which is better href="#" or href="javascript:void();" which is better href="#" or href="javascript:void();" in terms of all aspects  like user experience, speed etc  is it necessary to put a 0 inside void when using href="javascript:void();" .like this href="javascript:void(0);"
Which "href" value should I use for JavaScript links, "#" or "javascript:void(0)"? The following are two methods of building a link that has the sole purpose of running JavaScript code. Which is better, in terms of functionality, page load speed, validation purposes, etc.?    function myJsFunc() {     alert("myJsFunc"); } &lt;a href="#" onclick="myJsFunc();"&gt;Run JavaScript Code&lt;/a&gt;     or    function myJsFunc() {     alert("myJsFunc"); }  &lt;a href="javascript:void(0)" onclick="myJsFunc();"&gt;Run JavaScript Code&lt;/a&gt;
How do I print bold text in Python? How does Ascending work? The next upgrade I am likely to get is "ASCENSION" for Amenhotep.  The mouseover text on the ability says it restarts the game and gives a single hero soul.  Does anything else cross over (heroes, gold, achievements, etc.)?  When would be the best time to Ascend?  Should I wait until I unlock all the heroes or buy all their abilities?  I'm on level 85 now (not yet beat the boss), should I wait until a specific level? Find the sum $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$ Please help me calculate the following sum     $$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$$ Grub rescue after installing Ubuntu on a external HDD from USB Does the review queue page show only reviews requiring your action? Does it differ, and when does it? This confusion came up in comments .  On the site I use most actively (RPG.SE), I've noticed that the counts for each review queue appear to only indicate the reviews available to me requiring my action. Just to be clear, I'm talking about this number here:        Since I'm a 10k user, that's pretty easy to tell: even after doing all my reviews, and the review page saying there's 0 items in each queue, I might still have that 10k notice in my top bar saying there's [5] reviews pending, or some other number. So they must still be there, just waiting on other users to finish them, and it just doesn't show them to me. This is fine, I like this behaviour.  However, on Meta SE about an hour ago, there was 1 item in the close vote review queue. I reviewed that item. The count stayed at 1 on the review page. When I clicked to re-enter the queue in case there were new items, I was presented with this:        Another user in that meta question I linked claims this is normal behaviour for them on their own site (Christianity.SE I think?).  What's going on here? Are the review queues supposed to be customised just to what I personally can actually review at the time? Is there a bug, are sites customised differently, or is there just some kind of caching thing leading to a time delay on the review count getting updated? (As they say, it is always caching, but is caching really the culprit here too?) Remote Desktop Settings Missing I'm trying to setup remote connection to my Windows 10 machine. During the setup I noticed that my installation is missing some settings all the tutorials have. Can anyone tell me why I'm stupid and what I should do? Below you can find a screenshot of what I'm talking about. The first image is how it should be. The second picture is a screenshot on my machine.      Edit: After popular request, my windows version: Book: Handful of children kept in a school as political mythological prisoners $\gcd(a,b)=1, x^a = y^b\Rightarrow x = n^b$, $ y = n^a$ for an integer $n$. If $ a$, $ b$, $ x$, $ y$ are integers greater than $1$ such that $ a$ and $ b$ have no common factor except $1$ and $ x^a = y^b$ show that $ x = n^b$, $ y = n^a$ for some integer $ n$ greater than $1$.  I started this way: $x^a=y^b$ $=&gt;$ $x=y^{b/a}=(y^{1/a})^b$. Thus $x$ is an integer, and so is $y^{1/a}$, so $y=n^a$ for some integer $n$ greater than $1$. Then from the given relation, we get $x=n^b$.   I am not very convinced with my solution, as I feel I did some mistake. Can anyone point out where the mistake is? Any other solution is also welcome :) Input field for decimal property My USB drive used to be originally 8GB when I bought it.  I'm trying to reformatted in Windows 7 by right clicking on the drive and selecting Format.... But the capacity only shows 250MB.  Is there something I can do to get the original size back? Maybe it got partitioned in a weird way?   The  flash drive is a SanDisk Cruzer Micro 8GB. I got it from Wal-Mart but this is the same drive. Is zero input impedance good or possible when considering impedance matching? For RF applications, we are usually considering of impedance matching to choose load impedance as the conjugate of the source input impedance for the maximum transferred power. But how about design the source having zero input impedance? Is it even possible or good? A naive thought is that all the power will be on the load if so. I might think it is true for DC case and for amplifier design to have a small output impedance and big input impedance. I lack the knowledge to find the missed chain to put all the knowledge together. How do you think? Thanks! Input field added through javascript not in PHP $_POST variable. How to fix this? Merging Raster Datasets of Different Resolutions using ArcGIS Desktop? I am using ArcGIS Desktop.  I have two DEMs in .tif format and I am looking to merge the two. The main problem with merging these are that they are different resolutions (one is 10m and one is ~20m). Also, another problem with merging the two is that they overlap and I am not sure of the most effective way of dealing with that overlap. Obviously I want to keep the values of the better resolution.   So what are the best methods here?   Is there a python script available that works best with merging rasters? I would like to automate this with a script if possible. Which process of resampling is best for DEM data? Bipolar, Cubic... ? What is the best way to merge overlapping DEMs? I think this is a very recent design change, because I've not seen it before:    Where before a small box with the bounty value dislocated the title a couple of characters to the right, now the additional text saying how long until it expires dislocates the title by about a quarter of the width of the column.  I find that this makes it much harder to read the title, and is counterproductive to the goal of attracting attention. Boot drops to a (initramfs) prompts/busybox How do I increase the render distance? Keyboard shortcuts for chat Yes, I know there's a script for that. But I think it's useful enough to be inserted into the live chat for everyone.  The following keyboard shortcuts are incredibly useful:   CTRLUp and CTRLDown - For replying to messages (each Ctrl+Arrow should change the :[message ID] in the chat input. CTRLSpace - For retrying a failed message, or ESC to cancel it. There should also be a keyboard shortcut to edit a failed message as per the newest functionality (Maybe the Down key?)   Using a chat with only the keyboard is very useful, it will be especially appreciated on chats with a more technical nature. What is the source for defaulting to plain English readings of non-game terms? Health effect of electromagnetic and acoustic waves on brain function Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible.