Add language support files for validationengine jquery dynamically
Validationengine jquery not working properly for localization
I need a robust and simple way to remove illegal path and file characters from a simple string. I've used the below code but it doesn't seem to do anything, what am I missing?  using System; using System.IO;  namespace ConsoleApplication1 {     class Program     {         static void Main(string[] args)         {             string illegal = "\"M&lt;&gt;\"\\a/ry/ h**ad:&gt;&gt; a\\/:*?\"&lt;&gt;| li*tt|le|| la\"mb.?";              illegal = illegal.Trim(Path.GetInvalidFileNameChars());             illegal = illegal.Trim(Path.GetInvalidPathChars());              Console.WriteLine(illegal);             Console.ReadLine();         }     } } $\nabla U=0 \implies U=\mathrm{constant}$ only if $U$ is defined on a connected set? System.Windows.Threading.Dispatcher and WinForms? Show that if $p$ is prime then $\Bbb Z_p$ is a field Check my proof please.  Let $\Bbb Z_p:=\Bbb Z/p\Bbb Z$ be the quotient ring modulo $p$. I want to prove that if $p$ is prime then $\Bbb Z_p$ is a field.  Known facts about the quotient rings of the kind $\Bbb Z_n$ (I dont list all the characteristics of a ring, just some ones):   They are commutative rings with unity. $[0]=[n]$ is the identity of addition or zero. Then $[n]+[a]=[a]$ and $[n]\cdot[a]=[n]$ for all $[a]\in\Bbb Z_n$. $[1]$ is the multiplicative identity or unity. Then $[1]\cdot[a]=[a]$ for all $[a]\in\Bbb Z_n$. The addition is defined as $[a]+[b]=[a+b]$, and the multiplication is defined as $[a]\cdot[b]=[ab]$. $[a]=[b]$ means that exists $z\in\Bbb Z$ such that $a=b+nz$. $|\Bbb Z_p|=p$.   A ring $\Bbb Z_p$ dont have zero divisors, i.e. doesnt exist $[a],[b]\in\Bbb Z_p$ distinct of $[p]$ such that $[a]\cdot[b]=[p]$. Proof by contradiction: suppose that exists such $[a],[b]$ distinct of $[p]$ that are zero divisors. Then exist some $z_j\in\Bbb Z$ such that:  $$(a+pz_1)(b+pz_2)=pz_3\iff ab+pz_4=pz_3\iff ab=pz_5$$  but this is a contradiction with the assumption that $a,b\neq p$ and $p$ prime, so dont exist divisors of zero on any $\Bbb Z_p$.$\Box$  For any $[a]\in\Bbb Z_p$ distinct of $[p]$ exist $[b]$ such that $[a][b]=[1]$. Because $\Bbb Z_p$ dont have zero divisors and is finite for some $[a]\neq[p]$ we have that if $[b]\neq[c]$ then $[a][b]\neq[a][c]$. Proof by contradiction: if $[a][b]=[a][c]$ in the above conditions then  $$[a][c]=[ac]=[ab]=[a][b]\iff ac+pz_1=ab+pz_2\iff a(c-b)=pz_3$$  for some $z_j\in\Bbb Z$. But $pz_3\in[p]$, then $[a][c-b]=[p]$ but because there are no zero divisors in $\Bbb Z_p$ then or $[a]=[p]$ or $[c-b]=[p]$ what contradicts the above conditions.$\Box$  Because $[a][b]\neq[a][c]$ for $[a]\neq[p]$ and $[b]\neq[c]$ then for any $[a]\neq[p]$ we have that  $$[a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1]$$  In other words: every element of $\Bbb Z_p$ but $[p]$ have a multiplicative inverse as stated above.$\Box$  Because $\Bbb Z_p$ is a commutative ring with $[1]\neq[0]$ and have multiplicative inverse for all their elements but $[p]=[0]$ then $\Bbb Z_p$ is a field.$\Box$ Elegantly determine if more than one boolean is "true" Evaluate $\sum\limits_{k=1}^{n} \frac{k}{2^k}$ Evaluate $$\sum\limits_{k=1}^{n} \frac{k}{2^k}$$ How to prove floor identities? Switch statement fall-through...should it be allowed? For as long as I can remember I have avoided using switch statement fall-through. Actually, I can't remember it ever entering my consciousness as a possible way to do things as it was drilled into my head early on that it was nothing more than a bug in the switch statement. However, today I ran across some code that uses it by design, which got me immediately wondering what everyone in the community thinks about switch statement fall-through. Is it something that a programming language should explicitly not allow (like C# does, though it supplies a workaround) or is it a feature of any language that is powerful enough to leave in the programmer's hands? Edit: I wasn't specific enough to what I meant by fall-through. I use this type a lot:     switch(m_loadAnimSubCt){         case 0:         case 1:             // Do something             break;         case 2:         case 3:         case 4:             // Do something             break;    }  However, I'm concerned about something like this.    switch(m_loadAnimSubCt){         case 0:         case 1:             // Do something, but fall through to the other cases             // after doing it.          case 2:         case 3:         case 4:             // Do something else.             break;    }  This way whenever the case is 0, 1 it will do everything in the switch statement. I've seen this by design and I just don't know if I agree that switch statements should be used this way. I think the first code example is very useful and safe. The second seems kind of dangerous. What is the closed formula for finding      $$ \sum_{i=0}^{k} \binom{n}{2i}=?