I have one user on WP website. I want to give him editor role because he need to edit pages and as I see -  requires user to be at least editor. The problem is that i want to limit list of pages and posts that he can edit. How it can be done?
We currently have about 50 pages, each of which I want a user (eg, bob, rob, smith) to be able to edit only 1 page.  For example, bob &amp; smith each have their own page.  I do not want bob to be able to edit smith's page.  I want bob to ONLY be able to edit bob's page.  I don't care if he can see other pages. Looking through the user roles, I don't see a way to currently to do this- I only see how to allow restrict access on a global scale.  Are their any plugins to help restrict edit access per user, or roles restricted to editing specific pages and I can just add 1 user per role?  Or was there a way to do this with default settings I missed.
Stop execution of Javascript function (client side) or tweak it Understanding shape and calculation of confidence bands in linear regression I am trying to understand the origin of the curved shaped of confidence bands associated with an OLS linear regression and how it relates to the confidence intervals of the regression parameters (slope and intercept), for example (using R):  require(visreg) fit &lt;- lm(Ozone ~ Solar.R,data=airquality) visreg(fit)     It appears that the band is related to the limits of the lines calculated with the 2.5% intercept, and the 97.5% slope, as well as with the 97.5% intercept, and the 2.5% slope (although not quite):  xnew &lt;- seq(0,400) int &lt;- confint(fit) lines(xnew, (int[1,2]+int[2,1]*xnew)) lines(xnew, (int[1,1]+int[2,2]*xnew))     What I don't understand are two things:   What about the combination of 2.5% slope &amp; 2.5% intercept as well as 97.5% slope and 97.5% intercept? These give lines that are clearly outside the band plotted above. Maybe I don't understand the meaning of a confidence interval, but if in 95%  of the cases my estimates are within the confidence interval, these seem like a possible outcome? What determines the minimum distance between the upper and lower limit (i.e. close to the point where the two lines added above intercept)?    I guess both questions arise because I don't know/understand how these bands are actually calculated.   How can I calculate the upper and lower limits using the confidence intervals of the regression parameters (without relying on predict() or a similar function, i.e. by hand)? I tried to decipher the predict.lm function in R, but the coding is beyond me. I'd appreciate any pointers towards relevant literature or explanations suitable for stats beginners.  Thanks. What's wrong with this equal probability solution for Monty Hall Problem? I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.   Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.   Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)? Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.   Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.   What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks. Why are the Degrees of Freedom for multiple regression n - k - 1? For linear regression, why is it n - 2? I'm reading a textbook and I see this question:    So there are 200 women, and the DF is 196, implying that the equation for DF is $n - k - 1$. There are 3 variables: bp, age, and type so $k == 3$. What's the intuition behind this?  Also, why is the degrees of freedom for linear regression n - 2? Which media player has seek preview with thumbnails? I was looking for a good video player to replace VLC, something that is as good as vlc but has seek preview. I am tired of fast forwarding in vlc  EDIT: I meant seek preview with thumbnails Is it correct to say "I wrote him" or "I wrote to him"?  My Mother was a stickler for English grammar and would say "I wrote your Uncle..." rather than "I wrote to your Uncle..." Use dynamic variable names in JavaScript Prove $\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{e}^{-2\sqrt{ab}}$ This is Example 8.4.1 from Chapter 8 (The Normal Integral) of  by Boros and Moll. The authors outlined a solution method, which I will provide in full here. My question: is there another way to obtain this result?  Here is the solution of Boros and Moll with their notation.  Let \begin{equation} \mathrm{L}(a,b) := \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x \label{eq:iic8-1} \tag{1} \end{equation} Making the substitution $t=x\sqrt{a}$ yields \begin{equation} \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty}  \mathrm{exp}(-t^{2}-\frac{ab}{t^{2}}) \mathrm{d} t \label{eq:iic8-2} \tag{2} \end{equation} Letting $ab=c$ we call the integral in equation \eqref{eq:iic8-2} $f(c)$, \begin{equation} f(c) = \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \mathrm{d} t \label{eq:iic8-3} \tag{3} \end{equation} so that  \begin{equation} \mathrm{L}(a,b) = \frac{f(ab)}{\sqrt{a}} \label{eq:iic8-4} \tag{4} \end{equation} In equation \eqref{eq:iic8-3} we let $y=\sqrt{c}/t$ \begin{equation} f(c) = \sqrt{c} \int\limits_{0}^{\infty} \mathrm{exp}(-y^{2}-\frac{c}{y^{2}}) y^{-2} \mathrm{d} y \label{eq:iic8-5} \tag{5} \end{equation}  Combining equations \eqref{eq:iic8-3} and \eqref{eq:iic8-5}, we have \begin{equation} f(c) = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \left(1+\frac{\sqrt{c}}{t^{2}} \right) \mathrm{d} t \label{eq:iic8-6} \tag{6} \end{equation} Now we let $s = t - \sqrt{c}/t$ \begin{equation} f(c) = \frac{\mathrm{e}^{- 2\sqrt{c}}}{2} \int\limits_{-\infty}^{\infty} \mathrm{exp}(-s^{2}) \mathrm{d} s = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \lim_{z \to \infty} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \label{eq:iic8-7} \tag{7} \end{equation} Combining equations \eqref{eq:iic8-1}, \eqref{eq:iic8-4}, and \eqref{eq:iic8-7} yields our result. Is my ESTA valid for a second trip? I visited the USA in May this year, on an ESTA which I applied for in April. It was approved and I was allowed entry into the country.    