I need to create a user which can SSH to a server, but can not run any commands or view anything, for the purposes of tunneling the connection without risk of revealing credentials. How do I do this? I looked on google but found no sources.  Edit: Would it also be possible to show a certain message to the user upon login? Instead of the usual 'Last logged in from xxx at yyy'?
How can I restrict a user on the SSH server to allow them only the privileges for ? i.e. So they cannot run commands even if they log in via SSH.  My Linux servers are Ubuntu 11.04 and OpenWrt.
How can I download an entire website? How does a calculator calculate the sine, cosine, tangent using just a number? Sine $\theta$ = opposite/hypotenuse   Cosine $\theta$  = adjacent/hypotenuse   Tangent $\theta$  = opposite/adjacent      In order to calculate the sine or the cosine or the tangent I need to know $3$ sides of a right triangle. $2$ for each corresponding trigonometric function. How does a calculator calculate the sine, cosine, tangent of a number (that is actually an angle ?) without knowing any sides? What would be better practice when giving a function the original variable to work with:  unsigned long x = 4;  void func1(unsigned long&amp; val) {      val = 5;             } func1(x);   or:  void func2(unsigned long* val) {      *val = 5; } func2(&amp;x);   IOW: Is there any reason to pick one over another? How to use SQLDeveloper to connect to embedded Derby database Let $L$ denote the set of all lines in the Euclidean plane $\Bbb R^2$. We say that two lines $L$ and $M$ are parallel if either $L = M$ or $L$ and $M$ have no points in common.  Using Euclid’s Parallel Postulate prove that parallelism is an equivalence relation on the set $L$.  I need some help determining if I have the right idea about this.  $L=L$ which means that $L$ is parallel to itself. So this is reflexive.  Suppose we have $L || M$. Then Either $L=M$ or $L$ and $M$ share no points. So Then $M=L$ or $M$ and $L$ share no points. Then $M||L$. So this is symmetric.  Suppose we have $L || M$ and $M || N$. So then either $L=M$ or $L$ and $M$ share no points and either $M=N$ or $M$ and $N$ share no points. So we have that either $L=N$ or $L$ and $N$ share no points. So $L || N$. So then this is transitive.  So from all of this, parallelism is an equivalence relation. I'm not sure if this is correct or not. Any help would be appreciated. How can I repair grub? (How to get Ubuntu back after installing Windows?) Geometry with Trigonometry From the given figure, prove that      $$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$     I could not even start to the first point. Please help me to solve this.  Thanks in advance What translation occurs when writing to a file that was opened in text mode that does not occur in binary mode? Specifically in MS Visual C.  unsigned char buffer[256]; for (int i = 0; i &lt; 256; i++) buffer[i]=i; int size  = 1; int count = 256;   Binary mode:  FILE *fp_binary = fopen(filename, "wb"); fwrite(buffer, size, count, fp_binary);   Versus text mode:  FILE *fp_text = fopen(filename, "wt"); fwrite(buffer, size, count, fp_text); while working with webservice quotes are not inserting may i know what i have to do. my code is     for(integer l=0;l&lt;reqquotes;l++){          qu = new Quote();          qu.Name = qName;          //opp.CloseDate =system.today();           qList.add(qu);             insert oppList;          insert qList;     } Prove that $ \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z} $ Show that for $|z|&lt;1$, one have $$ \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots+\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots = \frac{z}{1-z} $$ and $$ \frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots+\frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots = \frac{z}{1-z} $$ Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots+2^k$. How to draw neat polygons around scatterplot regions in ggplot2 How can I copy the path to a folder or a file? The Info Dialog doesn't allow me to copy the information that stands there ...    Is there an easy way to do this? ng-options with simple array init Beamer - place text of footnote in footline I must place footnotes and footcites at the bottom of the frames, like shown in the picture:   The line above the footnotes must always be there, and it must have a minimum distance to the bottom of lets say 2.5em.  So I see two possible solutions:   Add the footnote via \setbeamertemplate to the footline. Change the format of the footnoteline and let it always appear, even if no footnote was used.   For the last hours, I tried to accomplish point 1, but I had absolutely no success. Can you please help me? Otherwise I have to use *whisper* Powerpoint. ;)    EDIT: Do you know the phenomenon that only after you ask someone, you can get the right ideas? ;)  Here is a MWE of my current solution:  \documentclass[xcolor={dvipsnames}]{beamer}  \usepackage[absolute, overlay]{textpos} \usepackage{environ}   \usetheme{default} \setbeamertemplate{navigation symbols}{}%remove navigation symbols   \newcommand{\makemyrule}{%     \setlength{\unitlength}{14pt}%      minimum distance to bottom     \ifdim\footnoteheight&gt;\unitlength \setlength{\unitlength}{\footnoteheight} \fi%     check which distance is bigger     \begin{textblock*}{\linewidth}(0pt,\dimexpr\textheight-\unitlength)%    place with use of textpos         \textcolor{blue}{\rule{\paperwidth}{.3ex}}     \end{textblock*} }   \let\oldfootnote\footnote%      redefine the footnote command \renewcommand{\footnote}[1]{\Addheight{#1}\oldfootnote{#1}}  \newlength{\footnoteheight} \newsavebox{\mytext} \newcommand{\Addheight}[1]%     measure height of the text         {%                 \savebox{\mytext}{\small\strut\parbox{\textwidth}%                 {\hspace*{2em}\thefootnote#1}\strut}%       approximated by eye, not perfect!                 \addtolength{\footnoteheight}{\ht\mytext}%                 \addtolength{\footnoteheight}{\dp\mytext}%         }     \NewEnviron{myframe}[1]{%     \begin{frame}{#1}%     \setlength{\footnoteheight}{0pt}%     \BODY%     \makemyrule%     \end{frame}%     }    \renewcommand\footnoterule{}  \begin{document} \begin{myframe}{} \frametitle{Beamer example} using\footnote{Y LUSPIN, et al} footnotes\footnote{P. GHOSEZ, et al and a lot more information to create a linebreak} excessively \footnote{You do not need to read this, this is only blindtext to create even more and more text. Blindtext will go on forever, nobody can stop the allmighty blindtext. I think I am a blindtext author.