I'm learning vue.js now on laracasts.com and there jeffrey way shows 2 way to display data on web page.  One is to display data from Laravel foreach and second is to display data with vue.js. I have one question: is there any difference by performance to display data with Laravel or display data with vue.js?
For the following pretty straightforward task: query a list of products from a DB and present it on a webpage, consider 2 setups:  Setup1: PHP script queries. All content is built on the server and the entire page is served back to the client.  Setup2: Static HTML "page skeleton" requesting content using AJAX. Received content is parsed on the client side using Javascript and rendered using innerHTML or similar.  Of course the second setup only makes sense when you have pages, categories and tags for the client user to choose from.  I need to compare these two, at least by means of:   time it will take content to be served user experience (setup1 is delivered as a whole, setup2 is delivered in "two parts") scalability - how do the setups compare when I have 100,000 queries daily   Any thoughts on the issue will be much appreciated.
If I have a sequence of .obj (or any other readable format) files created from my own simulation code (say cloth simulation), can I use the rendering capabilities of blender to generate a video?   Is there a way to set the material properties for frame 1 and use the same or rest of the frames while the corresponding meshes are imported afresh from that ones I provide? Zooming in to a selection of vertices with the period key does not work as expected I am following through the exercises in BlenderBasics_4thEdition2011.pdf and have gotten to the section on texturing.  The book goes through the creation of a Lighthouse and now I am texturing it.  I want to have a closeup view of the roof of the lighthouse after partitioning it as the lesson describes on page 54.    In order to get a close up view I selected the roof and want to center the screen on the vertices of it so I pressed the Numpad . key.  When I do that the view zooms so that the roof is off the screen.    What am I am doing wrong?    I tried Ray's answer and was able to move the origin but pressing the Numpad . key still did not work.  The centering was still on the bottom of the lighthouse, as if the origin was still there. Why do snap-packages exist - is there a real need? Assumptions  Honestly, I do not know much about  snap-packages - but this is irrelevant for this question - see below. I assume the system is significantly different from the existing one.  Does the change make sense?  Is there an actual need, strong enough? That is - is there a new use case, that is important enough to develop a new format - and the associated infrastructure?  Was it not feasible to change the current method to cover the new use cases too?  Or am I missing the point?  It is possible that what I see is mostly marketing - new names and presentation for minimal technical change, to get a chance that any body assumes it as "new and better" and it may be actually used. Also, it may be that the new packages are so closely based on the existing format that it is mostly a change in presentation to the user. That could be a good solution, of course. In this case, this question would have not much relevance.  Then, I would hope that is still useful enough to answer to side aspects.  Just let me know if the question is not helpful or confusing to new users, I'm happy to delete it.  So, why do they exist?    Background  My first reaction was "That makes no sense!"  This is similar to a situation where, on the physics site, someone asked, unhappy, why nobody discusses his brilliant new ideas in answers. It looked strongly like crackpot ideas; So far from actual physical knowledge that it was hard to find a point to even start. I wrote an answer that did not touch his ideas with a single word, but explained why one would just not discuss assuming crackpot ideas - not the first case of that. The answer was actually hitting the point, I think.    If my assumptions were right, this case is similar.    But then, maybe not - let's see. How can I get the node id the comment is attached to in hook_comment_insert?  I tried the following code:  function mymodule_comment_insert(Drupal\Core\Entity\EntityInterface $entity) {   if ($entity-&gt;bundle() == 'comment_custom' &amp;&amp; $entity-&gt;get('entity_type')-&gt;value == 'node') {      // Below returns empty, because "$entity-&gt;get('entity_id')-&gt;value" is empty.     $node = Node::load($entity-&gt;get('entity_id')-&gt;value);      // No luck with this one too.     $node = \Drupal::routeMatch()-&gt;getParameter('node');    } } Suppose you want to ask someone to do something, say to tell you the way to a station. Which is more polite, "Would you tell me the way to the station?" or "Could you tell me the way to the station?"? When I learned about the Minkowski Space and it's coordinates, it was explained such that the metric turns out to be $$ ds^{2} = -(c^{2}dx^{0})^{2} +(dx^{1})^{2} + (dx^{2})^{2} + (dx^{3})^{2}   $$ where $ x^0,x^1,x^2,x^3 $ come from $ x^{\mu} : \mu = 0,1,2,3 $, and $ c $ is the speed of light. The first resource I had access to--I have to do a bit of digging for the exact paper--of course addressed that this invariant takes the tensor form: $$ ds^2 = g_{\mu \nu}dx^{\mu}dx^{\nu} $$ as well. This two things I've seen in all of my texts and online resources regarding resources. The element in question that varies between authors, is the time coordinate, $ x^0 $. When it was first explained to me, it was using the standard Cartesian representation for the spatial portion of the coordinates, and the time coordinate was labeled as $ x^0 = ict $. Squaring gives $ (x^0)^2 = -c^2t^2 $, and applying differential calculus, we get $ (dx^0)^2 = -c^2dt^2 $. Sensible and expected to have the tensor formula spit out the Minkowski Metric. The author then later explicitly states the coordinates are $ x^0 = ict, x^1=x,x^2=y,$ and $x^3 = z$.  My questions is then, why is it most authors on the subject omit the imaginary unit on the time coordinate? For example, .   The only reason I can fathom omission is if the author is using metric signature $ [+,-,-,-] $, where I started off learning the theory with signature $ [-,+,+,+] $ which may be the reason seeing the time coord with no imaginary unit seems dissonant to me. All help appreciated!  Edit: After reading the other answers, my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign. Why there isn't a feature that lets any user on SO to request other user to answer their question? When I browse through other questions to find my answer, sometimes I'll find answers similar to what I'm looking for, and though they don't exactly answer my question, the author might be able to do so. Could we have a feature on SO that will let me request that another user answer my question, in case I find someone who could potentially answer my question? Disable App Update Notifications on OSX? 'longtable' caption without incrementing 'table'? For reasons both complex and confusing, I would like to caption a longtable without incrementing the table counter in the process.  I want to have two consecutive longtable environments to be captioned with identical table numbers; i.e. one longtable captioned "Table x.y: Data Set (Part A)" and one captioned "Table x.y: Data Set (Part B)", with x.y being identical for both tables.  Apparently however, even \caption*{} command bumps table. I am a bit at a loss here.  Note: I know that the xtab package would enable me to have one long table with switches of headers and captions mid-table. However I got a (different) problem with that package, akin to  but sadly not solved by egreg's answer there; so this question should be considered longtable-only as xtab, supertabular etc. don't float my boat right now. :-( For each $x,y,z∈\mathbb N$, if $x<y$ then $x+z<y+z$ Deriving of formula for finding the length of median In the below image    $AD$ is the median of $\triangle ABC$  We know that $m_A = \frac 1 2 \sqrt{2b^2 + 2c^2 - a^2} $    But can someone tell me how it's derived !! I am just unable to think of it !!   But I assume it muste be linked with the Appolonius Theorem !! Change Legend Right to Left Reading in QGIS I am a new QGIS user and i have a challenge to convert legend from right to left reading in QGIS like ArcGIS. The words in the node of tree I want to plot this graph, but I don't know how to code the words in the nodes.     