nested

问题 Closed . This question is opinion-based. It is not currently accepting answers. Want to improve this question? Update the question so it can be answered with facts and citations by editing this post. Closed 4 years ago . If I need to create a couple of nested DOM elements, I know one way to do it, is to write them as long string and then place them in the document using a suitable jQuery function. Something like: elem.html( '<div class="wrapper"> <div class="inner"> <span>Some text<span> <

问题 I have a mixin that accepts an argument that I want to pass into a variable. @mixin my_mixin($arg) { background-color: $state-#{$arg}-text; } 回答1: Interpolation of variable names is currently not possible in SASS. Here is the issue that discusses this https://github.com/nex3/sass/issues/626 However, you may use interpolation of placeholders: %my-dark-styles { background-color: #000; } %my-white-styles { background-color: #FFF; } @mixin my_mixin($arg) { @extend %my-#{$arg}-styles; } .header {

问题 This question is the inverse of this question. Given a hash that has an array for each key like { [:a, :b, :c] => 1, [:a, :b, :d] => 2, [:a, :e] => 3, [:f] => 4, } what is the best way to convert it into a nested hash like { :a => { :b => {:c => 1, :d => 2}, :e => 3, }, :f => 4, } 回答1: Here's an iterative solution, a recursive one is left as an exercise to the reader: def convert(h={}) ret = {} h.each do |k,v| node = ret k[0..-2].each {|x| node[x]||={}; node=node[x]} node[k[-1]] = v end ret

问题 I need to remove all occurrences of a bb style tag from a string. The tags can be nested, and this is where I am failing. I also need to relocate each tag and contents to the end of the string, and replace the tag with an HTML element. I have tried to play with regex and preg_replace_callback, but I have only been so far unsuccessful. I also tried to modify the following, and have also had no luck: Removing nested bbcode (quotes) in PHP and How can I remove an html element and it's contents

问题 This question already has answers here : list.extend and list comprehension (6 answers) Closed last year . I'm not experienced in Python, and I often write code that (simplified) looks like this: accumulationList = [] for x in originalList: y = doSomething(x) accumulationList.append(y) return accumulationList Then after my test passes, I refactor to return [doSomething(x) for x in originalList] But suppose it turns out a little different, and my loop looks like this: accumulationList = [] for

问题 <h:dataTable value="#{SearchingBeans.list}" var="entry"> <h:column> <f:facet name="header"> <h:outputLabel>Photo</h:outputLabel> </f:facet> </h:column> <h:column> <f:facet name="header"> <h:outputLabel>Pseudo</h:outputLabel> </f:facet> <h:outputLabel value="#{entry.pseudo}"></h:outputLabel> </h:column> <h:column> <f:facet name="header"> <h:outputLabel>Description</h:outputLabel> </f:facet> <h:outputLabel value="#{entry.description}"></h:outputLabel> </h:column> <h:column> <f:facet name=

问题 I have two lists of lists that have equivalent numbers of items. The two lists look like this: L1 = [[1, 2], [3, 4], [5, 6]] L2 =[[a, b], [c, d], [e, f]] I am looking to create one list that looks like this: Lmerge = [[1, 2, a, b], [3, 4, c, d], [5, 6, e, f]] I was attempting to use zip() something like this: for list1, list2 in zip(*L1, *L2): Lmerge = [list1, list2] What is the best way to combine two lists of lists? Thanks in advance. 回答1: >>> map(list.__add__, L1, L2) [[1, 2, 'a', 'b'], [3

问题 I'm looking for a PHP (with MYSQL) nested sets class with all needed functions. For example: createLeftNode, createRightNode,createRootNode, createSubNode,deleteNode and moveTree . Not only 1 left, 1 right, 1 up and 1 down but also a part of a tree in a nother tree. Thanks! 回答1: Well nested sets are great if you are working with hierarchical data. It's much more complex to implement it only with php arrays especially if you want to save these information in a database. you could try this on.

问题 I'm writing Perl for quite some time now and always discovering new things, and I just ran into something interesting that I don't have the explanation to it, nor found it over the web. sub a { sub b { print "In B\n"; } } b(); how come I can call b() from outside its scope and it works? I know its a bad practice to do it, and I dont do it, I use closured and such for these cases, but just saw that. 回答1: Subroutines are stored in a global namespace at compile time. In your example b(); is

问题 I would like to do the equivalent of the following: #define print_max(TYPE) \ # ifdef TYPE##_MAX \ printf("%lld\n", TYPE##_MAX); \ # endif print_max(INT); Now the #ifdef or any nested preprocessor directive is not allowed as far as I can see in a function macro. Any ideas? Update: So it seems like this is not possible. Even a hack to check at runtime seems unachievable. So I think I'll go with something like: #ifndef BLAH_MAX # define BLAH_MAX 0 #endif # etc... for each type I'm interested in