I have javascript code into xslt which should check image dimension. If width is greater than height to rotate img.  &lt;script language="javascript"&gt;     var a = 3;     var b = 1;     if (a &gt; b) {       document.getElementById("img1").style.transform = "rotate(90deg)";     }    &lt;/script&gt;     I tried:  1) &amp;gt;  2) &lt;![CDATA[ --code --]]&gt;  3) disable-output-escaping="yes"   But nothing works. In console see an error message    Thanks
i have a tiny little problem with xslt, js and html entities, eg. within a template:  &lt;script type="text/javascript"&gt;     &lt;xsl:value-of select="/some/node"/&gt;     for (var i = 0; i &amp;lt; 5; i++) {         //            ^^^ js error     } &lt;/script&gt;  &lt;script type="text/javascript"&gt;     &lt;xsl:value-of select="/some/node"/&gt;     for (var i = 0; i &lt; 5; i++) {         //            ^ xslt error     } &lt;/script&gt;  &lt;script type="text/javascript"&gt;     &lt;xsl:value-of select="/some/node"/&gt;     // &lt;![CDATA[     for (var i = 0; i &lt; 5; i++) {         //            ^ becomes &amp;lt;     }     // ]]&gt; &lt;/script&gt;   &lt;script type="text/javascript"&gt;     &lt;xsl:value-of select="/some/node"/&gt;     for (var i = 0; i &lt;xsl:value-of disable-output-escaping="yes" select="string('&amp;lt;')"/&gt; 5; i++) {         // works of course     } &lt;/script&gt;   does anyone have an idea where my problem could come from? i always thought the xslt processor would leave the content of a &lt;script/&gt; element unescaped when using the html output method ...  i run libxslt2 version 1.1.24 on OSX which was installed using macportsports ...
In a multi server architecture, is it possible to find out from which machine a user logged into current machine using ssh?  For example, user1 logged into host1. From there, the user logged into host2 as sruser using ssh.  In above example, would it be possible to trace the sruser session of host2 back to user1 session in host1?  Would it be possible to get the hostname or terminal of user1 session? How to reduce space between image and its caption? Are there consequences to stamps in my passport that are not from border control? Integral vanishes on all intervals implies the function is a.e. zero I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions? When is it ok to remove the intercept in a linear regression model? What is the relationship between redshift and conformal time ?   For example in a paper i found:   taking $z_e = 3234 $ at the time of radiaton-matter equality yields the conformal time $\frac{\eta_e}{\eta_0} = 0.007$ and taking $z_E=0.39$ at matter-$\Lambda$ equality yields $\frac{\eta_E}{\eta_0}=0.894$ and setting redshift at decoupling $z_d=1089$ yields $\frac{\eta_d}{\eta_0} = 0.0195$  where $\eta_0$ is the present decoupling time.  Further some cosmological parameters are given as :  $\Omega_r = 8.36 \times 10^{-5}, \Omega_m = \Omega_b + \Omega_{dm} = 0.044 +0.226, \Omega_\Lambda = 0.73,  H_0=0.72$  Now how can i calculate all those $\eta_e, \eta_E, \eta_d, \eta_0$ from given redshift values and/or above parameters ? I searched whole of my text books trying to find an explicit relation for conformal time $\eta$ but all i got was $ \mathrm{d}t=a(t)\mathrm{d}\eta$. Any help would be very helpful. What do two question marks together mean in C#? Symmetric encryption construction (meet-in-the-middle attack) This is a question I had from an exercise session.  First we do DES in CBC encryption mode using a key $K_1$ and $IV_1 = O^n$. Then we do DES in ECB decryption mode using $K_1$. And then DES in CBC decryption mode with a secret $IV_2$ and key $K_2$.  Is this construction safe against meet-in-the-middle attack?  Wouldn't the first two operations negate themselves because $C_1$ would decrypt into $P_1$? example of a non differentiable manifold Take the manifold: The graph of $|x|$ on $(-1,1)$, with the induced topology from $\mathbb{R}^2$. This is a topological manifold, which is homeomorphic to $(-1,1)$ by projection. Is it a differentiable manifold? I believe it is, because we can take an atlas with only the projection, and then we will have only one transition map which is the identity, and therefore differentiable. But I'm not sure I get the definition of a differentiable manifold right...  Thanks! There seems to be a preference for to have the    18 mm - 55 mm 55 mm - 250 mm   lenses, but no mention of why they're better than a single 18-200mm. Could someone please explain why the two separate are better than the one combined, and if anyone has quantified this difference?  (I'm curious about Canon in particular.) jQuery animate() $(\Bbb Z\oplus \Bbb Z)[X] \not\cong \Bbb Z[X]\oplus \Bbb Z[X]\,$? Well i tried to prove it but failed. They are both commutative and have the same set of units. So i can't see a property of one ring which the other ring does not possess. Thank you for any help. Searching through ODT documents without opening them? Let's say I have 2000 .ODT files. Their names are random numbers. How do I go about searching for, let's say, "pricing list"? How do I find the document I need without opening all of them one by one and checking to see if it's the right one?  