Consider the following stacked divs:    .page {   height: 100%; }  .top {   height: 100%;   background-color: red; }  .bottom {   height: 100%;   background-color: grey; }  .options {   height: 100%;   overflow: scroll; } &lt;div class="page"&gt;   &lt;div class="top"&gt;     &lt;h1&gt;TOP&lt;/h1&gt;   &lt;/div&gt;   &lt;div class="bottom"&gt;     &lt;h1&gt;Options Menu&lt;/h1&gt;     &lt;div class="options"&gt;       &lt;h1&gt;Option 1&lt;/h1&gt;       &lt;h1&gt;Option 2&lt;/h1&gt;       &lt;h1&gt;Option 3&lt;/h1&gt;       &lt;h1&gt;Option 4&lt;/h1&gt;       &lt;h1&gt;Option 5&lt;/h1&gt;       &lt;h1&gt;Option 6&lt;/h1&gt;       &lt;h1&gt;Option 7&lt;/h1&gt;       &lt;h1&gt;Option 8&lt;/h1&gt;       &lt;h1&gt;Option 9&lt;/h1&gt;       &lt;h1&gt;Option 10&lt;/h1&gt;     &lt;/div&gt;   &lt;/div&gt; &lt;/div&gt;     How can I change it to allow only the options to scroll, not the page?
I am working on a web application where I want the content to fill the height of the entire screen.  The page has a header, which contains a logo, and account information. This could be an arbitrary height. I want the content div to fill the rest of the page to the bottom.  I have a header div and a content div. At the moment I am using a table for the layout like so:  CSS and HTML    #page {     height: 100%; width: 100% }  #tdcontent {     height: 100%; }  #content {     overflow: auto; /* or overflow: hidden; */ } &lt;table id="page"&gt;     &lt;tr&gt;         &lt;td id="tdheader"&gt;             &lt;div id="header"&gt;...&lt;/div&gt;         &lt;/td&gt;     &lt;/tr&gt;     &lt;tr&gt;         &lt;td id="tdcontent"&gt;             &lt;div id="content"&gt;...&lt;/div&gt;         &lt;/td&gt;     &lt;/tr&gt; &lt;/table&gt;     The entire height of the page is filled, and no scrolling is required.  For anything inside the content div, setting top: 0; will put it right underneath the header. Sometimes the content will be a real table, with its height set to 100%. Putting header inside content will not allow this to work.  Is there a way to achieve the same effect without using the table?  Update:  Elements inside the content div will have heights set to percentages as well. So something at 100% inside the div will fill it to the bottom. As will two elements at 50%.  Update 2:  For instance, if the header takes up 20% of the screen's height, a table specified at 50% inside #content would take up 40% of the screen space. So far, wrapping the entire thing in a table is the only thing that works.
Can't get rid of header X-Powered-By:Express Available US State or National business (or employer) database lists? This question dovetails with this post on .  I am familiar with some of larger, paid-for business registry's like  and .  While the detailed information their business listings database include (like # of employees, annual sales, industry code, etc) is fantastic, if you are on a limited budget, sometimes the cost of purchasing the data outweighs the value you can get from it.  If you have geocoding capabilities in house, one could potentially create a reasonable surrogate by spatially-enabling (geocoding) a publicly available business listing on your own.    If you were looking to take on this type of project, where would you start your research? Do you know of US agencies at a State or National level that maintain this type of information, AND are publicly available for download or minimal purchase? Historical reason behind different line ending at different platforms Why did DOS/Windows and Mac decide to use \r\n and \r for line ending instead of \n? Was it just a result of trying to be "different" from Unix?  And now that Mac OS X is Unix (-like), did Apple switch to \n from \r? I have read several questions regarding how to fix an already enabled desktop cube or the problems it had in 11.04. But in 11.10, how can I correctly enable desktop cube in Unity 3D? In any finite graph with at least two vertices, there must be two vertices with the same degree Show that, in any finite graph with at least two vertices, there must be two vertices with the same degree.  HINTS ONLY! Controlling the direction of curve extrusion Is there an easy way to control how the extruded width of a curve object is oriented, for example like this     or this     The curve in the images above has only two control points, and I got the effect by rotating one of the points. However, for a curve with lots of control points it would be somewhat difficult or at least tedious to achieve this. Is there a simpler way? Subscribe to tags I am interested in? How do I subscribe to the tags that I am interested in?  I like to get notified via email or so, when someone asks new questions or if someone responds to my questions. Irreducibility check for polynomials not satisfying Eisenstein Criterion. Making a log4j console appender use different colors for different threads How to replace a video on YouTube With English not being my first language, I usually need support for at least two languages when using features like auto-correct and dictation.  Is it possible to toggle the language which is used for dictation via a shortcut?  As of now, I always have to manually navigate through the menus in the System Preferences which if far from convenient. I've tried to see if there are any pre-defined actions in Automator, but found none. Recursive matching with regular expressions in Javascript Why $(a+d)(b+c)$ equals $-16$? If $a,b,c$ and $d$ satisfy the equations:$$a+7b+3c+5d=0$$$$8a+4b+6c+2d=-16$$$$2a+6b+4c+8d=16$$$$5a+3b+7c+d=-16$$ then $(a+d)(b+c)$ equals $-16$.   I can't understand why $(a+d)(b+c)$ equals $-16$? Electrolytic Capacitor Voltage Rating vs capacitance I have always believed that it was ok to replace a capacitor with a slightly higher voltage rated one. eg 16v with 25v.Going higher than this would risk the capacitance value being out at the working voltage. However I just read the following quote from this       Electrolytic capacitors are non-ferroelectric with a very low dielectric constant. Their capacitance is derived from a very high surface area and nanometer thick dielectric layers. Their capacitance is not a function of applied voltage.‐Design Solutions For DC Bias In Multilayer Ceramic Capacitors (August 2010 Electronic Engineering Times)   So does this rule not apply to aluminum electrolytics? Eg will I have no issues replacing an electrolytic 25v 100uf with a 50v 100uf assuming both are low esr, similar footprint and good quailty? Is there a word or phrase for two nouns or adjectives joined by a conjunction (usually "and") in a fixed sequence?        alive and well           fast and furious   hat and gloves   pen and pencil   law and order   wit and wisdom   salt and pepper   bacon and eggs   horse and carriage   Adam and Eve   ladies and gentlemen   loud and clear   heads or tails   yes or no   dead or alive I am getting a NoClassDefFoundError when I run my Java application.  What is typically the cause of this? When is it appropriate to create an entity instead of a new content type? What is the benefit to creating new entity types over just building a new content type?    It seems a bit overkill to do all of the custom coding that is required to create a new entity when you have all of the CRUD and Views functionality already built into content types. How to get less spacing in math mode LaTeX commands like this:  Here again, four different simulations are run on networks of size  $C^{G} = 25 \times\ 25 = 625$, $C^{G} = 50 \times\ 50 = 2500$, $C^{G} = 250 \times\ 250 = 62500$, and $C^{G} = 500 \times\ 500 = 250000$...   When compiled generate the following:    The spacing in math mode really irritates me. There's too much white space around the equal sign, and the multiplication sign has more right padding than left padding. Is there any way to reduce this spacing? Proving that $\sum\limits_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$ I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated How does Windows 7 licensing work for running the OS as Virtual Machines? I plan on buying Windows 7 Ultimate Edition to run on an old Pentium 4. I also want to run Windows 7 virtual machines inside Windows 7. My question is two fold:   Do I need a license for each Virtual Machine? Does the answer change if I use a third party virtualization platform (e.g. , , , etc) as opposed to ?