} \end{myframe}{} \begin{myframe} \frametitle{Beamer example} using\footnote{Y LUSPIN, et al} footnotes\footnote{P. GHOSEZ, et al} \end{myframe}{} \begin{myframe} \frametitle{Another example} without footnotes\\ \end{myframe} \end{document}   What this does: At the beginning of each myframe-environment, the length \footnoteheight is reseted to zero. When a \footnote is placed, the height of the text is added to this length. At the end of each frame, via \makemyrule the bottom rule is created, which is above the footnote texts (my solution is not perfect) or, if there are no footnotes, at least 14pt above the bottom.  My wishes (not real problems, but I want it to work better) are   The height of the footnotetext is only approximated. I would like a solution where you can measure the height of the formatted footnote. The frame environment needs to be renamed to myframe. I would like a solution which works without changing the document. But it all failed, because I could not execute \makemyrule before \end{frame} automatically. What's the best counter strategy against battlecruisers? How should I prepare my army when attacking someone who has 12+ battlecruisers? (Protoss/Zerg/Terran answers are more than appreciated) A couple of months ago, I suggested a set of badges for . The responses were... mixed. The  can be summarized:  In practice, all this would do is encourage people to ask yet more stupid, worthless questions and we have quite enough of that as it is.  We believe that one of the main things holding back many of our beta sites is that when people have questions they aren't in the habit of asking on the site. Even our flagship site, Stack Overflow, doesn't get many questions from its most knowledgeable users. Perhaps they don't have questions, but far more likely they find answers without needing to ask. Our subtle bias against asking (questions =&gt; clueless) means we lose out on what might be our best source of excellent questions. So we went back to the drawing board and designed a badge set that should not encourage worthless questions:  Asked a well-received question on X separate days, and maintained a positive question record.  In order to award badges, we need to define our terms:  positive question record   (total questions - negative questions - closed - deleted)/total questions &gt;= 0.5  It triple counts a question that is downvoted, closed, and deleted. If you ask one that hits that trifecta, you'll need to ask three better questions to dig yourself out. My own record of asking does not meet the criteria:   (44 - 0 - 12 - 12)/44 = 0.4545...  My  looks a lot better; I had a bunch of questions from the beta period that have been subsequently (and rightfully) deleted. With care, I could bring my record into positive territory with 4 questions. The takeaway is that people can't sweep past indiscretions under the rug by deleting them. And they can recover from a bad start as long as they don't ask more bad questions.  well-received question on X separate days A well-received question is one that's open, not deleted, and has a score &gt; 0. If you ask at least one well-received question in a  and none of your questions that day are deleted, downvoted, or closed, you get one more day of credit toward the badge. Whenever the badge criteria is checked, previous asking days are re-evaluated so that there's an incentive to fix downvoted and/or closed questions. The no-deleted-questions criteria is to prevent people from asking several questions at once and deleting any that aren't upvoted or answered to ask another day. (Though this is a losing strategy in the long run.)   My proposed badge names and levels:  Asked a well-received question on 5 separate days, and maintained a positive question record.  Asked a well-received question on 30 separate days, and maintained a positive question record.  Asked a well-received question on 100 separate days, and maintained a positive question record. In addition, since some outstanding individuals ask on more than 100 days, the gold badge may be awarded multiple times. Over on English, Oishi-sama (now a ) would have a remarkable . Bronze and silver badges are only awarded once. We've decided to hold off on multiple awards for now. See the . Neither the names nor the levels nor even the descriptions are set in stone. Please play around with  to get an idea of how the badge works using our public data. Since the query does not account for deleted questions, it's a bit more generous than the actual badges would be. This version of the badge set should encourage a habit of well-received questions over time and encourage people to fix poorly received questions.  I'd like to express my gratitude to the Parenting community for participating in our . Is there a word for four times as much, analogous to once, twice, and thrice? Basically, I acidentally spawned too many ArmorStands due to a wrong code writing , resulting in the game crashing, and wanted to know how to delete those armorstands from the World save folder. They hava names and I belive they also have a scoreboard set, if that helps.   I have already tried to open the world but it crashes, I tried to open a server (Since it worked the last time I tried it) and killing it all from the console but, different from the last time, it crashed before killing them and now I just tried MCedit but it crashes as well. Where are the default settings saved for the Command prompt defaults? I want to put a table and a figure side-by-side. The table should fill two third of the text width, and the figure the last third.  So I'd like to achieve one of the following effects (which are equivalent in my case):   One table with 3 columns. The first two columns contain text, the third column contains a lengthy image. The height of the image should be determined by the height of the whole table, if that's possible Or, a divided page like in newspapers. On the left side I have the table, on the right side the image.   What I have tried:  \begin{multicols}{2} \begin{tabular}{ l l } Text &amp; Text \\ Text &amp; Text \\ ... &amp; ... \\ \end{tabular} \vfill \begin{figure}[H]     \includegraphics[]{images/image} \end{figure} \end{multicols}   But this places the picture directly on top of the table.  I also tried using \usepackage[export]{adjustbox} and \includegraphics[right]{images/image} but the caption is still in the center of the page and therefor on top of the table.