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How to override parent functions in child themes? I've been reading around and trying to figure out how to do this, but for some reason I can't seem to override parent functions in my child theme.  I'm using TwentyTen as a parent - can anyone tell me why this function in my child theme isn't overriding the parent function please?  // Override read more link function osu_twentyten_continue_reading_link() {  return ' &lt;a href="'. get_permalink() . '"&gt;' . __( 'Read on &lt;span class="meta-nav"&gt;&amp;rarr;&lt;/span&gt;', 'twentyten-child' ) . '&lt;/a&gt;'; } function osu_twentyten_auto_excerpt_more( $more ) {  return ' &amp;hellip;' . osu_twentyten_continue_reading_link(); } remove_filter( 'excerpt_more', 'twentyten_auto_excerpt_more' ); add_filter( 'excerpt_more', 'osu_twentyten_auto_excerpt_more' );   I thought you had to remove the filter/action etc. before re-adding it right?  Thanks,  osu The 10th and last round of negotiations between Iran and the six world powers was held in Vienna in a bid to strike a final deal. However, after the sides failed to agree on a final deal they decided to extend the negotiations until / by July 1, 2015.   In the original text the preposition until has been written. I am, however, wondering if we could use by instead, if so, is there any difference semantically?  Any help would be greatly appreciated I'm trying to prove the following statements in Folland's book.  Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)&lt;\infty$).  The first one is already done, but the second one is giving me trouble.   My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)&lt;\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.   The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.  Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon &gt; 0$ there is some $E\subset X$ measurable, such that $\mu(E) &lt;\varepsilon$ and $f_n\to f$ uniformly on $E^c$.  Thank you! Prove $aba^{-1}b^{-1}\in{N}$ for all $a,b$ I am trying to prove the above with the conditions that N is normal to G and $G/N$ is abelian.  So then for $Na,Nb\in{G/N}, NaNb=NbNa$.  But $NaNb=Nab=Nba$, so then we know $ab=ba$, so G is abelian.  So if $ab\in{G}, N=abN(ab)^{-1}=abN(ba)^{-1}=baNa^{-1}b^{-1}....$ and i'm getting lost... Example of non-isomorphic structures which are elementarily equivalent I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).  I am using the notaion of David Marker's book "Model theory: an introduction". what does ∇ (upside down triangle) symbol mean in this problem Given $f(x) = \frac{1}{2}x^TAx + b^Tx + \alpha $   where A is an nxn symmetric matrix, b is an n-dimensional vector, and alpha a scalar. Show that  $\bigtriangledown _{x}f(x) = Ax + b$   and  $H = \bigtriangledown ^{2}_{x}f(x) = A$   Is this simply a matter of taking a derivative with respect to X, how would you attack this one? Do reroll and dice-adding abilities work on critical hits and critical failures? How do you use Excel server-side? A client wants to "Web-enable" a spreadsheet calculation -- the user to specify the values of certain cells, then show them the resulting values in other cells.  (They do NOT want to show the user a "spreadsheet-like" interface. This is not a UI question.)  They have a huge spreadsheet with lots of calculations over many, many sheets.  But, in the end, only two things matter -- (1) you put numbers in a couple cells on one sheet, and (2) you get corresponding numbers off a couple cells in another sheet.  The rest of it is a black box.  I want to present a UI to the user to enter the numbers they want, then I'd like to programatically open the Excel file, set the numbers, tell it to re-calc, and read the result out.  Is this possible/advisable?  Is there a commercial component that makes this easier?  Are their pitfalls I'm not considering?  (I know I can use Office Automation to do this, but I know it's not recommended to do that server-side, since it tries to run in the context of a user, etc.)  A lot of people are saying I need to recreate the formulas in code.  However, this would be staggeringly complex. There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that  \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post  I don't see that...  My approach to this is:  Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing  $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.  But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...  I would be really thankful, if someone could give me quite a basic proof that even I understand... How does Incense work? Escaping WP_Query tax_query when term has special character(s) Here's my issue. I have a custom post type that I'm attaching a custom taxonomy to. This taxonomy will allow the admin to provide search suggestions that will ensure the post is found when a user searches. This taxonomy is also used on the front-end to autocomplete the search field with these suggestions.  The system works fine, except when the term contains a forward slash, a space or other "special" characters.  &lt;?php $keyword = 'Ski-in%2FSki-out';//Submitted via $_GET $keyword = urldecode($keyword); // (string)'Ski-in/Ski-out'  $taxQuery = new WP_Query(array(     'post_type' =&gt; 'ml_properties',     'tax_query' =&gt; array(         'relation' =&gt; 'OR',         array(             'taxonomy' =&gt; 'mc_tax_lifestyle',             'field' =&gt; 'name',             'terms' =&gt; array($keyword),             'operator' =&gt; 'IN'         )     ) )); ?&gt;   I've looked into using the escaping features provided by WordPress, but haven't had any luck there. I thought this would be a simple problem, but either my Google-fu is not strong, or I'm just not having any luck.  I've also verified that that Ski-in/Ski-out is the exact string stored in MySQL. I've looked into using esc_attr(), $wpdb-&gt;esc_like() and esc_sql() but none have an effect.  Perhaps I'm missing something obvious? Where are temporary files in Chrome stored? Photoshop: How to draw a guide with an exact distance from another guide Before drawing, I want to draw all guides for my layout. For example, I have a guide line named A already. After that, I want to have another guide named B  parallel with guide A and at 50px distance from it.   I have three solutions:   Draw line B and by "some way", I can see the current distance directly when I'm drawing, so I can make the distance exactly what I want it to be.  Draw line B. And then I use the Ruler tool to see the distance between the two guides. This method has disadvantages: maybe the ruler line that I draw isn't perpendicular with the guides and I'm not sure when the distance is enough. There are other clever ways, that help me get this job done fast and easily, but that I don't know yet. I think Photoshop should support this feature.   What is the best way to achieve this? As we know, whenever school ends, Harry is sent back to live with the Dursleys, where he suffers from abuse. He would much prefer to live with the Weasleys, and often visits there before returning to school. It stands to reason that they probably don't live that far from each other, since:   Both Harry and the Weasleys use Platform 9&frac34; to get to school Ron was able to take Harry to the Burrow in a flying car within a reasonable amount of time   It's heavily hinted at that the Weasleys are poor. However, it's not like Harry would live there permanently. Three months isn't that bad, not to mention he's pretty well-off from his parents' money. There would be little to no extra expense on the Weasleys.  The only explanation is that Dumbledore goes out of his way to ensure Harry lives with the Dursleys. If so, why is this? How to call child component controller function/helper from Parent Component Controller Javascript add leading zeroes to date SQL Server nvarchar(max) vs nvarchar(n) affects performance What could cause a microcontroller to unexpectedly reset? One particularly irritating variety of bug in a microprocessor-controlled system is for the microprocessor to unexpectedly reset. An important tool for debugging this kind of problem is a list of possible causes. What could cause a microcontroller to unexpectedly reset? As homework, I have to prove that     $\forall n \in \mathbb{N}: n^3-n$ is divisible by 6   I used induction  1) basis: $A(0): 0^3-0 = 6x$ , $x \in \mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?  2) requirement: $A(n):n^3-n=6x$  3) statement: $A(n+1): (n+1)^3-(n+1)=6x$  4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$  So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $\forall n \in \mathbb{N}$  Did I do something wrong or is it that simple? H2G2 — Teleportation Also Results in Time Travel?