children

I have a UniformGrid with a number of Button 's as Children . Each Button has a Tag with an ID, e.g. (dumbed down code): MyUniformGrid.Children.Add(new Button { Margin = new Thickness(5), Tag = Query.GetUInt32("id"), Width = 200 }); How can I select the child Button object with an ID of 87? (as a for instance) Intellisense isn't popping up with the Linq methods when I type MyUniformGrid.Children. (after adding using System.Linq; ). Here you go: var MyButton = MyUniformGrid.Children. OfType<Button>(). Single(Child => Child.Tag != null && Child.Tag == 87); Linq can't be run directly on

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 see: news video audio If user is on the "Blog" page, he should see: About author Favourite colors red

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 The signatures of findChild and findChildren are different in PySide/PyQt4 because there is no real

使用element ui组件库实现一个table的两棵树的效果 效果如下，左边树自动展开一级，右边树默认显示楼层，然后可以一个个展开 代码如下 <el-table :data="relativeData" :fit="isFit" height="700px" :row-style="showTr" :row-class-name="tableRowClassName" :header-row-class-name="tableRowClassName" size="small" highlight-current-row> <el-table-column fixed width="310" show-overflow-tooltip style="max-width:310px;"> <template slot="header"> <div class="monitor_header"> <div class="monitor_item">分项名称</div> <div class="line_item"></div> <div class="building_item">建筑空间</div> </div> </template> <template slot-scope="scope"> <template v-if="spaceIconShow()"> <span v

const data = [{ id: 1, children: [{ id: 2, children: [{ id: 3, }, { id: 4, }], }], }, { id: 5, children: [{ id: 6, }], }]; let nodes = []; function getParentNodes(id, tree) { _getParentNodes([], id, tree); return nodes; } function _getParentNodes(his, targetId, tree) { tree.some((list) => { const children = list.children || []; if (list.id === targetId) { nodes = his; return true; } else if (children.length > 0) { const history = [...his]; history.push(list); return _getParentNodes(history, targetId, children); } }) } 　　要找到一颗树中指定id的那个节点很简单。如果要找到指定的所有父节点，转换一下思路就是将深度遍历的每条顺序都记录下来，直到找到了指定id的节点时

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 suggestions? Thanks! Update: I'm now using the following, according to answer suggestions. Is this the

I'm trying to loop through childNodes like this: var children = element.childNodes; children.forEach(function(item){ console.log(item); }); However, it output Uncaught TypeError: undefined is not a function due to forEach function. I also try to use children instead of childNodes but nothing changed. Does anybody know what's going on? The variable children is a NodeList instance and NodeList s are not true Array and therefore they do not inherit the forEach method. Also some browsers actually support it nodeList.forEach ES5 You can use slice from Array to convert the NodeList into a proper