HTTP/1.1 200  X-Application-Context: fss-proxy Date: Tue, 03 Apr 2018 11:38:13 GMT X-Content-Type-Options: nosniff X-XSS-Protection: 1; mode=block Cache-Control: no-cache, no-store, max-age=0, must-revalidate Pragma: no-cache Expires: 0 X-Frame-Options: DENY Access-Control-Allow-Credentials: true Access-Control-Allow-Methods: GET, POST, OPTIONS, PUT, DELETE, PATCH Access-Control-Allow-Headers: Authorization,Content-Type,If-Modified-Since,headers,If-Modified-Since Content-Type: application/json;charset=UTF-8 Transfer-Encoding: chunked Connection: close  18  activeSessionsCount=1  0   I have JSON here. And I want to grab only integer that may vary from 0 to like 200 or even more using regex in the current string "activeSessionsCount=1". Is it possible? I'm completely stuck here.
I don't really understand regular expressions.  Can you explain them to me in an easy-to-follow manner?  If there are any online tools or books, could you also link to them?
Cannot install R-forge package using install.packages Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even? Not booting from USB or CD (SYSLINUX Message) If sup A $\lt$ sup B show that an element of $B$ is an upper bound of $A$ (a) If sup A &lt; sup B, show that there exists an element of $b \in B$ that is an upper bound for $A$.  I have argued that if sup A $\lt$ sup B, then choose an $\epsilon&gt;0$ such that sup A +$\epsilon \in B$. Since $a \le $ sup A for all $a \in A$, it follows $a \lt $ sup A + $\epsilon$ hence sup A + $\epsilon $ is an upper bound of A as well as an element of B.  (b) Give an example to show that this is not always the case if we only assume sup A $\le$ sup B.  I am having trouble extending my argument into an example which makes me think that it may not be correct.   Question: Is my argument for (a) correct? If not how would one show this fact? Can you give an example asked for in (b)? Thanks in advance.  I need to understand how I am assuming what I am supposed to prove in my approach. Can someone elaborate on that please? The other question doesn't really address this issue. How can I check for Python version in a program that uses new language features? If I have a Python script that requires at least a particular version of Python, what is the correct way to fail gracefully when an earlier version of Python is used to launch the script?  How do I get control early enough to issue an error message and exit?  For example, I have a program that uses the ternery operator (new in 2.5) and "with" blocks (new in 2.6).  I wrote a simple little interpreter-version checker routine which is the first thing the script would call ... except it doesn't get that far. Instead, the script fails during python compilation, before my routines are even called. Thus the user of the script sees some very obscure synax error tracebacks - which pretty much require an expert to deduce that it is simply the case of running the wrong version of Python.  I know how to check the version of Python. The issue is that some syntax is illegal in older versions of Python. Consider this program:  import sys if sys.version_info &lt; (2, 4):     raise "must use python 2.5 or greater" else:     # syntax error in 2.4, ok in 2.5     x = 1 if True else 2     print x   When run under 2.4, I want this result  $ ~/bin/python2.4 tern.py  must use python 2.5 or greater   and not this result:  $ ~/bin/python2.4 tern.py    File "tern.py", line 5     x = 1 if True else 2            ^ SyntaxError: invalid syntax   (Channeling for a coworker.) At what stage should acknowledgements be inserted into the paper if review is not double-blind? When the review process is double-blind, I guess (although I have not read this anywhere) that acknowledgements should not be written in the submitted paper, as this might break anonimity. They should only be inserted into the camera-ready version, after the paper is accepted.  What is the custom when the review process is not double-blind? Should I write the acknowledgements in the submitted version, or wait to the camera-ready version? I am trying to write an if statement to test whether there are any files matching a certain pattern. If there is a text file in a directory it should run a given script.  My code currently:  if [ -f /*.txt ]; then ./script fi   Please give some ideas; I only want to run the script if there is a .txt in the directory. How do I protect random people pointing their domain to my server? What's the intuition behind difference of $\neg(S\land E)$ and $\neg S \land \neg E$? Let $S$ represent the English statement "Sales will go up."And let $E$ represent the English statement "Expenses will go up." Now my problem is that I have trouble telling the difference between these two statements: $\neg(S\land Q)$ and $\neg S \land \neg E$. When these two statements are translated into English they sound,to me, the same. However, when you look at their truth tables they are different.   Here is my best translation of these two statements:  1) $\neg(S\land E)$ represented in English, "Both sales and expenses will not go up."(Correct me if I translated it wrong)  2)$\neg S \land \neg E$ represented in English, "Expenses will not go up and sales will not go up."(Again, correct me if I'm wrong)  I still do not understand the difference between these two statements in the English representation. They both sound the same thing to me.  Once more, I have the same trouble I have with understanding with the or-connective( $\lor$ ). Again let E and S be the same statements as before.Tell me if I might have translated them wrong.  1) $\neg (S \lor E)$ represented in English is "Neither sales nor expenses will go up."  2)$\neg S \lor \neg E$ represented in English is "Sales will not go up, or Expenses will not go up."  Yet I still have trouble understanding the difference between the English translation.  