I'm re-visiting again this November for two weeks. Am I able to use the same ESTA again or do I need to re-apply for a new one? I know it says it's valid for two years, but is that valid for every use within those two years, or I have to use it within two years, and once I have visited on that visa, it's invalid? I tried to install wine using these commands: sudo apt-get install -y wine  And: sudo apt-get install wine  But terminal says that: Reading package lists... Done Building dependency tree        Reading state information... Done Package wine is a virtual package provided by:   winehq-staging 2.16.0~zesty   winehq-stable 2.0.2~zesty   winehq-devel 2.16.0~zesty   wine-development 2.0-3ubuntu1 You should explicitly select one to install.  E: Package 'wine' has no installation candidate  So there are 4 different version of wine. What is the difference between these versions. Which one is better? I wanted to install Internet Explorer from it and I installed winehq-stable. But I couldn't see that among applications in Winetricks. Do the other versions of wine have more applications to install? What is the difference? Error-Handling in Swift-Language I want to run a command when there are only 10 entities left. I know you can use /execute unless entity to detect when all matching mobs have died, but it doesn't work if you want to run a command when there are for example 10 entities left.  A command like /execute if entity @e[limit=10] does not work, it also triggers if there is only one entity.  Such a counting system could also be used to determine once there is only one player left in an arena, meaning they have won a battle or similar. Undefined Symbol ___gxx_personality_v0 on link "Above"/"below" before/after a noun Allow moderators to see delete/undelete activity on users' profiles We get auto flags for 'possible vandalism' related to deletes, and these can be helpful at times in finding 'rage quit' users, or other potential problems.  But often, I would like to know if there is a prior pattern with the user doing this. While we can see posts of theirs which have been flagged, and perhaps find previous auto flags this way, it would be nice to see an overall view of their self-deletions and undeletions; mostly just to identify patterns.  Such a feature would be nice if it was activity centered, rather than post-centered. That is; each delete/undelete action should be shown. So we could easily see, for example, if someone went back-and-forth on a single post.  The "activity" tab in the profile would be a logical place to show these events.  I think this would be best to be limited to instant self-deletions, not deletion votes. Difference in output between SAS's proc genmod and R's glm I'm trying to replicate a model fitted in SAS in R but the fit I'm getting gives me slightly different coefficients and standard errors.  Data:  testdata &lt;- data.frame(matrix(c("f","Test", 1.75,   16, 0,  16, 0,  1,  1, "m",    "Test", 1.75,   15, 1,  16, 6.25,   1,  0, "f",    "Test", 2.75,   4,  12, 16, 75, 1,  1, "m",    "Test", 2.75,   9,  6,  15, 40, 1,  0, "f",    "WHO",  1.75,   15, 1,  16, 6.25,   0,  1, "m",    "WHO",  1.75,   14, 2,  16, 12.5,   0,  0, "f",    "WHO",  2.75,   2,  13, 15, 86.6667,    0,  1, "m",    "WHO",  2.75,   3,  13, 16, 81.25,  0,  0 ), ncol=9, byrow=TRUE)) names(testdata) &lt;- c("sex", "vaccine", "dose", "not_p", "para", "n", "pct",                       "vacnum", "sexno")   SAS:  proc genmod data=model_data; class sex; model para/n  = dose sex vacnum    /dist=bin    link=logit   type3; run;  Analysis Of Maximum Likelihood Parameter Estimates  Parameter   DF Estimate Std Error Wald 95% Conf Lim Wald Chi-Square Pr &gt; ChiSq  Intercept   1 -9.4020   1.6220    -12.5810  -6.2230  33.60           &lt;.0001  dose        1  3.9208   0.6460      2.6546   5.1870  36.83           &lt;.0001  sex f       1  0.5574   0.5184     -0.4587   1.5735   1.16           0.2823  sex m       0  0.0000   0.0000      0.0000   0.0000    .              .  vacnum      1 -1.3221   0.5483     -2.3967  -0.2475   5.81           0.0159  Scale       0  1.0000   0.0000      1.0000   1.0000        R:  testdata$sexno &lt;- as.factor(testdata$sexno)     a &lt;- contr.treatment(2, base = 1, contrasts = TRUE)  contrasts(testdata$sexno) &lt;- a  fitreduced &lt;- glm(para/n ~ dose + as.factor(sex) + vacnum,                    family=quasibinomial(link="logit"), data=testdata)  coef(summary(fitreduced))                    Estimate Std. Error   t value    Pr(&gt;|t|) (Intercept)     -9.4013750  1.7613982 -5.337450 0.005935450 dose             3.9173794  0.7001133  5.595351 0.005007179 as.factor(sex)1  0.5704671  0.5568436  1.024466 0.363525300 vacnum          -1.3336100  0.5887552 -2.265135 0.086189704   I believe I have the right contrasts to give me a type III SS but there is a small discrepency in values, have a missed something here? If $|G| = pq$,  $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems? "State-of-the-art" and "technology" Accessing class members on a NULL pointer I was experimenting with C++ and found the below code as very strange.   class Foo{ public:     virtual void say_virtual_hi(){         std::cout &lt;&lt; "Virtual Hi";     }      void say_hi()     {         std::cout &lt;&lt; "Hi";     } };  int main(int argc, char** argv) {     Foo* foo = 0;     foo-&gt;say_hi(); // works well     foo-&gt;say_virtual_hi(); // will crash the app     return 0; }   I know that the virtual method call crashes because it requires a vtable lookup and can only work with valid objects.   I have the following questions    How does the non virtual method say_hi work on a NULL pointer? Where does the object foo get allocated?   Any thoughts? If I seem tired, can I say: "I haven't slept last night"? If not, why have I been told that we use present perfect for actions that have present effects?