Removing $() jQuery wrapper to just get raw JS element
How to get a DOM Element from a jQuery selector?
jQuery - setting the selected value of a select control via its text description I have a select control, and in a javascript variable I have a text string.  Using jQuery I want to set the selected element of the select control to be the item with the text description I have (as opposed to the value, which I don't have).  I know setting it by value is pretty trivial. e.g.  $("#my-select").val(myVal);   But I'm a bit stumped on doing it via the text description. I guess there must be a way of getting the value out from the text description, but my brain is too Friday afternoon-ed to be able to work it out. We have an Always On "cluster" that consists of 2 servers (there will be more), so one is PRIMARY and other(s) is(are) SECONDARY. The idea was to dedicate SECONDARY as a readonly replica, so it would be a somewhat search-server.  I do know how to set ConnectionString in the C# so it will use SECONDARY replica - just add ApplicationIntent=ReadOnly.  But I don't understand how to address the SECONDARY from stored procedure if I'm using distributed query. Is there a way to set some parameters to the query (like WITH statements or something), so query would use only replica and not the PRIMARY node?  The thing is, obviously from the start server1 is PRIMARY and server2 is SECONDARY, but when server1 fails then server2 is PRIMARY and server1 will be SECONDARY after it get fixed. So I cannot just use the static server2's name.  So far I've managed to get current replica's server name as a variable and use it in EXEC:  -- find first available replica DECLARE @replicaServer nvarchar(50)  SELECT TOP 1     @replicaServer = RCS.replica_server_name FROM     sys.availability_groups_cluster AS AGC     INNER JOIN sys.dm_hadr_availability_replica_cluster_states AS RCS           ON RCS.group_id = AGC.group_id     INNER JOIN sys.dm_hadr_availability_replica_states AS ARS           ON ARS.replica_id = RCS.replica_id     INNER JOIN sys.availability_group_listeners AS AGL           ON AGL.group_id = ARS.group_id WHERE ARS.role_desc = 'SECONDARY'       AND connected_state = 1  -- and query it DECLARE @cmd nvarchar(MAX) = 'SELECT * FROM [' + @replicaServer + '].[somebase].[someschema].[sometable]' EXEC (@cmd)   But that's a shame, I guess. Should Dalvik Cache be wiped in Android OS that supports ART? When switching custom ROMs (in ART), is it necessary to wipe the Dalvik Cache?  Wiping the Cache is necessary, but should wiping Dalvik (which doesn't exist in Android Lollipop and afterward) be necessary?  What impact will it cause if we wipe Dalvik (something not present)?  In ART, what does the Dalvik cache contains? (is it an empty folder?) Let $V=\{f : S \to \mathbb{R} \}$ be the set of all the functions $f : S \to \mathbb{R}$. Let $S=\{ 1,2,...,n$}, now my train of thought was that the basis of V must be a linearly independent set that Span V so somehow show that there is a set of functions on S that can compose as a linear combination and map to any number in $\mathbb{R}$. Any idea how to get a general answer for basis for dimension of V? Proving that all entire & injective functions take the form $f = ax + b$? I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?  Also, more specific questions about the different cases:   Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple? Essential singularity case: Why exactly is $f(\{|z| &gt; r \} \cap f(\{|z|&lt;r\}) \neq \emptyset$? $f(\{|z| &gt; r \})$ is dense, but how does $f(\{|z|&lt;r\}$ being open guarantee that their union is non-empty? I am trying to solve:  $$\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx$$  and show that it equals $\pi e^{-a}$ for all $a &gt; 0$.  I translate it to its complex equivalent:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$  where C is a half circle with radius R, above the real axis, and $\gamma$ is only the curved (top part) of the half circle.  I've shown that $$\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$ tends to 0 as $R \rightarrow \infty$.  So now:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz$$  By residue theorem,   $$\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f$$  as $ai$ is the pole of $f$ that lies above the real axis.  $$2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}$$  By L'Hospital's rule, this is equivalent to:  $$2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}$$  But how is $\pi i \sin{ai}$ equivalent to $\pi e^{-a}$? Did I make a mistake? What sci-fi novel has the moon split into 7 pieces by an asteroid, threatening life on earth? I am looking for the title of a novel that I only caught part of on audio. It takes place within our current time frame and deals with the Moon getting struck and fragmented into (initially) 7 pieces by a high-speed asteroid that they called "The Agent".  