How to assign words with spaces as a whole name to variable in shell script? I need to make some directory of Author name. It's easy when doing them alone. I can simply  mkdir "Mario Luis Garcia"   This will create one directory named "Mario Luis Garcia"  But when trying do it massively with shell script, I failed.  for i in "Mario Luis Garcia" "etc.";     do          mkdir $i     done;   This will create 4 separate directories instead of 2. I tried to use {}, ()to enclose the quoted text in. Doesn't work.  How to do that?
Why does my shell script choke on whitespace or other special characters? Or, an introductory guide to robust filename handling and other string passing in shell scripts.  I wrote a shell script which works well most of the time. But it chokes on some inputs (e.g. on some file names).  I encountered a problem such as the following:   I have a file name containing a space hello world, and it was treated as two separate files hello and world. I have an input line with two consecutive spaces and they shrank to one in the input. Leading and trailing whitespace disappears from input lines. Sometimes, when the input contains one of the characters \[*?, they are replaced by some text which is is actually the name of files. There is an apostrophe ' (or a double quote ") in the input and things got weird after that point. There is a backslash in the input (or: I am using Cygwin and some of my file names have Windows-style \ separators).   What is going on and how do I fix it?
I know that like is a preposition but why not using it as conjunction?  Examples:        It's as if I'm walking on air   It's like I'm walking on air      What is the difference? How do thrown weapons interact with the Extra Attack feature?  Can you throw more than one weapon per turn?  How about two-weapon fighting?  Do you need the Dual-Wielder feat to throw two weapons every turn? Select longest continuous sequence I am trying to construct a query in PostgreSQL 9.0 that gets the longest sequence of continuous rows for a specific column.  Consider the following table:  lap_id (serial), lap_no (int), car_type (enum), race_id (int FK)   Where lap_no is unique for each (race_id, car_type).  I would like the query to produce the longest sequence for a given race_id and car_type, so it would return an int (or long) that is highest.  With the following data:  1, 1, red, 1 2, 2, red, 1 3, 3, red, 1 4, 4, red, 1 5, 1, blue, 1 6, 5, red, 1 7, 2, blue, 1 8, 1, green, 1   For car_type = red and race_id = 1 the query would return 5 as the longest sequence of the lap_no field.  I found a similar question  however my situation is a bit more straightforward.  (I would also like to know the longest sequence for a given car_type for all races, but was planning to work that out myself.) Let's say I have Windows installed on drive C, however I install all of my programs on the D drive, and I keep my user profile on the D drive as well (things like My Documents). How can I have Windows use those folders instead of the ones on the C drive? Which Ubuntu to use alongside Windows 7? Short circuit or disconnect to bypass a compressor creating model mesh and face explosions on Panda3d and Blender I'm doing a masters course at a different university than I did my undergrad (both in maths). I did well in my undergrad but now I'm finding it very difficult. The shock of the change is hard to cope with.  I'm not just finding it very difficult, but I get zero feedback: we don't have any tests and I was not able to make friends so I don't know how others are finding it. I'm starting to think that I might be an idiot.   How can I build some confidence in my abilities at the same time as I rush and struggle to keep up with the massive rate of new material I need to learn? How might I get myself some reassurance that I can actually pass, or stop worrying about this so much? (I feel terribly guilty using so much money from my family to be here) It's possible that I don't really have a chance at all but knowing that would be fine. I'm going to submit form using cURL, where some of the contents is come from other file, selected using sed  If param1 is line matching pattern from other file using sed, below command will works fine:  curl -d param1="$(sed -n '/matchpattern/p' file.txt)" -d param2=value2 http://example.com/submit   Now, go to the problem. I want show only text between 2 matching pattern excluding the matching pattern itself.  Lets say file.txt contains:  Bla bla bla firstmatch It is a long established fact that a reader will be distracted by the readable content of a page when looking at its layout. secondmatch The point of using Lorem Ipsum is that it has a more-or-less normal distribution of letters, as opposed to using 'Content here, content here', making it look like readable English.   Currently, lots of "beetween 2 matching pattern" sed command won't remove firstmatch and secondmatch.  I want the result to become:  It is a long established fact that a reader will be distracted by the readable content of a page when looking at its layout. Java equivalent of unsigned long long? In C++, I enjoyed having access to a 64 bit unsigned integer, via unsigned long long int, or via uint64_t. Now, in Java longs are 64 bits, I know. However, they are signed.  Is there an unsigned long (long) available as a Java primitive? How do I use it? Who else knew that Lord Voldemort was Tom Marvolo Riddle? We know that some people were aware of the fact that Lord Voldemort was Tom Marvolo Riddle prior to his return in Goblet of Fire; at the very least his close soon-to-be-Death-Eater &quot;friends&quot; and Albus Dumbledore:  “Very few people know that Lord Voldemort was once called Tom Riddle. I taught him myself, fifty years ago, at Hogwarts...