The issue is that I'm trying to select  elements with $("a") and its affect only few of the  tags in the document, my code  $(document).ready(function() {     $.ajax({         type: 'GET',         url: 'http://localhost/queryProjects',         success: function(data) {             var el = $('&lt;div&gt;&lt;/div&gt;');             el.html(data);             var ProjectsObject = $("b:contains('projectName')", el).parent().next();             var ProjectsArray = $.makeArray(ProjectsObject);             var DescriptionsObject = $("b:contains('description')", el).parent().next();             var DescriptionsArray = $.makeArray(DescriptionsObject);             for (var i = 0; i &lt; ProjectsArray.length; i++) {                 $("tbody:last-child").append('&lt;tr&gt;&lt;td&gt;' + (i + 1) + '&lt;/td&gt;&lt;td&gt;&lt;a id="link" href="#"&gt;' + ProjectsArray[i].textContent + '&lt;/a&gt;&lt;/td&gt;' + '&lt;td&gt;' + DescriptionsArray[i].textContent + '&lt;/td&gt;&lt;td style="margin-right=40px;"&gt;&lt;a href="#"&gt;Run&lt;/a&gt;&lt;/td&gt;&lt;/tr&gt;');             }         }     });     $("a#link").click(function() {         $("table").hide();     }); });   I have tried with an ID and without, like $("a") and $("a#link").  the AJAX is irrelevant but just so you see how I inject my info to the document.  Thanks!
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.
True for every formula. For a formula $ \gamma = r \iff (p_1 \vee p_2) $ there is true that: $\rho \models \gamma \iff \rho(r) = \max(\rho(p_1), \rho(p_2))$  Is it a true for every formula r? I kust realized that some other PC is also getting assigned a IP address (DHCP Lease) from my modem. I am sure this is not one of the PCs in my home.  My modem was using WEP security. I know WEP is far less secure and WPA is better, but this actually is a borrowed modem, and I did not know that WEP was used.  I immediately did the following : On My Modem    Changed the encryption mode from WEP to WPA   Changed the password   After this I do not see the intruder PC on the DHCP list.  On my Mac:     Changed the permissions on the public folder to "everyone - no access"   Turned off "File Sharing" in System Preferences (But I would want to turn it back on sometime)   My mac is the one that is always on, Both the admin user and the user I normally use are password protected.  Is the above sufficient? Should I do something more?  Also, is WPA security sufficient? Should I take other steps? I installed or upgraded to Ubuntu 16.04, but I'm having some problems with my graphics card.   The display is blank The display flickers Items on the screen do not display correctly Certain programs complain of graphics issues The computer boots to low-graphics mode Your specific issue may not be here, but try the solutions described in the answers anyway.   This question is very broad, but each one of these issues likely has one solution.    If you are experiencing a login loop, check the last part of my answer.    If you experience one of the listed issues and have your own solution (mine didn't work, or there's a faster way), feel free to post another POLITE answer, describing your problem and how you fixed it.    Those attempting to install Ubuntu, please refer to the section titled: If you are unable to enter a TTY or are trying to install Ubuntu. How to manipulate 2D plots? If no spell can reawaken the dead, how did the resurrection stone work? Dumbledore himself said that no spell can reawaken the dead. Isn't that the main objective of the resurrection stone? If one can't reawaken the dead then what is the use of the stone? Surely then one of the Hallows is flawed? Recall the following relevant definitions. We say that   $b$ is divisible by $a$ in $R$, or $a\mid b$ in $R$, if $b = r a$ for some $r\in R$. $a$ and $b$ are associates  in $R$ if $a\mid b$ and $b\mid a$, (or, equivalently, if $aR = bR$). $u\in R$ is a unit if it has a multiplicative inverse (a $v\in R$ such that $uv=vu=1$). $a$ and $b$ are unit multiples in $R$  if $a = ub$ for some unit $u\in R$.   Given these definitions, my question is,     If $R$ is a commutative ring with unity and $a,b\in R$ are associates in $R$, are $a$ and $b$ unit multiples in $R$?   I was told that this not always true. But I encountered some difficulties in finding a counterexample. Java: Detect duplicates in ArrayList? How could I go about detecting (returning true/false) whether an ArrayList contains more than one of the same element in Java?  Many thanks, Terry  Edit Forgot to mention that I am not looking to compare "Blocks" with each other but their integer values. Each "block" has an int and this is what makes them different. I find the int of a particular Block by calling a method named "getNum" (e.g. table1[0][2].getNum(); How can edge rings be selected in a predictable way? When trying to select multiple edge-rings Shift-Alt-RMB sometimes loops get selected in the one (green) way or the other (red) way.  