Prefix index length impact for VARCHAR?
MySQL - varchar length and performance
Stack Overflow is a site of passers-by They says that Stack Overflow is a community-driven web-site. But, as a matter of fact, it is not.  There is no community at all. It's more like an imageboard, as a number of participants makes them pretty much like anonymous to each other despite of registration.   Stack Overflow is a site of passers-by.  Passers-by who want to solve their silly typo problem and passers-by who eager to share their vast knowledge in spotting silly typos. And - oh - sometimes to perform a civic duty and vote to close a question they have no idea what to answer to and thus calling it "not a real question".  So, as a funny consequence, whatever question can get   answered upvoted downvoted closed deleted   depends not on its quality but only on the fluctuation happened to "community" at the given moment, at a pretty random rate.  It is not only quite discouraging. It makes the very idea of collecting and improving knowledge quite impossible. How can I pad an integer with zeros on the left? How do you left pad an int with zeros when converting to a String in java?  I'm basically looking to pad out integers up to 9999 with leading zeros (e.g. 1 = 0001). My PC has 8 Gb of ram. Is there any way to make ubuntu use most of it? I mean rarely drop caches and keep once opened programms in ram longer, preload apps on boot, etc. I was wondering if there is any way to escape a CDATA end token (]]&gt;) within a CDATA section in an xml document. Or, more generally, if there is some escape sequence for using within a CDATA (but if it exists, I guess it'd probably only make sense to escape begin or end tokens, anyway).   Basically, can you have a begin or end token embedded in a CDATA and tell the parser not to interpret it but to treat it as just another character sequence.  Probably, you should just refactor your xml structure or your code if you find yourself trying to do that, but even though I've been working with xml on a daily basis for the last 3 years or so and I have never had this problem, I was wondering if it was possible. Just out of curiosity.  Edit:  Other than using html encoding... Conditional Rendering inside Visualforce pages Do you have to take the second leg of a domestic flight? Clone Method not working I am trying to Clone a lead when certain condition is arrived using before trigger but i am not able to see cloned record.   for( Lead ld : Trigger.new ) {                          if(ld.Integration_Source__c == 'Hubspot' &amp;&amp; Trigger.isUpdate){         Lead LeadBeforeUpd = Trigger.oldMap.get( ld.Id );         if(ld.Program__c != LeadBeforeUpd.Program__c ){            Lead ClonedLead = new Lead();            ClonedLead = LeadBeforeUpd.clone(false, false, false, false);         }     }  } Prove $-(-a)=a$ using only ordered field axioms I need to prove for all real numbers $a$, $-(-a) = a$ using only the following properties:     I tried by starting with P12 written using symbolic logic:  $\begin{align} \forall a,b \in P \to &amp; a \cdot b \in P \Leftrightarrow True \\ \forall a \in P \land \forall b \in P \to &amp; a \cdot b \in P \Leftrightarrow True \\ \lnot \left[ \left( \forall a \in P \right) \land \left( \forall b \in P \right) \right] \lor &amp; a \cdot b \in P \Leftrightarrow True \\ \exists a \notin P \lor \exists b \notin P \lor &amp; a \cdot b \in P \Leftrightarrow True \\ \end{align} \\ $  and played with evaluating it with $b = -1$, $-a$ and so on. But, felt this wasn't really a solution because the last symbolic logic statement above is ambiguous.  Not to mention--other than P12--I haven't used any of the axioms listed above whatsoever. ( And I've not bothered to prove any of the theorems of symbolic logic I have used. )  Can anyone give me a start on this problem?  Many thanks in advance.  Edit  Thank you, everyone! .  Best regards! Are token creatures considerd as a creature spell? Are token creatures considerd creature spells? For instince when I use  tap effect combined with  would it allow me to draw a card? If I have the following code:  MyClass pClass = new MyClass(); pClass.MyEvent += MyFunction; pClass = null;   Will pClass be garbage collected? Or will it hang around still firing its events whenever they occur? Will I need to do the following in order to allow garbage collection?  