&quot; ... underwent so many dangerous, magical transformations, that when he resurfaced as Lord Voldemort, he was barely recognizable. Hardly anyone connected Lord Voldemort with the clever, handsome boy who was once Head Boy here.” (Professor Dumbledore to Weasleys, Chamber of Secrets, CHAPTER EIGHTEEN, Dobby’s Reward)  and  “Voldemort,” said Riddle softly, “is my past, present, and future, Harry Potter…” He pulled Harry’s wand from his pocket and began to trace it through the air, writing three shimmering words: TOM MARVOLO RIDDLE Then he waved the wand once, and the letters of his name rearranged themselves: I AM LORD VOLDEMORT “You see?” he whispered. “It was a name I was already using at Hogwarts, to my most intimate friends only, of course. ...  (Tom's Diary manifestation in the Chamber of Secrets).  QUESTION: Who - based on canon (books/interviews/pottermore) - knew about the fact that Lord Voldemort was Tom Marvolo Riddle prior to Ginny being taken to the Chamber of Secrets? Obviously, Albus Dumbledore and Tom himself. Who else? I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$ I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$    Differentiability Definition  When we say a function is differentiable at $x_0$, we mean that the limit:  $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  Continuity Definition  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.  Proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.  My Question  The proof seems to execute the following steps:   Assume the function is continuous at $x_0$ Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.   What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.  Hopefully my confusion is clear. Any help will be greatly appreciated. Ubuntu 18.04 opening system setting logs me out I've recently installed Ubuntu 18.04 in a dual-boot configuration with Windows 10. I noticed today that, when trying to open system settings, Ubuntu seems to log me out to the lock screen. When I log back in, all previously-running applications are now closed. Very unsure what to do from here, any help much appreciated. Why don't electromagnetic waves require a medium? Should training data in each batch size be resample only one time or at each epoch when using mini-batch I saw some related question regarding to the fact is one should use sampling with resampling or without when using minibatch.  However my question is different.  Let's assume that I use sampling witouth replacement. Meaning that for each epoch one training example can only belong one 'batch size' group.  I want to know if one should only resample the training data (into the different batch size) only 1 time or before to start to update the parameters before each epoch.  For instance suppose I have 20 'batch size' group. Suppose at the first iteration I have sampled picture A in group 2 and picture B in group 4.  If so should picture A always stay in group 2 and picture B in group 4. Or should I resample it before to start a new epoch (e.g. at epoch 2 put picture B in group 6 and picture B in group 20). Combine or rebase an arbitrarily large number of commits How do HttpOnly cookies work with AJAX requests? JavaScript needs access to cookies if AJAX is used on a site with access restrictions based on cookies. Will HttpOnly cookies work on an AJAX site?   Edit: Microsoft created a way to prevent XSS attacks by disallowing JavaScript access to cookies if HttpOnly is specified. FireFox later adopted this. So my question is: If you are using AJAX on a site, like StackOverflow, are Http-Only cookies an option?  Edit 2: Question 2. If the purpose of HttpOnly is to prevent JavaScript access to cookies, and you can still retrieve the cookies via JavaScript through the XmlHttpRequest Object, what is the point of HttpOnly?  Edit 3: Here is a quote from Wikipedia:     When the browser receives such a cookie, it is supposed to use it as usual in the following HTTP exchanges, but not to make it visible to client-side scripts.[32] The HttpOnly flag is not part of any standard, and is not implemented in all browsers. Note that there is currently no prevention of reading or writing the session cookie via a XMLHTTPRequest. [33].   I understand that document.cookie is blocked when you use HttpOnly. But it seems that you can still read cookie values in the XMLHttpRequest object, allowing for XSS. How does HttpOnly make you any safer than? By making cookies essentially read only?   In your example, I cannot write to your document.cookie, but I can still steal your cookie and post it to my domain using the XMLHttpRequest object.  &lt;script type="text/javascript"&gt;     var req = null;     try { req = new XMLHttpRequest(); } catch(e) {}     if (!req) try { req = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e) {}     if (!req) try { req = new ActiveXObject("Microsoft.XMLHTTP"); } catch(e) {}     req.open('GET', 'http://stackoverflow.com/', false);     req.send(null);     alert(req.getAllResponseHeaders()); &lt;/script&gt;   Edit 4: Sorry, I meant that you could send the XMLHttpRequest to the StackOverflow domain, and then save the result of getAllResponseHeaders() to a string, regex out the cookie, and then post that to an external domain. It appears that Wikipedia and ha.ckers concur with me on this one, but I would love be re-educated...  Final Edit: Ahh, apparently both sites are wrong, this is actually a . IE6 &amp; 7 are actually the only browsers that currently fully support HttpOnly.  To reiterate everything I've learned:   HttpOnly restricts all access to document.cookie in IE7 &amp; and FireFox (not sure about other browsers) HttpOnly removes cookie information from the response headers in XMLHttpObject.getAllResponseHeaders() in IE7. XMLHttpObjects may only be submitted to the domain they originated from, so there is no cross-domain posting of the cookies.   edit: This information is likely no longer up to date. How do I know if I'm working on a Virtual Machine or not? Is there a way to know if the Windows machine I'm working on is virtual or physical? (I'm connecting with RDP to the machine. If it's a virtual machine it is working and handled by VMWare). Can current be induced in a superconductor?