I have a page with a list box that gets it data from mysql. The list box allows the user to delete a item. I want to load the results of the deletemanagement.php in a div without reloading the main page. I also want to reload the list box so the delete item is not there anymore and finally the user should be able to delete another item from the (updated) list box. So far I managed to have the list box delete records from my database, load the php into a div and update the list box (by reloading the div). However when I try to delete a second item, the page fully reloads and and does not delete the selected item. After clicking submit again it works and the div is updated again. Why is the page reloading when trying to delete a second item and how can I prevent that?     &lt;div id=&quot;deleteMangement&quot;&gt;      &lt;form id=&quot;test&quot;&gt;         &lt;h2&gt;Delete management:&lt;/h2&gt;         &lt;p&gt;           &lt;label for=&quot;deleteMangementSelect&quot;&gt;Select from list:&lt;/label&gt;           &lt;br&gt;           &lt;select id=&quot;deleteMangementSelect&quot; name=&quot;deleteMangementSelect&quot; required&gt;             &lt;option value=&quot;&quot;&gt;Select&lt;/option&gt;             &lt;?php               $i=0;               while($row = mysqli_fetch_array($result)) {             ?&gt;               &lt;option value=&quot;&lt;?=$row[&quot;management&quot;];?&gt;&quot;&gt;&lt;?=$row[&quot;management&quot;];?&gt;&lt;/option&gt;             &lt;?php               $i++;               }             ?&gt;           &lt;/select&gt;         &lt;/p&gt;         &lt;p&gt;           &lt;input type=&quot;submit&quot; value=&quot;Delete&quot;&gt;         &lt;/p&gt;       &lt;/form&gt;     &lt;/div&gt;    &lt;div id=&quot;output&quot;&gt;    &lt;/div&gt; &lt;script&gt; $(&quot;#test&quot;).submit(function(event){   event.preventDefault();   $.post( &quot;deleteManagement.php&quot;, $(&quot;#test&quot;).serialize(), function(data){     $(&quot;#output&quot;).html(data);     $(&quot;#deleteMangement&quot;).load(location.href + &quot; #deleteMangement&quot;);   }); }); &lt;/script&gt;
I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.
Eisenstein Criterion with a twist As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$? Is there an irrational number $a$ such that $a^a$ is rational? It can be proved that there are two irrational numbers $a$ and $b$ such that $a^b$ is rational (see ) and that for each irrational number $c$ there exists another irrational number $d$ such that $c^d$ is rational (see ).  My question is: Is there an irrational number $a$ such that $a^a$ is rational (and how could you prove that)? Celebrating the same new year several times over? What's a concise way to check that environment variables are set in a Unix shell script? If a topology contains all infinite subsets, then it is the discrete topology I am solving question $8$ of exercises in section 1.1. of chapter 1 in "Topology without tears". The question reads as follows.     Let $X$ be an infinite set and $\tau$ be a topology on $X$. If every infinite subset of $X$ is in $\tau$, prove that $\tau$ is the discrete topology   My idea was to show that every singleton set is in the topology and thereby the topology is discrete.  Choose an infinite set $A$ from $X$ such that $A^c$ is also infinite. $A$ and $A^c$ belong to $\tau$ since they are infinite sets. Now consider any $x \in X$. Note that $A \cup \{x\}$ and $A^c \cup \{x\}$ belong to $\tau$ since they are infinite sets. Hence, their intersection which is nothing but $\{x\} \in \tau$ for every $x \in X$.  The problem I have is I don't know how to prove the first sentence in the previous paragraph. These are my line of thoughts to prove them.   If $X$ is a countably infinite set, then I can list the elements are let the odd numbered elements fall into $A$. This guarantees $A$ and $A^c$ are both infinite. If $X$ is uncountable, then I can choose a countably infinite subset and call it $A$.   Are the above arguments rigorous? I am not especially happy with my second argument of choosing a countably infinite subset from an uncountable set since I do not give an explicit procedure of constructing the set $A$.  Also, is there any other simpler way of answering the original question? Another meta question identified a bug that was marked as designed. To summarize the bug, the little number that appears next the Review link in the header shows an incorrect value most of the time.         I would like to propose that the feature be dropped. Honestly, I would prefer it be fixed, but the response to the question above gives me little faith that it will be. Half of the time, I click that little brown box and find out that I was lied to. There's nothing to review.  The little number no longer does the job it was created to do.        Oh sweet! Three things I can review!        Just kidding... Did she judge him "wrong" or "wrongly"? Which one is the correct use?  