Within a few weeks of this episode, scientists figure out the the Moon will continue to break down in a cascade reaction over the next couple of years until the planet gets bombarded by so much debris that it sterilizes Earth. The governments of the world frantically start to build a series of ships using swarming concepts to evacuate the planet as much as possible... and that's all I have heard so far.  Thanks for any info. $g\in G$ maximal order in $G$ abelian then $G=\left\oplus H$ If $g\in G$ is an element of maximal order in a finite abelian group $G$ then exists $H\leq G$ such that $G=\left&lt;g\right&gt;\oplus H$  Attempt: Using fundamental theorem I know that $G=C_{n_1}\times\cdots\times C_{n_k}$ where $n_i|n_{i+1}$. With some work I proved $|\left&lt;g\right&gt;|=n_k$, so $C_{n_k}\cong \left&lt;g\right&gt;$. Then $G\cong H\times \left&lt;g\right&gt;$, but this is not an equality. Also how do LEFT JOIN, RIGHT JOIN and FULL JOIN fit in? Working with objects created with Activator.CreateInstance Do lenses have focal planes or focal spheres/ellipsoids? 80's pop song with lyric "It's the story of my life if you wanna know" I'm trying to identify a song by a female 80's pop group.   Here are the lyrics I remember:     It's the story of my life if you want to know   a play where everything is meant to be   you're the story of my life and I can't let you go I saw this bug mentioned in image below, where in I was about to mark the question as duplicate..    Has you can see there is no radio buttons to select and go further.. And I am stuck..  Is this a KNOWN bug..  I am using Mozilla Firefox Browser..  Edit:-  As guided by @AzizShaikh, This kind of behaviour is observed when you have already flagged a post. In my opinion, I think the page as I mentioned above should not have come in first place, I mean don't allow the user to enter it doesn't belong here, or it is a duplicate section if his vote has been posted successfully for that section rather than showing go nowhere Im lost Screen ..  Can we do something about this...?? How can I get more of this macro photo in focus? If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is  $\bf{My\; Try::}$ Given $11z^{10}+10iz^9+10iz-11 = 0\Rightarrow \displaystyle z^9 = \frac{11-10iz}{11z+10i}.$  Now Put $z = x+iy\;,$ we get $\displaystyle (x+iy)^9 = \frac{11-10i(x+iy)}{11(x+iy)+10i} = \frac{(11+10y)-10ix}{11x+i(10+11y)}$  Now i did not understand how can i solve it,  Help Required  Thanks Which is Most TimeMachine-like Backup Program for Ubuntu? I know this  but there is no comparison to OSX's TimeMachine.  I would like to find a backup software which is about the same as TimeMachine or better.  I would like to have such a software which can use some format of HDD which is suitable for big files (50 GB) in Linux.  I noticed that   OSX format of HFS+ is not supported by Linux distros, since no maintainer of HFS in Linux since 2006 NTFS format made by Debian 8.1 is not supported by older Linux distros   If Windows support, good.   Try with the Backups (Deja Dup) software  I did start the backup first time but I wanted to go to lunch after backup was running already two hours of just 10 GB files.  So it gave me this after the lunch    which is really insane, reported now , since the software is not designed to have pauses in doing the backups. How much money did Charlie Harper have? This is quite tricky to figure out, and I am not looking for a specific number, but for a approximation.  As you know, Charlie (nearly) always had a lot of money to spend, but a few hints suggest that he actually did not have money at all (talking about absolute, not cash flow)   he went bancrupt after a few months without income he told Alan that there is no money, and everything just a house of cards he worries about Bertha giving away a few clothes he has three mortgages on his house when he died, there was no inheritage to Alan or anyone, and they had to sell the house ...   However, considering the lifestyle and spendings, Charlie must have had a substantial amount available.   driving the latest foreign expensive cars owning a malibu beachhouse paying the property tax for said house discussing retirement at 45 spending endless money on booze, betting and women ...   So all things considered, did he have money, or was he just a scam? When should I use GET or POST method? What's the difference between them? Is it possible to call a function whose name is stored in a string in vbscript? When I have .txt files or .py files, I can Quick Look those files in Finder. I have .jl files that are codes written in the Julia Language. Of course they are text files. They will open in a text editor like Sublime Text. How can I Quick Look these .jl files in Finder?