I couldn't derive a general rule.   How can two or more face-rings following one direction be selected ?  (Easy to reproduce with a default torus mesh.) Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through the point $P=(12,3)$ I was solving a few problems from a textbook and I came across this one:     Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through   the point $P=(12,3)$   I could find the tangency points $$ but my solution was completely messy (I had four equations that I had to go back-and-forth with.  Is there an elegant/simple solution to that?  Tangency points:  $$P_1 = (0,3)$$ $$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$ I'm on Ubuntu 16.04.1 at the moment, and system windows like the Files window, Terminal or System Software open behind the Unity launcher, which means they are partially hidden behind it. I can move them with Alt+F7, but I'd like a more permanent solution to this problem. Kernel panic after update to 4.4.0-108-generic Active Directory: Is it required that the "A" record for a domain point to a Domain Controller? If we had an airtight container that was large enough, say extended out to the Karman line, would there still be a pressure gradient? If we assume that the pressure at the bottom was $1\ \mathrm{atm}$, would it have the same exact pressure gradient at 6 miles up on the outside? This is assuming that the gravity is still $9.8\ \mathrm{m/s^2}$. Let $z_1, z_2, z_3,$ ...be a sequence of independent random variables s.t. $P(z_i=1)=P(z_i=-1)=\frac 1 2$. Does this sequence converge almost surely?  My intuition is telling me that the sequence will approach the random variable Z? i.e. $z_n-&gt;Z$ as $n-&gt;\infty$ and Z is just another RV s.t. $P(Z=1)=P(Z=-1)=\frac 1 2$.  Here is my confusion, if a sequence converges almost surely, then $z_i$ should be different for each term s.t. it approaches Z in the end; but in this sequence $z-i$ is the same for every term. So I dont udnersntand. Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $ I need to compute the improper integral $$ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $$ using contour integration and possibly principal values.  Trying to approach this as I would normally approach evaluating an improper integral using contour integration doesn't work here, and doesn't really give me any clues as to how I should do it. This normal approach is namely evaluating the contour integral $$ \oint_{C}{\frac{z^2}{\cosh\left(z\right)}\mathrm{d}z} $$ using a semicircle in the upper-half plane centered at the origin, but the semicircular part of this contour integral does not vanish since $\cosh\left(z\right)$ has period $2\pi\mathrm{i}$ and there are infinitely-many poles of the integrand along the imaginary axis given by $-\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ and $\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ for $n \in \mathbb{Z}$. The residues of the integrand at these simple poles are $-\frac{1}{4}\mathrm{i}\pi^{2}\left(1 - 4n\right)^{2}$ and $\frac{1}{4}\mathrm{i}\left(4\pi n + \pi\right)^{2}$, so that even when we add up all of the poles, we have the sum $4\pi^{2}\mathrm{i}\sum_{n = 0}^{\infty}\,n$, which clearly diverges.  Any hints would be greatly appreciated. I dual booted ubuntu 11.04/11.10 a while ago with a USB stick and i want to uninstall it.   I would like to completely uninstall the old version and get 12.04.  I have a 64 bit laptop. Please help! Express  $$\forall\ n\in\mathbb N\ \exists\ m \in\mathbb N: \ n^4 = m^2$$  in words without using the symbol $\mathbb N$. clerics have the ability  (PHB, p. 62):     When a creature within 5 feet of you that you can see hits you with an attack, you can use your reaction to cause the creature to make a Dexterity saving throw. The creature takes 2d8 lightning or thunder damage (your choice) on a failed saving throw, and half as much damage on a successful one.   It doesn't specify what the DC of the saving throw is.   I assume it uses the spell save DC, even though it technically isn't a spell?     Spell save DC = 8 + your proficiency bonus + your Wisdom modifier How to put a '+' sign inside a square like xor What ticket reservation do I need for a Schengen visa? I have a visa appointment this Friday 19 at the Italian Consulate in Philadelphia and I have a question about the flight reservation requirement: According to the Visa requirements "Confirmed flight reservation for a round trip showing complete itinerary."  Does this mean I have to reserve the tickets, pay and print the itinerary?  In this case if my visa gets denied then I will lose the money I paid or can I just show the planned itinerary (written not booked)? in this case the travel agency would not give me the Travel Insurance coverage until I book and pay for the reservation.   I am a US green card holder and applying from North Carolina.