Is there some program to search through the words in the documents without opening them? Unwanted faint lines connecting dashed double line zsh scripting - How do I NOT match a pattern in filename generation? I'm currently trying to loop through a set of directories, but only the ones I care about.   $ ls -d */ cfdpost_statefiles/     p050L0260/  p050L0510/  p060L0390/  p070L0260/  p070L0510/  p092L0390/  p097L0260/  p110L0390/  p131L0325/  p150L0510/  profiles/ ERCOFTAC_TestSettings/  p050L0390/  p060L0260/  p060L0510/  p070L0390/  p084L0510/  p094L0325/  p100L0510/  p125L0510/  p136L0260/  p162L0260/  settings/   I want to match all the p###* directories except for the ones beginning with p125 and p150.  If I just do patterns with a pipe, I get:  $ for dir in ./p(125|150)*(/); echo $dir ./p125L0510 ./p150L0510   However, if I try and use the ^ like it shows :  $ for dir in ./p^(125|150)*(/); echo $dir zsh: no matches found: ./p^(125|150)*(/)   Any ideas how to proceed? Serana disappeared during dawnguard questline. center align equation in ‎\documentclass[11pt,twoside,fleqn]{book}‎ I noticed that after joining a table to a shapefile the headers of each fields in the table used for the join are changed by QGIS and they are attributed to a different name.  For instance, I had a shapefile with vegetation_codes and a table with vegetation_codes and veg_definition (these were the headers). After joining the table to the shapefile and also saving the shapefile and creating a new one, the vegetation_classes header was changed into vegetati_1.  Is there a way to keep the original headers?  I am using QGIS Dufour on Windows 7 and the table that I join to the shapefile is CSV (tried with dbf as well and the problem is the same). How to run grep and show x number of lines before and after the match grep returns just the line where it matched the regex and often what I want to really see is a few (say 2) lines above and below the matched one. Is there a simple way to achieve it?  EDIT: OS: Ubuntu based Bodhi Linux.  As mentioned in comments, -C does not work in vanilla but its GNU grep in my case. Does wide angle equivalent in crop sensor skew image? I see many posts on both this forum and elsewhere the discuss the use of a wide angle on a crop sensor.  For example, they discuss how at 14mm lens is 14mm on a 35mm film or full frame, but with a 1.6x crop sensor it is an effective 28mm.  My question is, is a 14mm on a crop sensor the same image as a 28mm on a full frame.  Or does "effective" include some other conotation.  I read that a 14mm can create distortion on a full frame and some vignette, so people seem to say using it on a crop takes out some of those edge problems. My confusion is how the lens is actually taking in the extra image.  Reading kenrockwells blog I learned that wide angle keeps lines straight, while fish eye actually skews the lines.    So with this as my understanding, I am still a bit confused how a lens that is curved in order to get a greater angle of view, will be the same as a cropped version of a less curved lens.  Or does "effective" only refer to the objects to the sides but not to the distortion of the image.  Or, if it is different, is it so small that for photography it does not matter?  A technically description would be welcome as long as a basic understanding of the situation.  Thanks!  Edit: To be more clear after the comments,  I am specifically talking about the end resulting 2d projection from 3d space.  I have read both those previous answers.  the first one is closer to what I am talking about.  However, it is still confusing for me.  I have experience with 3d modeling and projection matricies, which might be confusing me with this.  But for example, i dont understand the images having a 50 mm lens being drawn Starting from the sensor, so it shows a different fov for crop and full frame . If the lens is the same, it takes the same amount of information from the world, which makes a ray from a certain degree off the last lens projected into a smaller space, there is no projection from the sensor, so drawing the line from the sensor doesnt make sense to me.  The distance of the sensor to the rear of the lens must have some impact on the light projected but is not represented in the images.  Further, again from that first answer it says cropping is the same as zoom, but from what i understand from how perspective works, it is different, since a wide angle will project lines differently than a small fov to the same size, so cropping the center of a wide angle and zooming into something is very different.  This is easily recreated in an opengl application with varying fov projection matrix, i imagine in painting class people learn this as well.  A fisheye would be a good example, since if cropping was the same as zooming, a fish eye lens would keep the center exactly like a normal lens and then a gradient weighted towards the outside would rapidly create a warped perspective, but from what I see, it is even.  To me those images just look like its comparing cropped with full frame in regards to orthogonal projections.