children

how to display part of a menu tree?

自闭症网瘾萝莉.ら 提交于 2019-11-29 21:16:41
问题 I'm trying to deal with Wordpress 3.0. It's rather cool thing but I can't get on with one problem. For example, I have such menu tree. Menu tree is constructed from pages. Home news video audio Blog About author Favourite colors red blue green My car wheels tires The Idea is: main menu consists of root elements: home, blog, my car On the left side I would like to display children elements of current active root element. For exammple if person is on the "home" page, on the left part he should

Suitable environment for a 7 year old [closed]

萝らか妹 提交于 2019-11-29 19:26:24
My 7 year old would like to learn, how to program? (his idea not mine, and he does things in the outside world. So, I am not too worried from that point of view. He already went so far as to take a game programming book out of my office to read at bed time.) The other day we sat down and wrote a very simple number guessing game (you pick 8 and it is correct, anything else it is wrong). It went OK but there were a number of questions he had based on the syntax of the language. (I happened to pick Java as I had the IDE opened at the time.) I teach post-secondary introductory programming courses

Suggestions on starting a child programming [closed]

允我心安 提交于 2019-11-29 18:41:47
What languages and tools do you consider a youngster starting out in programming should use in the modern era? Lots of us started with proprietary Basics and they didn't do all of us long term harm :) but given the experiences you have had since then and your knowledge of the domain now are there better options? There are related queries to this one such as " Best ways to teach a beginner to program? " and " One piece of advice " about starting adults programming both of which I submitted answers to but children might require a different tool. Disclosure: it's bloody hard choosing a 'correct'

How to get all child components of QWidget in pyside/pyqt/qt?

放肆的年华 提交于 2019-11-29 13:39:08
问题 I am developing a desktop application using pyside(qt), I want to access(iterate) all line edit components of QWidget. In qt I found two methods findChild and findChildren but there is no proper example found and My code shows error, 'form' object has no attribute 'findChild'. Here 'form' is Qwidget form consist components lineEdit, comboboxes, Qpushbuttons etc. Code: lineEdits = form.findChild<QLineEdit>() //This is not working lineEdits = form.findChild('QLineEdit) //This also not working

get text of an element without children in javascript

心不动则不痛 提交于 2019-11-29 09:35:22
How do I get the text of an element without the children? Neither element.textContent nor element.innerText seem to be working. HTML: <body> <h1>Test Heading</h1> <div> Awesome video and music. Thumbs way up. Love it. Happy weekend to you and your family. Love, Sasha </div> </body> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.4/jquery.min.js"></script> <script type="text/javascript"> fool("body"); </script> and here's the fool function: jQuery.fn.justtext = function(text) { return $(this).clone() .children() .remove() .end() .text(); }; function fool(el) { reverse(el); function

Python multiprocessing and independence of children processes

旧时模样 提交于 2019-11-29 04:07:43
From the python terminal, I run some command like the following, to spawn a long-running child process: from multiprocessing.process import Process Process(target=LONG_RUNNING_FUNCTION).start() This command returns, and I can do other things in the python terminal, but anything printed by the child is still printed to my python terminal session. When I exit the terminal (either with exit or CTRL + D ), the exit command it hangs. If I hit CTRL + C during this hang, the child process is terminated. If I kill the python terminal process manually (via the posix kill command), the child process is

Angular 2 routing redirect to with child routes

孤街醉人 提交于 2019-11-29 01:39:21
问题 I am a newbie in Angular 2. I want to create isolated modules for every part of my app. For example I created the AuthModule with default component - AuthComponent which contain a router-outlet for his child components (SignIn or SignUp). So I want to realise the following scenario: When navigate to / - root off app - redirect to /auth After redirect to /auth - load AuthComponent with router outlet After AppComponent loaded - load default sign-in component via redirecting to /auth/sign-in But

jQuery实现多级列表(树形菜单)

无人久伴 提交于 2019-11-29 00:37:26
<style> ul{ list-style:none; margin:0px; padding:0px; } ul li{ margin-left:15px; } ul li a{ display:block; height:30px; width:80px; line-height:30px; text-decoration:none; padding-left:15px; } .no{ display:none; } .yes{ display:block; } .page{ background:url("../img/page.png") no-repeat left center; } .plus{ background:url("../img/plus.png") no-repeat left center; } .minus{ background:url("../img/minus.png") no-repeat left center; } </style> <script type="text/javascript" src="../jq/jquery-1.12.4.js"></script> <script> $(function(){ $("li").each(function(){ if($(this).children("ul").size()>0)/

BootstrapVue框架集合

拥有回忆 提交于 2019-11-28 21:56:24
1.导航栏navbar 鼠标滑过既有下拉列表 效果图 以下代码段,toPage(pathName)是跳转页面的方法, <b-navbar id="nav" toggleable="sm" type="light" variant="light"> <b-navbar-toggle target="nav-text-collapse"></b-navbar-toggle> <b-collapse id="nav-text-collapse" is-nav> <b-navbar-nav> <b-nav-item class="d-sm-none" v-for="item in navList" href="#"> <span style="color: black" @click.native="toPage(item.path)"> {{$t(item.text)}} </span> </b-nav-item> <div style="display: flex;flex-direction: column"> <div class="row" style="display: flex;flex-direction: row;"> <b-nav-item-dropdown class="d-none d-sm-block" v-for="item in navList" right>

How do I make this loop all children recursively?

末鹿安然 提交于 2019-11-28 21:43:00
问题 I have the following: for (var i = 0; i < children.length; i++){ if(hasClass(children[i], "lbExclude")){ children[i].parentNode.removeChild(children[i]); } }; I would like it to loop through all children's children, etc (not just the top level). I found this line, which seems to do that: for(var m = n.firstChild; m != null; m = m.nextSibling) { But I'm unclear on how I refer to the current child if I make that switch? I would no longer have i to clarify the index position of the child. Any