I'm adding some code(!with js code included!) dynamically.  &lt;div class="r"&gt;&lt;/div&gt; &lt;script&gt;     //Here is code that dinamiccaly add another div into div.r     //Example:     //&lt;div class="w"&gt;&lt;img src="a.png"&gt;&lt;div class="i"&gt;Some Text&lt;/div&gt;&lt;/div&gt;     $('.r').on('click', $('.w'), function (e) {         console.log($(this).children($('.i')).text());     }); &lt;/script&gt;   Problem that this click event works anythere I click in div.w and returns text of ALL div.w combined in one string without spaces (sometext1sometext2).  Sorry if I wrote something wrong. I'm bad in english.
I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.
How do I programmatically get the field type of a specific field? GCD in a PID persists in extension domains Let $R$ be subring of integral domain $S.$ Suppose $R$ is $\text{PID}.$ Let $a\in R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 \in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$   Hints will suffice, thank you. In my document, I will write something like this (using natbib, but that shouldn't matter too much):  \citet{reference} show that .....   which would render as     Jones et al (1990) show that....   (\citet in \natbib creates an 'author-list (year)' reference style)  Of course if {reference} was a single individual, I should have written  \cite{reference} shows that ....   rendering as     Jones (1990) shows that....   Since I'm lazy and can't take the extra 5 seconds to search the reference to see if it's a single author or not, is there conceivably an automated way to do pluralization ? Or some relatively simple macro that would look like   \IfRefIsPlural{reference}{show}{shows} Creating an index of quality from multiple variables to enable rank ordering How many seconds count until bomb detonates in Counter Strike? In Counter-Strike, is there any count (in seconds/minutes) until the bomb explode when it's first set-up? And can the timing be modified? Creating Layer in GeoServer using REST As part of a possible solution to   I want create a layer in GeoServer via REST that supplies a SQL statement along with a userid. I think i would  and publish it. I need to do this all in REST. Does anyone have any code examples? Usage of "shall we?" What does it mean and where would I use it? How to Drag and Drop files to open application? Since Ubuntu 18.04 I can't drag a file e.g. image file to an open application such as Inkscape or  email message window. In Ubuntu 17.10 I was able to select the image file by clicking on it, and while holding down left mouse button I would hit the supper key displaying all open windows where for there I could then drop the file in any open app or folders. This is no longer possible/default behaviour since Ubuntu 18.04. Is there a workaround that would allow me to achieve this or even a way to recreate the same behaviour like in 17.10? Is the vector cross product only defined for 3D? Complete userChrome.css specification in Firefox Adding Facebook Online Account - "application you are trying to use doesn't exist or has been disabled" pronoun agreement in neither nor Which one is correct?   Neither my friend nor I feel my best in the morning. Neither my friend nor I feel her best in the morning. Neither my friend nor I feel our best in the morning.    I already know that the verb agrees with the subject closer to the verb:   Neither my friends nor my brother understands me. (correct) Neither my friends nor my brother understand me. (wrong)   AND   Neither my brother nor my friends understand me. (correct) Neither my brother nor my friends understands me. (wrong) What is the use of the yield keyword in Python? What does it do? For example, I'm trying to understand this code1: def _get_child_candidates(self, distance, min_dist, max_dist):     if self._leftchild and distance - max_dist &lt; self._median:         yield self._leftchild     if self._rightchild and distance + max_dist &gt;= self._median:         yield self._rightchild    And this is the caller: result, candidates = [], [self] while candidates:     node = candidates.pop()     distance = node._get_dist(obj)     if distance &lt;= max_dist and distance &gt;= min_dist:         result.extend(node._values)     candidates.extend(node._get_child_candidates(distance, min_dist, max_dist)) return result  What happens when the method _get_child_candidates is called? Is a list returned? A single element? Is it called again? When will subsequent calls stop?   1. This piece of code was written by Jochen Schulz (jrschulz), who made a great Python library for metric spaces. This is the link to the complete source: [Module mspace][1]. C++11: defining function on std::array If I want my app to behave differently on a jailbroken iPhone, how would I go about determining this? I have a Multisite Wordpress installation with 7 different blogs. I want to show an aggregate of the summaries of posts from those 7 blogs in the main blog. Do you know a good way to do this? This is something I'm supposed to be able to prove for an upcoming test, but I can't find anything to help me prove this in my notes or the chapter, which is on cosets and Lagrange's theorem. If all I start with is the group having no proper subsets, then that means any subset is the whole group. By Lagrange's theorem, that means the index is $1$, but that doesn't get me anywhere. Where do I start? Matrices $B$ that commute with every matrix commuting with $A$ There have been many questions in the vein of this one, but I can't find one that answers it specifically.     Suppose $A,B\in M_n(\mathbb C)$ are two matrices such that, for any other matrix $C\in M_n(\mathbb C)$, $$AC=CA\implies BC=CB.$$ Prove that $B=p(A)$ for a polynomial $p\in\mathbb C[t]$.    Nothing is assumed about $A$ being diagonalisable or having distinct eigenvalues or being invertible.  Edit: after further work, this might not actually be true. If it's not, a counterexample would be great. Time complexity of nested for-loop Central Pennsylvanian English speakers: what are the limitations on the "needs washed" construction?