She judged him wrong. She judged him wrongly.  Or, are both correct, but have slightly different meanings? Defining macros with arguments With newcommand I can define customized command like \ABC. Is there any way to define a command with arguments. For example I want \brat{A}{B} be the same as [\![A,B]\!]. I have the low battery alarm disabled in the BIOS settings on my Lemur laptop, but the alarm still sounds when the battery is low.  The sound is a very loud and annoying beep that repeats every couple seconds. The laptop will last another 20 minutes without being plugged in so the battery level isn't properly being detected anymore anyways. I found that if I mute my audio then the beep stops but I obviously cannot listen to anything anymore. If i turn the volume back on even to just 1% the battery alarm is audible again at full volume.  I remember there used to be a mixer somewhere that could control the internal speaker and so if I could find that again then maybe i could mute that and fix this problem.   I have Ubuntu 14.04.  Note that I'm not talking about a low battery notification, which is something through gnome; as opposed to a low battery ALARM which is this very loud beep every couple seconds coming through the internal speaker.  Nevertheless, here are my gnome power settings:  $ gsettings list-recursively org.gnome.settings-daemon.plugins.power org.gnome.settings-daemon.plugins.power button-power 'interactive' org.gnome.settings-daemon.plugins.power critical-battery-action 'nothing' org.gnome.settings-daemon.plugins.power percentage-low 1 org.gnome.settings-daemon.plugins.power priority 0 org.gnome.settings-daemon.plugins.power lid-close-suspend-with-external-monitor false org.gnome.settings-daemon.plugins.power idle-dim false org.gnome.settings-daemon.plugins.power button-hibernate 'hibernate' org.gnome.settings-daemon.plugins.power sleep-inactive-ac-type 'suspend' org.gnome.settings-daemon.plugins.power button-sleep 'suspend' org.gnome.settings-daemon.plugins.power button-suspend 'suspend' org.gnome.settings-daemon.plugins.power sleep-inactive-battery-timeout 0 org.gnome.settings-daemon.plugins.power time-low 1200 org.gnome.settings-daemon.plugins.power lid-close-ac-action 'nothing' org.gnome.settings-daemon.plugins.power notify-perhaps-recall true org.gnome.settings-daemon.plugins.power percentage-critical 1 org.gnome.settings-daemon.plugins.power percentage-action 1 org.gnome.settings-daemon.plugins.power sleep-inactive-battery-type 'suspend' org.gnome.settings-daemon.plugins.power time-action 120 org.gnome.settings-daemon.plugins.power lid-close-battery-action 'nothing' org.gnome.settings-daemon.plugins.power idle-brightness 30 org.gnome.settings-daemon.plugins.power sleep-inactive-ac-timeout 0 org.gnome.settings-daemon.plugins.power time-critical 300 org.gnome.settings-daemon.plugins.power active true org.gnome.settings-daemon.plugins.power use-time-for-policy false $ With slapd.conf you could globally disable anonymous binding and require authentication with the following static directives:  disallow bind_anon require authc   How can I achieve the same global settings, but using the new cn=config live configuration method? Short story where the Earth has been kept inside a force field for millennia The question asks us to classify all distributions in $\mathbb{R}^2$, say $u$, such that $$(xy)u(x,y) = (x^2-y^2)u(x,y)= 0.$$ I am genuinely stuck, any help is extremely appreciated. Automatic text wrapping in ArcMap layout text element? How to create batch file in Windows using "start" with a path and command with spaces I need to create a batch file which starts multiple console applications in a Windows .cmd file.  This can be done using the start command.  However, the command has a path in it.  I also need to pass paramaters which have spaces as well.  How to do this?  E.g. batch file  start "c:\path with spaces\app.exe" param1 "param with spaces" how to change the distance between tick mark and tick label What tools (ideally affordable ones) would folks recommend for creating a website wireframe? Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance. Here, we suppose that $x,y\in\mathbb{R}$ and that $x^n=y^n$, where $n$ is odd. I want to prove that $x=y$.  Maybe we can use that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$  So, it suffices to show that if $x^n=y^n$ then $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0$  Any hint? Delete the entire table rendered from different pages using Javascript Solving Keplerian two-body problem numerically? I'm trying to solve the two-body problem numerically, setting up G, m1 and m2 =1.0.  The masses are placed at the positions -10 and 10 respectively along the x-axis and gave them both 0 on the y-axis. I am having some real issue with the initial conditions fitting for a keplerian orbit with a given eccentricity (e=0.9)! Also how would the initial condition change for different mass ratios?