In short, can you give me some alternative example that better explains the difference, something intuitive that gives me a clear picture of the difference of the meaning between the statements I showed with the connective-or and connective-and. Thank you. Cannot use polyglossia/babel with biblatex in LuaTeX: etoolbox Toggle undefined This is an example test.tex document. Both babel and polyglossia failed and produce the same error: \documentclass{book}  %\usepackage[vietnamese]{babel} \usepackage{polyglossia} \setdefaultlanguage{vietnamese}  \usepackage{biblatex}   \title{Sth}  \begin{document}  \maketitle  \end{document}   is the full log. And my config for VS Code: &quot;latex-workshop.latex.recipes&quot;: [     {         &quot;name&quot;: &quot;lualatex-&gt;biber&quot;,         &quot;tools&quot;: [             &quot;lualatex&quot;,             // &quot;biber&quot;,         ]     } ], &quot;latex-workshop.latex.tools&quot;: [     {         &quot;name&quot;: &quot;lualatex&quot;,         &quot;command&quot;: &quot;lualatex&quot;,         &quot;args&quot;: [             &quot;--shell-escape&quot;,             &quot;-synctex=1&quot;,             &quot;-interaction=nonstopmode&quot;,             &quot;-file-line-error&quot;,             &quot;-pdf&quot;,             &quot;%DOC%&quot;         ]     },     {         &quot;name&quot;: &quot;biber&quot;,         &quot;command&quot;: &quot;biber&quot;,         &quot;args&quot;: [             &quot;%DOCFILE%&quot;         ]     } ],  I installed all Vietnamese-related package I found using tlmgr search, notably babel-vietnamese and vntex, to no avail. When I use English language, it works. When I paste the exact document to Overleaf, it also works. So it's something on my part. How can I fix this? Given two input layers  The clip layer - an arbitrary layer (though in my case it's a float 1 band geotiff) The mask layer - a paletted raster in which every pixel is either NO DATA or a single &quot;color&quot; value, same for every pixel.  I want to create an output raster whose overall extent matches the extent of the mask layer and each pixel is either NO DATA if the corresponding pixel of the mask layer is NO DATA or the value of the corresponding pixel from the clip layer if the corresponding pixel of the mask layer is not NO DATA. I know I can mask the clip layer with a polygon, but the mask layer is a raster. Polygonizing the raster layer is possible in principle, but would result in a polygon with a lot of very short edges. Is there a direct way to do this in QGIS? I suppose there's probably a path that involves creating a new layer and using the raster calculator, but I'm hoping there's something more direct. Further complicating the matter is that the two rasters might not have the same resolution and origin. I believe this is equivalent to  in ArcMap. I'm not sure if this is the right place to ask this question. I realise that this maybe a borderline philosophical question at this point in time, therefore feel free to close this question if you think that this is a duplicate or inappropriate for this forum. Anyway, I'm an electrical engineer and I have some basic knowledge of quantum mechanics. I know that Schrödinger's equation is deterministic. However, quantum mechanics is much deeper than that and I would like to learn more. If this question is not clearly answerable at this point than can anyone point out some recognized sources that try to answer this question. I would appreciate it if the source is scientific and more specifically, is related to quantum theory. SSH login without password with kerberos Reproducible LaTeX builds - compile to a file which always hashes to the same value So, I understand that by using 'tee' I can redirect the output of a command such as 'ping' to stdout as well as a file.  For example:  &gt; ping google.com | tee somefile   This would display the ping statistics on terminal and write them to 'somefile'.  Now, if I wish to modify the output from 'ping', I can use 'cut' in such a way:  &gt; ping google.com | cut -d' ' -f 1   But, if I wish to use all the three commands together, I get a no output on stdout and an empty file.  &gt; ping google.com | cut -d' ' -f 1 | tee somefile   What am I doing wrong? Is there a better way to do this? Somehow I feel I am not using 'tee' properly. Any insight would be appreciated.  I'm using bash shell, if this is relevant. Distribution of the digits of Pi Can anything be stated about the distribution of the digits of Pi, i.e., if I were to sample n digits of Pi, can anything be said about the probability to observe certain digits, or is there any reason to assume that they would not be evenly distributed? This is purely a curiosity question. "Contact us"... and then what? So I had a dispute with the moderators at a specific SE site. As a result, I got suspended there. As per SE guidelines, I pressed the "Contact us" button and wrote to the higher powers to appeal.  That was 5-6 weeks ago. There was zero feedback, neither positive, nor negative, not even "it's in progress, please wait".  So what is this "contact us" thing supposed to do? Is it a placebo to make me feel better while being routed to dev/null?  What is the normal time it takes to process a "contact us" message?  Notes:   As far as I remember, I did not contact the team before. I'm a longtime user of multiple SE sites and never had any problems except for that one site. I'm pretty sure my form was submitted. I didn't make screenshots or anything, because I did not think I would need them in a post like this. I didn't expect a super-fast reply either, so I wasn't wondering after a week or two. A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$  I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$  Please suggest how to proceed further. I wanted to change my own password (as an admin) so I did the following command:  sudo passwd   and then it prompted for a password but when I tryed to relog, my password was not changed. I'm now wondering what password did I changed since it did prompted me to enter a password but I'm not sure what password did it changed exactly.  I know that if I do sudo passwd &lt;MY_USER&gt; it works, I'm just wondering without a user what happens.  Thank you What's the simplest SQL statement that will return the duplicate values for a given column and the count of their occurrences in an Oracle database table?  For example: I have a JOBS table with the column JOB_NUMBER.  How can I find out if I have any duplicate JOB_NUMBERs, and how many times they're duplicated?