\quad ,k\leq \lfloor{n/2}\rfloor $$   I need summation only up to some intermediate  $k$ where $k\leq \lfloor{n/2}\rfloor$ .  SAY I need a closed formula to calculate $\binom{8}{0}+ \binom{8}{2}+\binom{8}{4} +\binom{8}{6}$  HERE $n=8$ and $k=6$ Set Text Color for textView Android I want to clear a element from a vector using the erase method. But the problem here is that the element is not guaranteed to occur only once in the vector. It may be present multiple times and I need to clear all of them. My code is something like this:  void erase(std::vector&lt;int&gt;&amp; myNumbers_in, int number_in) {     std::vector&lt;int&gt;::iterator iter = myNumbers_in.begin();     std::vector&lt;int&gt;::iterator endIter = myNumbers_in.end();     for(; iter != endIter; ++iter)     {         if(*iter == number_in)         {             myNumbers_in.erase(iter);         }     } }  int main(int argc, char* argv[]) {     std::vector&lt;int&gt; myNmbers;     for(int i = 0; i &lt; 2; ++i)     {         myNmbers.push_back(i);         myNmbers.push_back(i);     }      erase(myNmbers, 1);      return 0; }   This code obviously crashes because I am changing the end of the vector while iterating through it. What is the best way to achieve this? I.e. is there any way to do this without iterating through the vector multiple times or creating one more copy of the vector? How to determine which Python install is being used by the interpreter? I installed Redhat 6 x86_64. I am using the Network connection screen to set a static IP address like below (I want two PC's in my house to see each other: one Redhat PC and one Mac)   192.168.0.5   255.255.255.0   192.168.0.1     When I run ifconfig it displays only lo and virbr0 information. I don't know what these items are (I don't really know much about network settings).   When I try ifconfig -a it displays eth0, lo, sit0 and virbr0. The information for  eth0 is as follows:   Link encap : Ethernet HWaddr 90:2B:34:74:05:30 BROADCAST MULTICAST MTU:1500 Metric:1 RX packets:192 errors:0 dropped:0 overruns:0 frame:0 TX packets:6 errors:0 dropped:0 overruns:0 frame:0 collisions:0 txqueuelen:1000 RX bytes 53811 (52.5 KiB) TX bytes:468 (468.0 b) Interrupt:29 Base address:0xc000   Could someone help me to point out if anything wrong with my setting or how to resolve this problem? $P(X<Y)$ where X and Y are exponential with means $2$ and $1$ How can I search within a manpage? The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead Remount drive without physically disconnecting? So I have unmounted a usb external hard drive but I went to re-mount it.  Is there a command that I can use to remount the drive?  Only way I know to re-mount a drive is to disconnect the usb cable from my Mac and reconnect it. I am pretty much sure that for native speakers the issue I am going to bring up might look as an uncalled question as they can easily figure out which form of a verbal part of speech should be used, be it a gerund, an infinitive, or a bare infinitive. However, it can be pretty much misleading for a foreigner because the same construction works differently with different words and I don’t see a logical explanation. For example:  The purpose of this article is to analyze the issue…  My hobby is reading. One may assume, and many foreigners really come to this conclusion, that it’s OK to switch the “gerund” and “infinitive” as the meaning will stay the same, that is:  The purpose of this article is analyzing the issue… (same as 1 above)  My hobby is to read. (same as 2 above)   However, as I was told, this assumption is erroneous. Another example: His task was watching after them. (not good) His task was to watch after them. Is there any principle which governs the choice of the right form in such cases? PhD duration in the USA I am trying to find out what the average duration of PhD studies are in the USA. Here in Spain it used to be 4 years, but it seems that Europe is moving towards a 3-year scheme since students are expected to do their MSc before they start. What is it like in the USA? Is some kind of shortening also occurring? Is there a way to "auto login" in PuTTY with a password? Is there a way to configure a password for a stored session in ?  I know there is the capability to specify an "auto-login username" (under Connection/Data), but is there a way to do the same with the password?