MyClass pClass = new MyClass(); pClass.MyEvent += MyFunction; pClass.MyEvent -= MyFunction; pClass = null; Below is experimental data for the ratio $$R=\frac{\sigma(e^+e^-\rightarrow hadrons)}{\sigma(e^+e^-\rightarrow\mu^+\mu^-)}$$ as a function of the centre of mass energy $\sqrt s$. I am interested in the peak at around $100GeV$ which corresponds to the resonance of the $Z^0$ boson. There are two ways of looking at this that I have in mind:  We can effectively ignore the electromagnetic process around this energy, and so each process should have it's own Breit-Wigner peak centred on the mass energy of the $Z^0$. The ratio of these is just a constant. Thinking in terms of Feynman rules and again ignoring the electromagnetic process around this region, each process has the same propagator and vertex factors (roughly - ignore quark mixing) and there are some extra factors due to different quarks being possible and colour degeneracy, but still the ratio should be a constant (one possible issue here is interference between Feynman diagrams which I have neglected?).  So my question is why does the peak exist in the data? I saw this in a  and am putting it here for posterity.  Problem: Let $A$ and $B$ by square matrices and set $C=AB-BA$. If $AC=CA$, prove $C$ is nilpotent. So I did a fresh install of Ubuntu 20.04 on a computer with the new RTL8125B Realtek network card and I cannot get the network to work. I was very suprised since this is supposed to be the new feature of the 5.4 kernel (I have 5.4.0-26-generic installed). Does anyone have any idea on how to get it working? Thank you. $ lspci -knn | grep Eth -A3 06:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8125 2.5GbE Controller [10ec:8125] (rev 04) Subsystem: ASUSTeK Computer Inc. RTL8125 2.5GbE Controller [1043:87d7] Kernel modules: r8169 07:00.0 VGA compatible controller [0300]: Advanced Micro Devices, Inc. [AMD/ATI] Ellesmere [Radeon RX 470/480/570/570X/580/580X/590] [1002:67df] (rev e7)  $ dmesg | grep -i r8169 [    0.911236] r8169 0000:06:00.0: unknown chip XID 641 Cannot set LC_CTYPE to default locale: No such file or directory Update: (Originally this was a comment, but I thought it was worth sharing here at the top.)   In the original question, I asked if there was a polite, socially-acceptable way to ask an Indian co-worker not to use the phrase "do the needful", as I didn't care for it.   In the years since I asked this question I've asked many people about the phrase. To the Indians I've asked in-person, it's not rude in any way. To Americans I've asked it varies anywhere from "I don't like it, but I don't mind it" to "It's very rude and makes me angry when I see it".   Through a long series of edits the question morphed and I wasn’t able to delete it because it had upvoted answers. The question, as it stands now, has no other answer.  I guess, though, if you're non-Indian and you find it rude, or if you're Indian and you've never realized someone might find it rude, this post may still have merit. Hopefully it does. It certainly has gotten a lot of views. Best wishes, folks.  End Update. See below for what's left of the actual question.    "Do the needful."   It's a phrase that I've only seen used in email, and I find it . . . presumptuous (maybe even rude). Regardless of prefacing with "please", one is commanding rather than asking for assistance.  I've only seen it used by those of Indian origin, so I've simply not mentioned it at all rather than worry about any cross-cultural offense that may come of bringing it up.  But still, I don't understand why it's used. Why not request rather than make two statements, one factual, one imperative?  For example, why would one use:        I'm told you have Jane's email address. Please do the needful.      instead of,        Would you send me Jane's email address? How to remove a filter that is an anonymous object? In my functions.php file I would like to remove the below filter, but I'm not sure how to do it since it's in a class. What should remove_filter() look like?  add_filter('comments_array',array( &amp;$this, 'FbComments' ));   It's on line 88 . How to change permissions of multiple files found with find command? Why is the change in Hamiltonian for an active infinitesimal canonical transformation defined the way it is? I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104,         $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.  Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.   In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?   Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive). I want to prove with mathematical induction that:  $$\sum_{i=1}^n i \cdot i! = (n+1)! - 1$$  So in the first step we define $n = 1$:  $$\sum_{i=1}^1 i \cdot i! = 1 \cdot 1! = 1 = 2! - 1$$  In the second step we define $n = n + 1$:  $$\sum_{i=1}^{n+1} i \cdot i! = \sum_{i=1}^n i \cdot i! + (n+1)(n+1)!$$ $$\implies (n+1)! - 1 + (n+1)(n+1)!$$  This can now be simplified to $(n+2)(n+1)! - 1$ but I do not see how...  We could do $(n+1)!(-1 + (n+1)) = (n+1)! \cdot n$ but still this will not be the same as above.  What did I miss?  Bodo Which of these sentences is correct?     Rupees are interchangeable with dollars   or      Rupees is interchangeable with dollars.