I have this html:  &lt;div class="parent"&gt;     &lt;div class="child"&gt;&lt;/div&gt;     &lt;div class="child"&gt;&lt;/div&gt;     &lt;div class="child"&gt;&lt;/div&gt; &lt;/div&gt;   At the event of a click I need to fire some actions. Problem is that I need to exclude the clicked element. Let's say that I click at the first .child then actions just be applied ONLY to the second and third .child elements.    $('.child').on("click", function() {     $(".parent").find(".child").DO_SOMETHING();   });
I have something like this:  &lt;div class="content"&gt;     &lt;a href="#"&gt;A&lt;/a&gt; &lt;/div&gt; &lt;div class="content"&gt;     &lt;a href="#"&gt;B&lt;/a&gt; &lt;/div&gt; &lt;div class="content"&gt;     &lt;a href="#"&gt;C&lt;/a&gt; &lt;/div&gt;   When one of these links is clicked, I want to perform the .hide() function on the links that are not clicked. I understand jQuery has the :not selector, but I can't figure out how to use it in this case because it is necessary that I select the links using $(".content a")  I want to do something like  $(".content a").click(function() {     $(".content a:not(this)").hide("slow"); });   but I can't figure out how to use the :not selector properly in this case.
How to make Gmail automatically label incoming emails I would like to set it up so that Gmail labels incoming mail based on which business email of mine the emails are sent to. I have two business emails that come to my Gmail account.  For instance, if someone emails me at example@firstbusiness.com, I would like the email to be automatically labeled first business. If someone emails me at my other email address (which also is forwarded to Gmail), example@secondbusiness.com, I would like that email to be automatically labeled second business.   Is there any way to do this? Automate fixes for issues found by Sonar Cancelling a Schengen Visa via the Italian Consulate Connecting two Ethernet PHY without magnetics? Disposing WPF User Controls How can I set the maximum amount of characters in a UITextField on the iPhone SDK when I load up a UIView? How can I edit/create new launcher items in Unity by hand? Will Unity allow making custom launcher icons from .desktop files or via menu editing system? (Right now the launcher doesn't give the option to "keep in launcher" on all programs.  For some programs I use, I have to make custom launchers or .desktop files.  For instance, daily blender builds are generally just folders with an executable.  In basic Gnome or KDE, I can make a new menu entry with the menu editing system. Then, I can also add it to Docky either from the menu or by dragging a .desktop file to it. Unity launcher doesn't support drag and drop, so thats not a bug or anything, but when I open a .desktop file, it has unpredictable results. Most time it will not have "keep in launcher". Sometime it will have a pinnable item without the .desktop's icon, and if I pin the item to the launcher, it will not call upon the program again after closing it. I've also gotten it to just work with a .desktop file for celtx. Parallel universes: are there an infinite amount of universes? Parallel universe: in every universe, you are presented with a different decision. Does that mean there are an infinite amount of parallel universes? I.e., if in one universe I choose to go left, I am then presented with different choices vs if i choose right, and that creates more universes with more decisions to be made.  Or does it mean that in the end of one's life we end up in the same place, meaning that its just one parallel universe that eventually collides with the other. Please be kind, I am an electronics nub.  This is in reference to getting an LED to emit photons.  From what I read (Getting Started in Electronics - Forrest Mims III and Make: Electronics) electrons flow from the more negative side to the more positive side.  In an example experiment (involving a primary dry cell, a SPDT switch, a resistor and an LED) it states that the resistor MUST be connected to the anode of the LED.  In my mind, if the electrons flow from negative to positive, wouldn't the electron flow run through the LED before the resistor; thereby making the resistor pointless? jQuery doesn't work after content is loaded via AJAX I'm having trouble understanding how to access each individual result returned from a subquery. The picture shows the code which:   takes fields and objects chosen by a user and assigns them to String variables, the string vars are used to query a dynamic query string, the query string is feed into a database.query and the result is stored in an array I can individually access each of the query results from this array (parent object results),  But I have not been able to individually access the subquery results.   If I feed the subquery results into an array...something like:  SObject[] c = p.get(i).getSObjects(childO);   I still can't get to each individual child record per parent. I can only see all child records per each parent. I need to get to each child record per adult, so that I can roll up results.  Can anyone help?  // These strings will be populated dynamically from a VF page String parentO  = 'Account'; String parentF  = 'Products_Purchased__c'; String childO   = 'Opportunity';     if(childO  == 'Opportunity'){ childO = 'Opportunities'; } String childF   = 'Product_Type__c';  // create dynamic SOQL query string String query = 'SELECT id, (select ' +childF+ ' from ' +childO+') FROM ' +parentO;  // query and store 33 results in variable p SObject[] p = Database.query(query);  for(Integer i = 0; i &lt; p.size(); i++){     system.debug('Parent result from SOQL query: ' +p[i]);     if(p.get(i).getSObjects(childO) &lt;&gt; null){         system.debug('Child result(s) from SOQL query: ' +p.get(i).getSObjects(childO));         system.debug('Child field: ' +c.get(i).get(childF));     } } Binomial Distribution Problem - Airline Overbooking Is it possible to install Ubuntu FROM a smartphone? So, I'm in a situation were I want to install ubuntu in my laptop (currently running windows 8) but I don't happen to have pendrives, CDs or the like available. The only "external" device I have with me is a samsung galaxy s2 with a micro SD-card in it.   Is it possible to use the phone as a pendrive to install ubuntu from it somehow? (without damaging the phone's OS and data) If $a\neq 1$, find $(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)$. If $a\neq 1$, find $$(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)$$   Or i.e.     If $a\neq 1$, find $\prod_{i=0}^n(a^{2^i}+1)$.   It really does seem like $$(a+1)(a^2+1)(a^4+1)\ldots(a^{2^n}+1)=a^{2^n+2^{n-1}+\ldots+2+1}+a^{2^n+2^{n-1}+\ldots+2+1-1}+a^{2^n+2^{n-1}+\ldots+2+1-2}+\ldots+a^2+a+1$$  Or i.e. $$\prod_{i=0}^n(a^{2^i}+1)=\sum_{i=0}^{2^n+2^{n-1}+\ldots+2+1}a^i$$  Is there an interesting proof of this? I'd appreciate any help. Can blender use the Alpha channel of an image as transparency on a plane? NOTE: Signing up or Logging in via OpenID is no longer officially supported from any of our sites. Currently,  is still on the internet, but we are in the process of completely removing it.    Stack Overflow was an early and strong supporter of . We built our sign up/log in flow around it. We were idealistic and had , but these hopes weren't realized. Over the years people have wondered if  is . We've had to remove support as OpenID providers  or .  The time has come to part ways. The reality is OpenID support has created a ton of complexity in our codebase and the number of users actively using OpenID simply don't justify that cost. Users have spoken with their actions. You prefer Google, Facebook and Stack Exchange (aka email/PW) based account auth.     Of our 9,813,747 accounts around one-tenth of a percent are actively using OpenID (defined as having visited a site in the past 12 months). If you include all the inactive accounts it is still less than 2.9% of all accounts.  So, what is the plan?   DONE - Remove the ability to create a new account with most OpenID providers DONE - Add the ability to  DONE - Convert Launchpad (used on  ) to OAuth DONE - Remove creating new accounts with Steam from  (Steam removed support for OAuth. Only 454 accounts that use Steam have visited Arqade in the past 6 months.) DONE - We will email all users that rely on deprecated OpenID credentials and request that they  DONE - Run an announcement banner for any user who signs in with deprecated OpenID credentials urging them to  DONE - End support for logging in via OpenID  In progress - Remove  (note: only 0.2% of authorized sites in the last year have been non-Stack Exchange sites.)   Obviously, if your account relies on OpenID, then make life easy and go now and .   If you've forgotten what you credentials you use to login, then  to refresh your memory. If you don't see one of the three account types shown in the image below (Google, Facebook or Stack Exchange) then click the "add more logins" button to add one of them to your account.    Remember, SE OpenID accounts are really email/pw accounts handled through complicated OpenID code path. They will be migrated to a simpler email/pw only code path as a part of this change. No extra work required on your part. Can creatures without hands (and as a consequence fingers) use rings? Pretty much what it says on a label. I have a monster for my next adventure that has a pretty interesting ring in his treasure. Since it is intelligent, and would be able to use the ring, I don't see why the ring should be lying there in a chest waiting to be looted.  Monster would use the ring, if it only had hands or even fingers. Hold a second, are fingers even necessary in 5e?  For reference, assume it only has grasping / manipulating tentacles like a  (although it is not one); Said tentacles are very flexible, capable of fine manipulation, and have varying thickness. querying Recently account viewed Why an R-module is an R/I-module precisely when it is annihilated by I? Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. Prove that the formula $(r + I) \cdot m = r \cdot m$ for $r + I \in R/I$, and $m \in M$ makes $M$ into an $R/I$-module if and only if $i \cdot m = 0$ for all $i \in I, m \in M$.   I don't quite understand why we must have $i \cdot m = 0$ instead of $I \cdot m \in M$. This is for proving the only if part. Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b&gt; 0$ and $c\geq 0$ ?   Is there a general proof for that relation ?