Selecting all childs text value
How do you parse and process HTML/XML in PHP?
Reference - What does this error mean in PHP? Examples and further results about the order of the product of two elements in a group chpasswd apply changes in the chroot directory When I tried to install Matlab using the Software Center, it prompted me for the Location of MATLAB installation(s). Since I am trying to install it, I don't know what to do. Find the integer solutions: $$a·b^5+3=x^3,a^5·b+3=y^3$$ This is the first problem of today's USAJMO (has finished),I only find a trival result that $x\equiv y \pmod6$ and $abxy≠0 \pmod 3$.  Thanks in advance! How can I reorder my divs using only CSS? Given a template where the HTML cannot be modified because of other requirements, how is it possible to display (rearrange) a div above another div when they are not in that order in the HTML? Both divs contain data that varies in height and width.  &lt;div id="wrapper"&gt;     &lt;div id="firstDiv"&gt;         Content to be below in this situation     &lt;/div&gt;     &lt;div id="secondDiv"&gt;         Content to be above in this situation     &lt;/div&gt; &lt;/div&gt; Other elements   Hopefully it is obvious that the desired result is:  Content to be above in this situation Content to be below in this situation Other elements   When the dimensions are fixed it easy to position them where needed, but I need some ideas for when the content is variable. For the sake of this scenario, please just consider the width to be 100% on both.  I am specifically looking for a CSS-only solution (and it will probably have to be met with other solutions if that doesn't pan out).   There are other elements following this. A good suggestion was mentioned given the limited scenario I demonstrated&mdash;given that it might be the best answer, but I am looking to also make sure elements following this aren't impacted. Proof that in a field $x=0$ is equivalent to $x=-x$ Let $F$ be a field and $x\in F$. If $x=-x$ and $1\neq -1$, then $$0=\frac{x+x}{1+1}=\frac{1+1}{1+1}x=x.$$ This means that the statement in the title is true if and only if $1\neq -1$. But how do we know that $1\neq -1$ for an arbitary field? Ubuntu 17.10 and later: can't change the input switching shortcut to alt+shift 16.04 VirtualBox VM (from vhd file) hangs at "non blocking pool is initialized" I'm trying to boot Canonical's Azure 16.04 .vhd file in a VirtualBox VM and it hangs at "non blocking pool is initialized".  It works in Hyper-V, so it may only be a bug in virtual box (5.0.20). Why does `Position` by default return a list of lists of positions instead of a simple list of positions? Geometric series for two-point function In deriving the expression for the exact propagator  $$G_c^{(2)}(x_1,x_2)=[p^2-m^2+\Pi(p)]^{-1}$$   for $\phi^4$ theory all books that i know use the following argument:  $$G_c^{(2)}(x_1,x_2)=G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}+G_0^{(2)}\Pi G_0^{(2)}\Pi G_0^{(2)}+...$$  wich is a geometric series so the formula for the exact propagator.Here  $$G_0^{(2)}$$   is the free propagator and   $$\Pi=X+Y+Z+...+W$$  is the sum of all irreducible diagrams.Here the irreducible diagrams is represented by $X,Y,Z,...,W$.  Using the path integral i can see, that  connected diagrams $D$ can be written in the form  $$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$  Question: But how to prove that there is no constant $C$ so that instead we have  $$D=G_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W$$  we would have  $$D=CG_0^{(2)}XG_0^{(2)}Z...G_0^{(2)}W~?$$  In the last case we would not have a geometrical serie. Can someone explain me i t please  or give me another way to derive it. How do I add custom validations to Webform? I have created the country and state fields with the Webform module, and I want to add a field validation for the state field when user select US as country.   How can I achieve this? Probably a really simple one this - I'm starting out with C# and need to add values to an array, for example:  int[] terms;  for(int runs = 0; runs &lt; 400; runs++) {     terms[] = runs; }   For those who have used PHP, here's what I'm trying to do in C#:  $arr = array(); for ($i = 0; $i &lt; 10; $i++) {     $arr[] = $i; } I try to delete Action.005 permanently.    But I don't know how. If I click on the X    then it is still in the list.    I already researched without success, I even managed to loose all my animations by attempting to apply a fix, so I want to be cautious now... How can I do this? A twice differentiable function $f$ satisfies $f′′(x)+f(x)=−xg(x)f′(x)$, $\forall x\ge 0$. Consider $f\in C^2$ so that $$f''(x)+f(x)=-x\,g(x)f'(x), \ \forall x\ge0 $$ where $g(x)\ge 0$. Then ($\forall x\ge 0$)         (A) $f(x)^2+f'(x)^2$ is non-increasing,   (B) $f(x)^2&lt;3f(0)^2+(2f'(0))^2$,   (C) $|f(x)|\le\alpha$, where $\alpha$ is a fixed real constant.   (D) $\lim_{x\to\infty} f(x)\sin\left(\dfrac{1}{x}\right)$ exists.        Answer Given: (A),(B),(C),(D)  What I have tried : I differentiated option (A) and then with the help of the given equation in the question I was able to show that  it is less than zero . Hence I was able to infer that option (A) is correct .   Now coming to option (B).  I assumed the function in option (A) to be $h(x)$.   $$ h(x) = f(x)^2 + f'(x)^2 $$  Since $h'(x)&lt;0$ ,therefore $h(x)$ is a decreasing function   hence for all $x\ge 0$  $h(x)&lt; h(0)$  which implies $(f(x))^2 + (f'(x))^2&lt;(f(0))^2 + (f'(0))^2$  From the above expression we can get the second option .   Is my method correct ?. If not , please show the right method .  I am unable to get the the 3rd and 4th option . Clean install of Windows 10 OEM: license key issues? We are in the combat phase:   I declared my attackers  My opponent declared his blockers, stating which attackers he was blocking I cast an instant, which automatically kills his blocker. There is nothing blocking my attacker now   Now we're in the damage phase. I said that my attacker now does damage and  reduces my opponent's hit points. My opponent said that because my attacker was blocked no damage could be done. I pointed out that the blocker was destroyed before the damage phase, so therefore he wasn't blocking, but he wasn't having it. (My attacker does not have trample)  Is this covered by a specific section in the rules? Book recommendation on plane Euclidean geometry I consider myself relatively good at math, though I don't know it at a high level (yet). One of my problems is that I'm not very comfortable with geometry, unlike algebra, or to restate, I'm much more comfortable with algebra than geometry. I think that's mainly because my geometry education was sparse through the years, lacking in consistency etc. So I'd like to revise (and learn more) all at once, catching the basic axioms, understanding why such is such, etc. Essentially, a moderately rigorous textbook in plain Euclidean geometry (nothing fancy). Please don't say "The Elements" - I have browsed it at the bookstore, it is quite good, but not really what I'm looking for right now. I'm basically searching for the opposite of putting all your eggs in one basket, where the risk is total failure because you did not hedge your efforts. I'm searching for a phrase that encompasses splitting your efforts and evokes the prospect of failing at both. A bird in the hand is worth two in the bush also encompasses a sense of having focused your efforts, but only alludes to total failure. To satisfy the criteria, the phrase must cover:  attempting two things (or multiple iterations of one thing) both of which you might succeed at, if you addressed them serially but both of which you fail at because you address them in parallel  Here are the idioms/phrases that do not sufficiently cover the idea and why:  &quot;A servant to two masters&quot;  This has the sense of competing priorities, but not the sense of failing to achieve your goals.  &quot;Jack of all trades. Master of none.&quot;  The sense is insufficiently pejorative. Being a jack-of-all-trades can be seen as an advantage; I'm trying to focus on the lack of mastery.  &quot;The man who fancied himself a wit, and was half-right.&quot;  From Christopher Hitchens. Hilarious, but lacking the sense of split effort between two things.  &quot;Walking and chewing gum at the same time.&quot;  An idiom for multi-tasking, not for failing to multi-task. Edit Addressing a few suggestions and another one that occurred to me:  &quot;Hedging your bets&quot; or &quot;Betting against yourself&quot;  In this case, you cannot win both bets, but you will almost certainly win one. You limit your upside, but also limit your risk. I'm looking in particular for total, catastrophic failure due to split attention/effort.  &quot;Can't walk and chew gum at the same time&quot;  Failing to do two fairly simple things that most everyone can do, not failing at two hard things because you tried to do both.  &quot;Robbing Peter to pay Paul&quot;  In this case, Paul gets paid. If you were to rob Peter to pay Paul, and get robbed yourself before you could get the money to Paul, that'd be closer. :)  &quot;Spreading yourself too thin&quot;  An actual English idiom, but missing the implied sense of failure.  &quot;Too many irons in the fire&quot;  &quot;A lot of irons in the fire&quot; can be viewed positively, so this is missing the sense of a strategic misstep.  &quot;Between a rock and a hard place&quot;  The sense is of needing to act but having no good alternative.  &quot;Falling between two stools&quot;  My favorite answer so far because it does appear to be an actual English idiom (not a proverb). In that sense, it's great, but some of the usages are closer to a failure of  than a failure of effort. And falling between two stools implies that the attempted action was simply sitting down.  &quot;Chasing two rabbits and catching none.&quot;  This is the exact sense I was looking for, but it appears to be a cross-cultural proverb rather than idiomatic English like &quot;putting all your eggs in one basket&quot;. I'll be accepting it, and using it in my work! But if there are other ideas, please let me know. register_nav_menus() won't register menus Please help with menu registeration:  I have followed this guide (http://codex.wordpress.org/Navigation_Menus) step by step.  Added the following code to functions.php:  function register_my_menus() {   register_nav_menus(     array('header-menu' =&gt; __( 'Header Menu' ) )   ); }  add_action( 'init', 'register_my_menus' );   But the guide says that this will make the menu appear in the Appearance -> Menus admin screen in the "Theme Locations" box as "Header Menu".  When I go to Appearance -> Menus everything is greyed out and there is no option of "Theme Locations"