Here is my code:  \documentclass{article} \usepackage{amsmath} \usepackage{titletoc} \usepackage{tikz} \usetikzlibrary{trees,positioning,fit}  \begin{document}{\allowdisplaybreaks[4]}  \tikzstyle{bag} = [text width=2em, text centered] \tikzstyle{end} = []  \begin{center} \begin{tikzpicture}[sloped] \node (a) at ( 0,0) [bag] {$\ln r_0$}; \node (b1) at ( 3,1.5) [bag] {$\ln r_{10} + \sigma_1\sqrt{\Delta t_1}$}; \node (b2) at ( 3,-1.5) [bag] {$\ln r_{10}$}; \node (c1) at ( 6,3) [bag] {$\ln r_{20} + 2\sigma_2\sqrt{\Delta t_2}$}; \node (c2) at ( 6,1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c3) at ( 6,0) [bag] {$\ln r_{20} + \sigma_2\sqrt{\Delta t_2}$}; \node (c4) at ( 6,-1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c5) at ( 6,-3) [bag] {$\ln r_{20}$}; \draw [-&gt;] (a) to node [below] {} (b1); \draw [-&gt;] (a) to node [above] {} (b2); \draw [-&gt;] (b1) to node [below] {} (c1); \draw [-&gt;] (b1) to node [below] {} (c2); \draw [-&gt;] (b1) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c4); \draw [-&gt;] (b2) to node [below] {} (c5); \end{tikzpicture} \end{center}  \end{document}       Update code use  \draw [-&gt;] (a) to node [below] {} (b1.south east); \draw [-&gt;] (a) to node [above] {} (b2.north east); \draw [-&gt;] (b1.south east) to node [below] {} (c1.west); \draw [-&gt;] (b1.south east) to node [below] {} (c2.west); \draw [-&gt;] (b1.south east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c4.west); \draw [-&gt;] (b2.north east) to node [below] {} (c5.west); I'm quite confused about the following regular expressions I found in a shell script:  ${0##*/} ${0%/*}   How do they work? Is it acceptable to write a thank you in a comment? There have been a number of cases where I would like to thank someone for providing me with a thorough response and give them feedback beyond the vote up. Often I wish people would tell me why they voted me up. Not that the thank you part is necessary but it sets it apart from the other comments related to answering the question itself.  In this type of situation is a "Thank you" acceptable in a comment?     For more information, see  in the . How to proof that $\forall n \ge 1: \sum_{i=1}^n (i(i+1))^{-1} = n(n+1)^{-1}$ using mathematical induction So for a better presentation:   I need to prove this using mathematical induction, so i started with a base of $n=1$  Since $i$ starts at $i = 1$ I got this:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{2} = \frac{1}{2}$$  So since this was true I went to the induction part: I chose $n = k$ and assumed this was true:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{i(i+1)} = \frac{k}{k+1}$$  but I got stuck here since I didn't know if I needed to increment $i$ or not and because I didn't know if I needed to add the summation symbol or not and needed to do a few things with it. Is it possible to have several different views of one scene and set them as named "presets"? Can every even integer be expressed as the difference of two primes? Taxonomy terms view sorting results numerically I have a taxonomy that's comprised entirely of numeric term names, 1 through 1200 (with a few numbers missing in between). I created a new "Taxonomy terms" view to show all terms in this vocabulary. When I look at the view results, I'd like to see the terms sorted numerically, however the view is displaying the terms alphabetically. I understand the reason for this is because term names are strings, but I'm hoping someone has an answer of how to make the query treat the term names as integers.  Here's an example of the view results as they are displayed now:   1 10 100 1000 1001 1005 101 1014 1015 1016 1019 102 1020 1024 103 1034 104   Normally in MySQL, you could write SELECT name FROM {taxonomy_term_data} ORDER BY CAST(name AS UNSIGNED). Is there a way to get this same type of CAST() happening in views? how to avoid ssh asking permission? We are attempting to speed up the installation of oracle nodes for RAC installation. this requires that we get ssh installed and configured so that it doesn't prompt for a password.  The problem is: On first usage, we are prompted for  RSA key fingerprint is 96:a9:23:5c:cc:d1:0a:d4:70:22:93:e9:9e:1e:74:2f. Are you sure you want to continue connecting (yes/no)? yes   Is there a way to avoid that or are we doomed to connect at least once on every server from every server manually?