before         yesterday}&amp; \tabnode{Yesterday} &amp;  &amp;  \\      &amp; &amp; &amp;  &amp; \tabnode{Today} &amp; \tabnode{Now}\\      \tabnode{Long time in the future}&amp; \tabnode{A few days on}&amp;     \tabnode{Day after tomorrow}&amp; \tabnode{Tomorrow} &amp;  &amp; \\      \end{tabular}    \end{minipage}   \begin{tikzpicture}[overlay] % Define the circle paths \draw [blue](1.north west) -- (1.north east) -| (7.north east) |- (7.south     west) |- (1.north west); \draw [red](8.south west) -| (2.north east) -| (2.south west) |- (8.south     west); \draw [yellow](9.south west) -| (3.north east) -| (3.south west) |- (9.south     west); \draw [green](8.south west) -| (3.north east) -| (2.south west) |- (8.south     west);   \end{tikzpicture}  \end{table} \end{document}  %%% Local Variables:  %%% mode: latex %%% TeX-master: t %%% End:
Query standard input of lightning-input I want to query standard HTML textarea which is wrapped by lightning-textarea. How can I get it?  &lt;lightning-textarea name=&quot;secret_field&quot; label=&quot;Secret&quot; value={secretFieldValue}&gt;&lt;/lightning-textarea&gt; Remove Indentation for a Single Paragraph Is there a way to remove the indentation at the beginning of a paragraph for a single paragraph.  I know I could do  \setlength{\parindent}{0cm} This is my paragraph.  Blah blah blah. \setlength{\parindent}{default}   Where default is whatever the default indentation is for the document class I am using (part of the problem is that I don't know what this default value is).  I feel as if there should be a better way to accomplish this than by the above method.  I am looking for a solution of the form:  \noindent This is my paragraph.  Blah blah blah.   Is there a way to do this, or will I have to resort to the former method?  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I already read whatever I could find about Missing bundle property errors, but none helped solving.  dpm($entity) returns  uid (String, 2 characters ) 70 name (String, 9 characters ) John pass (String, 55 characters ) $S$DUwPuOuDPiDL4nRTYXqc7a5uOfMKey7pyhOFUEKka1XM... mail (String, 30 characters ) john@example.com theme (String, 0 characters ) signature (String, 0 characters ) signature_format (String, 13 characters ) filtered_html created (String, 10 characters ) 1396286331 access (String, 10 characters ) 1397146661 login (String, 10 characters ) 1396513460 status (String, 1 characters ) 1 timezone (NULL) language (String, 2 characters ) fr picture (NULL) init (String, 30 characters ) john@example.com data (Array, 5 elements) roles (Array, 1 element) og_user_node (Array, 0 elements) message_subscribe_email (Array, 1 element) field_bio (Array, 0 elements) field_name_first (Array, 1 element) field_name_last (Array, 1 element) field_facebook_url (Array, 0 elements) field_linkedin_url (Array, 0 elements) field_twitter_url (Array, 0 elements) user_trusted_contacts (Array, 1 element) group_group (Array, 1 element) group_access (Array, 1 element) metatags (Array, 0 elements) rdf_mapping (Array, 3 elements) realname (String, 13 characters ) John Doe content (Array, 13 elements) entity_view_prepared (Boolean) TRUE privatemsg_disabled (Boolean) FALSE Rack Mount Servers - Build them or Buy them? 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In C++ you can declare variables pretty much anywhere (and declaring them close to first use is obviously a good thing) but the following still won't work:  switch (val)   {   case VAL:     // This won't work   int newVal = 42;     break; case ANOTHER_VAL:     ...   break; }     The above gives me the following error (MSC):     initialization of 'newVal' is skipped by 'case' label   This seems to be a limitation in other languages too.  Why is this such a problem? Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46. Attempted proof based on  and .  Given: $a,b$ elements of $G$, and $ab$ has finite order $n$.   Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$.   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