Checking if a url has http:// at the beginning & inserting if not
How to add http:// if it doesn't exist in the URL
Is there a way to make an Oracle query behave like it contains a MySQL limit clause? In MySQL, I can do this: select *  from sometable order by name limit 20,10  to get the 21st to the 30th rows (skip the first 20, give the next 10). The rows are selected after the order by, so it really starts on the 20th name alphabetically. In Oracle, the only thing people mention is the rownum pseudo-column, but it is evaluated before order by, which means this: select *  from sometable where rownum &lt;= 10 order by name  will return a random set of ten rows ordered by name, which is not usually what I want. It also doesn't allow for specifying an offset. What does "it" refer to in "it's raining"? Transpose/Unzip Function (inverse of zip)? I have a list of 2-item tuples and I'd like to convert them to 2 lists where the first contains the first item in each tuple and the second list holds the second item.  For example:  original = [('a', 1), ('b', 2), ('c', 3), ('d', 4)] # and I want to become... result = (['a', 'b', 'c', 'd'], [1, 2, 3, 4])   Is there a builtin function that does that? Regex to match only letters I have read that watchdog timer pushes microprocessor/controller into reset state whenever it experiences any software anomaly like executing in infinite loop. and after some time a very senior prof told us that after reset the processor resumes operation from the next instruction to the erroneous one.  when watchdog timer resets the processor, from where does the processor start its next execution?  I just wanted to know what is correct or if some other concept is employed? Why do the big nuts always remain at top? The Brazil-nut Effect Most of the time when I mixed different nuts in a bowl, I observed that the big Brazil nuts were always on top while the small peanuts settled near the base. No matter how you take it, if the big nuts are put at the bottom and then the small nuts are poured, then also after some time the big ones will come to the top. Will not the gravity attract those big nuts more and eventually will they not remain at the base? But it doesn't happen!   I googled it and found the term Brazil-nut Effect but couldn't find any proper explanation. What is the physical explanation for this effect? How to re-index the values of an array in PHP? Which OpenType Math fonts are available? Strange distance measurements in distance matrix I'm trying to create a distance matrix between two point shapefiles in QGIS 2.6. Both are in WGS84 UTM Zone 30N co-ordinates (and so is the project file), but the distance matrix returns values such as 5673346.938, instead of expected numbers of between 1 and 5km.  I saw a similar thread mentioning disabling 'on the fly projection changes' - I've done this, and no change. Threads already checked:     Here are samples of the two shapefiles I'm using, stripped down to just ID and location. If router is off for a few days, do you have an ip address as far as ISP is concerned? I have an ISP which assigns IP addresses dynamically. If I go out and turn my router off I am always assigned a new one once I turn it back on. Does that mean that...  1) my previous IP goes elsewhere when my router is turned off?  2) i dont actually have an IP address if my router is off for a day or more? Prove that $\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x$ is also $e$. Trying to make sense out of the idea that $100\%$ continuous decay is $\frac{1}{e}$, I thought about this:  You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$  And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$  Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar, and I can imagine (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.  However I'm clueless about limit proofs, how do you prove that?  EDIT: Just had a last second insight.  Can you say that $$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?$$   (not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you) The equivalence of sample correlation and R statistic for simple linear regression How to process all products defined in a project (ConTeXt) If I have  project.tex  \startproject project \product one \product two \endproject   and the corresponding product files, is there an easy way to process all product files, so I get  one.pdf  two.pdf  at once? For $x,y\in I:=[0,1]$ define the relation on $I$ as $x-y\in \Bbb Q$. How big (using cardinal number) is the cardinality of the equivalence class $[1/\sqrt2]$?  I have tried to solve it by finding the equivalence class but I'm not sure about the method that I can solve the question. Please, can you help me to solve it. "See somebody do" and "see somebody doing" Is there a way in Java to determine if a path is valid without attempting to create a file? How do I extract all the external links of a web page and save them to a file? Is true that $\sum_{k=1}^n\frac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$? Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$?  My work:  Let $x\in\mathbb{R}$ be given. For every $k=1,\dots,n$, define $i_k\in\{0,1,\dots,k-1\}$ such that  $$ [x]+\frac{i_k}{k}\leq x&lt;[x]+\frac{i_k+1}{k}\hspace{20mm} (*). $$  On one hand, we have that $[kx]=k[x]+i_k$ for every $k=1,\dots,n$.  So  $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}=n[x]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}=[nx]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}-i_n$  Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}\leq i_n\,.$$  On the other hand, from $(*)$ is direct that $i_k=[k(x-[x])]$. Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{[k(x-[x])]}{k}\leq [n(x-[x])]\,.$$  In other words, we can assume that $x\in[0,1)$.   Now the questions is:  Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in[0,1)$ and for every $n\in\mathbb{Z}^+$?  Any simpler approach? Any thoughts? Determine if Term has Grandparent/Great-Grandparent Infinity, is clearly not . But there is some dissent on whether or not zero is finite. I have seen authors use "finite" to indicate the value of $0$ is excluded as well as infinity.  Is there any rule about this?